Onall ty
:,
, , . tílmtsprightfu Scot a! Scots, frauglas,
that runs a-harseback up a hilt
pupendicular-William
Shake peare
Henry IV Part I
Act iI, cene IY
íisurg 5,1
hadows on a wall are projections
e**M,,
In this chapter, we will extend the noťon of orthogonď projection that we encountered
first in Chapter 1 and then again in Chapter 3. Until now we have discussed
only projection onto a single vector (or, equivalently, the one-dimensional subspace
spanned by that vector). In this section, we will see if we can find the analogous formulas
for projection onto a plane in R3, Figure 5.1 strows what happens, for óxample,
when parallel light rays create a shadow on a wa]l. A similar proces occurs when a
three-dirnensional object is displayed on a two-dimensional screen, such as a computer
monitor. Later in this chapter, we will consider these ideas in full generality.
To begin, let's take another look at what we already know about projecťons. In
Section 3.6, we showed that, in R2, the standard matrix of a projection onto the line
through the origin with direction vector u : [O'l
,
La,)'"
s.l! lntrltilustlllfi; fttstlw tltl ff tťall
D_ 1 [d? drdr1
I d? + dlLdrd, di J
dld\l(d? + dil1
d\I(d? + dil J
Hence, the projection of the y ctor rr onto this line is just Pv.
rmffi gm Show that P can be written in the equivalent form
(What does 0 represent here?)
ploblGm 2 show that p can also be written in the form p : uuT, where u is a unit
yector in the direction of d.
rgk eXrX & Using Problem 2, frnd P and then find the projection oť v :
onto the lines with the follo\,\ring unit direction yectors:
(a} u -::
P]ObtG^m { Using the form P : uuT, show that (a) PT : P (i.e., P is symmetric)
and (b) P2 : P (i,e,, P is idempotent).
i d?l(dť + dl)I
Larart@? + di)
p-tcos26 co0sin6l
[cos6 sin6 sinZ 0 J
|irÁ(b)u:Ll] (c)u:Li]
Ij]
Section 5.0 Introduction: Shadorv on a WalI
PlOblOm Explain why, if P is a 2 X 2 projection matrix, the line onto which it
projects vectors is the column space of P.
Now we will move into R3 and consider projections onto planes through the
origin. We will explore several approaches.
Figur9 5.2 shows one way to proceed, If 9 is a plane through the origin in R3 with
normal vector n and if v is a vector in R3, then p = ptojg (v) is a vector in 9 such that
r, - cn - p for some scalar c.
]I0urc s.2
Projection onto a plane
v*
firwffi wm S tJsing the fact that n is orthogonal to
cíl x p far ťt find an exprť iůnfor p in terms of v
P & e tJse the method of Problern 6 to find the p
t 1l
vŤ-| 0l
L -zl
every yectsr in P, solve
and n.
rojection of
onto the planes with the following equations:
(a) x+ y * z:0 (b) x - 22: O (c) Zx- 3y l z: 0
Another approach to the problem of finding the projection of a vector onto a
plane is suggested by Figure 5.3, We can decompose the projection of v onto 9 into
the sum of its projections onto the direction vectors for 9. This works only if the
direction vectors are orthogonal unit vectors. Accordingly, let u, and u2 be direction
vector for 9 with the property that
ll",ll ll"rll 1 and lli,ll2 -. 0
lsure .S
Chaptcr Orthogonality
By Problem 2, the projections ofv onto u1 and u2 are
pr : uruív and p2 = a2u!v
respectively, To show that pl * p, gives the projection of v onto 9, we need to show
that v - (pr + pr) is orthogonal to 9. It is enough to show that v - (p, * pr) is
orthogonal to both u1 and u2. (Why?)
plóuom 8 Showthatu,lr" ](p, * pz)) : 0 andu2. (v- (pr * pr)) : O.|Hint:
Use the alternative form of the dot product, x} = x. y, togpther with the fact that u,
and u2 are orthogonal unit vectors.]
It follows from Froblem 8 and the comment preceding it that the matrix of the
projectiort onto the subspace g of R3 spanned by orthogonal unit vectors u1 and u2 is
.Pl0[!0m 0 Repeat Problem 7, using the formula for P given by Equation (1),
Use the same v and use u1 and u2, as indicated below. (First, veri$, that u1 and u2 are
orthogonal unit vectors in the given plane.)
(a) Jť+ y+z= Owithu, and u,
(b) x.*2z:Owithu,*
(c) 2x*3y+g-Owithu, anď u2 x
PíOnbm l0 Show that a projection matrix given by Equation (1) satisfies properties
(a) and (b) ofProblem 4.
PlOblCn ll Show that the matrix P of a projection onto a plane in R3 can be
P=AAr
for some 3 X 2 matrixÁ , |Hint: Show that Equation (1) is an outer produď expansion. j
PíOtl8n t2 Show that if P is the matrix of a projection onto a plane in R3, then
rarrk(r; : 2.
In this cbapter, we will look at the concepts of orthogonality and orthogonal projection
in greater detail. Wb will see that the ideas introduced in this section can be
generalized and that they have many important applications.
In this section, we wil.l generalize the notion of orthogonality of vectors in R' from
two vectors to sets of vectors. In doing so, we will see that two properties make the
standard basis {e1, 2, , , . , er]r of Rí easy to work with: First, any two distinct vectors in
(1)
| -ztÝ61
:I r/Ý6 I
L I/\/6)
:,,u]
*-t'z*[;]
r 2/\/61
I r l\/6l
L-t /Ýa1
r L/\fr1
:L-l,#)
rnnl ,'o;,!.!,r ,- , . .___
Section 5.1 Orthogonality in R' S§S
the set are ort}rogonal. Second, each vector in t}re set is a unit vector. These two properties
lead us to the noťon of orthogonal bases and orthonormal bases-concePts that
we will be able to fruitfuny apply to a variety of applications.
0lthogonrl tnil 0iln0no]nel stts oíUcctoí$
The standard basis {e1, a2l , , , len} of R'is an orthogonal set, as is any subset of it. As
the first example illustrates, there are many other possibilities,
Show
sOlutl
is trur
Y3:
tR3
[0-
l1
L,.
fve
{vt,v2, yl} is an orthogonal set in
1-21 t
y1 :|
'l,y2:lL*r] l
We mu t show that eyery pair o1
ce
that
0íl
e, in
rs fro,
'if
l
],
3ctOec
Ll]
m this set is orthogonal. This
Yt 'Y2
Y2'Y3
Yl 'Y3
* 2(0) + 1(1) +
: 0(1) + 1(-1)
: 2(1) + 1(-1)
(*tXt) : 0
+ {tXt) : 0
+ (*rxl) * 0
5,1 are mutually perpen
ílsutgs,{
Án arthogcnal set oť vectcrs
0rsm .1
dicular, a
Ť
i]],r.jl,,1,1 If c1,I t.ldp&í scalarssuchthatCtyt +", * clvft: O,then
(crv, + .., + c;rv7r),yi: 0,v,* 0
or, equivalently,
cr(v1 . yi) +, l, + c,(vi, Vi) +, !, + c1o(vlr, yi) : 0
Since iv,, Y2l . . . r v1} is an orthogonal set, all ofthe dot products in Equation (1)
are zeío, except v; , v;. Thus, Equation ( 1 ) reduces to
ci(v; ,vi) : 0
Geometrically the yectors in xample
Figure 5,4 shows.
one of the main advantages of working with orthogonal sets of vectors is that
tlrey are necessarily linearly independent, as Theorem 5.1 shows.
(1)
C}rapter 5 Orthogonality
Now, v; .vl * a because vi * 0 by hypothesis, So we must have c; : 0. The fact that
tíris is true for all l = 1, . . . , /c implies ttrat {y1, v2, . . . , v7.} is a linearly independent set.
,&il,
n0mtlt Thanks to Theorem 5,1, we know that if a set of vectors is orthogonal, it
is automatically linearly independent. For example, we can immediately deduce that
the three vectors in Example 5.1 are linearly independent, Contrast this approach
with the work needed to establish their linear independence directly!
The vťctsr
Y1 = Y2 :* v3 -T
Li]Il]
Ii],
L*tJ
from Example 5.1 are orthogonal^and, hence, linearly independent. Since any three
linearlyindependentvectors in R3 form abasis for R3, bythe Fundamental Theorem
of Invertible Matric , it ťollows that {v1, y2, v3} is an orthogonal basis for ffi3
nOmOil In Example 5.2, suppose only the orthogonal vector v1 and v2 were
given and you were asked to find a third vector v3 to make {v1, v2, v3} an orthogonal
basis for R3. .one way to do this is to rememb", tirut in m3, ttre cLsr"product oitwo
vectors v1 and v2 is orthogonal to each of them. (see Exploration The cross product
in Chapter 1,) Hence we maytake
Y3 v' X vz : I i] X [l] :k
L-,j L,J
= -}
e sub-space
{[;]|l
"1 =Jil + ,l
-,1
L: .l=,Loj L?]
r21
L;j
Find an CIrthosonel basis for th 1ť of ffi3 given by
|z!/ . *y+2z
S0lutl0n Section 5.3 gives a general procedure for problems of this sort. For now,
1ve
witl nnd th1 orthogonal basis by brute force. The subspace Iť is a plaae through
the origin in R3. From the equation of the plane, we havei : y - 2z,'so Iť consists
ofvectors ofthe form
Note that the resulting vector is a mulťple of the vector v3 in Example 5.2, as it must be.
Section 5.1 Orthogonaliry- in Ru'
ltl | -z1
Itfollowsthatu: l i l*U": I O l*..abasisforl4/,butthey arenotorthogoLo.l
L ,]
nal. It suffices to finá another nonzero vector in I,1i'that is orthogonal to either one of
these.
[,l
Suppose w : l .y I is avector in l4lthatis orthogonalto u. Then x - y + 2z : 0,
L")
since w is in the plane W'. Since u. w, : 0, we also have x l y : 0. Solving the linear
ysterr\
x-y+2z:0
x+y -0
&,*+, we find that x : - z anď y : z. (Check this.) Thus, any nonzero vector w
| -,1
,n:
| "|
L z)
l--ll
will do. To be specific, we could take w : | ' |. O
',
easy to check that
L ,]
orthogonal set in I4land, hence, an orthogonal basis for I4l, since dim 14/
of the fo
{u, w} is
_r}
'
Another advantage of working with an orthogonal baďs is that the coordinates of
a v ctor with respect to such a basis are easy to compute. Indeed, there is a formula
for these coordinates, as the following theorem establishes,
.2
are given by
l'i:iil ],
il i.,]i. ii iiillli|ii.fi liilli]j l ilifi lllitlii.iiiíill
Ťi*l}íSince iv1, y21 . , , >v1} is a basis for 1ť, we know that tlrere are unique scalars
C1, . .,,c1 such that w : ClV1 * ... * cl,vl(from Theorem 3,29). To establish the
formula for c;, we take the dot product of this linear combination with v; to obtain
w.v,: (crv, +.,.+ cevfr).yi
: cl(vl ,yJ +",* c;(v;,vi) + ",* clvt,y,)
sincev;.vi = 0 íorj *;'r;".]" * 0,vi.vi * a.Dividingby.,,.rnu, we obtainthe
desiredresult. _,^.,,,',]ffi
1ť' Vi
C;: fori:1,.",rk' Vi'Vi
Section 5.1 Orthogonality in ffin
oťthe form
{u, w} is an
-?. A
l
Another advantage of working with an orthogonal basis is that the coordinates of
a vector with respect to such a basis are easy to compute. Indeed, there is a formula
for these coordinates, as the following theorem establishes.
l::i:::;l,i Since {vr, y2, . . , ,v1} is a basis for Iť, we know that there are unique scalars
c',,. ,c1 suchthatw: clv, * ",* c;v1,(fromTheorem3.29). Toestablishthe
formula for c;, we take the dot product of this linear combination with v; to obtain
*'"'
: iŤ;
-- :Ťx,v,) + + cl(vl,v;)
since v,. v; : 0 íor j * l. Since yi * 0, yi. v; * 0. Dividing by rr,, Ý, we obtain the
desired result. ,_.::.-lt*S
ltl | -z]
Itfollowsthatu: l i l*o": I o |*.abasisforlV',buttheyarenof orthogoLo]
L ,.l
nal. It suffices to finá another ,rorr""ro vector in I4lthat is orthogonal to either one of
these.
l-"l
Suppose,n : l y I i, uu..to, in lď'that is orthogonalto u. Thepx - y * 2z : 0,
l,^
sincewisintheplane Iť'. Sinceu.w = 0,we alsohavex * y:0. Solvingthelinear
system.
x-y+2z:0
x+y :0
&}**," we find that x : _ z and y = z. (Check this.) Thus, any nonzero vector w
f -r1
-:L
)
t--tl
will do, To be specific, we could tut . * : I r |.
It i. easy to check that
L r_]
orthogonal set in 1ť and, hence, an orthogonal basis for W, since dim 1ď
1
ThgOísm
Chapter 5 Orthogonality
'índthe ccardinates cfw *
*ť x*mples 5"1 and 5,3,
í-tl
|
'
I
with respect to tlre orthogonal basis 6
L,J
{o,, v2, y3}
ltlt sm lJsing Th*ore m ,2, w compute
1ť' Yt
' vr 'v,
W" Yž
t,f
- Y2'Y2
W' Y3
L.7
"' Y3'Y3
2+2 3
4 + 1+ 1
0 + 2 + 3
6
q
0+1+1
1 2+3
2
2
1+1+1
Thus,
I{ : cly1 * c2v2 *.ro, = *o, + |v, + lv,
(check this.) with the notation introduced in section 3.5, we can also write the above
equation as
[w]n :
compare the procedure in Example 5.4 with the work required to find these
coordinates directly and you should start to appreciate the vďue oforthogonal bases.
Ás noted at the beginning of this section, the other, property of the standard basis
in Rr is that each standard basis vector is a unit veďor. Combining this próperty with
orthogonaliry we have the following definition.
n0fťI,|f = {qr,..., {t} is anorthonormal setofveďors, then9r.9; = 0for
i * j and ll*ll = 1. The fact that eaclr q is a unit vector is equívalent to qi. q, : 1.
It follows that we can summarize the statement that is orthonormal as
{h.Ť *
{l
hcw ťJrat x {qr, qz} is an orthCIí}srmal set in R3 iť
9r:
Li]
ifi*j
ífi * j
|y,#1
LttÝ6 l
[-l,ffi1
L I/ )
ffi,-*b
nIruf; nn }\fu check that
Section 5.1 Orth*gonality in ffi"
1l\4s : ff{lr ' {l"z
{h'{lr x
Il s * zlx/ta +
9z'8z :
frnr
ince
emtog
{b,qg} is
Since any orthonormal set oťvectors is, in particular, orthogonal, it is linearly independent,
by Theorem 5.1. If we have an orthonormal basis, Theorem 5"2 becomes
even simpler,
P1O0í ApplyTheorem 5.2 andusethe factthat9,.q; : l íot i = 1,. . ., k,
,;..ffin
OrthOsonal mfiricc
Matrices whose columns form an orthonormal set arise frequentlyin applications, as
you will see in Section 5.5. Such matrices have several attracťve properties, which we
now examine,
9z'tlz-L/oT */o T l/o*r á
If we have an orthogonal set, we can easily obtain an orihonormal set from it We
tho
Wť
fret
t
he
1l'
}n tru
lHfisn
nlire th
ten {q1,
Cons
|Iltlll
maliz
Then
b&1,$/enor*
Tnsoígm .il
simply noťmffliy,e each vector"
,normal basis far ffi3 frcm the vectors in xampl* 5,
already know tlrat y11 y2l afid y1 &ť an orthossnal
9 :rho,_*[ i] : I T|i1lr llv, l L_lj L-'/vil
1 1rol r0lQz
-- rr rv).:-:,=lll:IrtÝžlIL ilv, ll ' l/zl,] LrtÝžl
1 1t1l ruÝ:1
{l::ň"3:l6L
l.j
:L
l,,#j
an orthonormal basis for ffi3.
Chapter 5 Orthogonality
The columns of an rn X
QrQ : In.
(a
Let qi denote the ith column of
entry of QrQ is the dot product
ťollows that
praoí we need to show that
(Q'Q)u - 9;' 9l'
by the definition of matrix multiplication.
Now the column Q form an orthonormal set iť and ontry if
fo iťi + j
9;'97: t, ífi* j
which, by Equation (2), holds if and only if
r0
(Q'Q)ry :
t,
This completes the prooť.
_"_;itW
If the matrix Q in Theorem 5.4 is a squaťe matrix, it has a speciaI name.
The most important fact about orthogonal matrices is given by the next theorem.
,Q)u
{l
rrrrunIi, l
Q (and, hence, the ith row sf Qr). Since thc (ť, j)
' oť the ith row CIf Qr and th* jth column of Q, it
(2)
ifi
Líl
TheOrsm
PIOOí By Theorem 5.4, Q is ortho_gonal if and only ir Q"Q = /. This is trrre ť and
onlyif QisinvertibleandQ-'= QT,bylh.orem3.13. , n
Show that the following matrices are orthogonal and find their inverses:
[o '
ol lcoso -sinolÁ:l0 0 1l and B:L.i.,o .oro.]
L, o o]
S*lt!íis:l The columns oíA arc just the standard basis vectors for R3, which are
clearly orthonormal, Hence, Á is orthogonď and
[o o tlll
A-I:e,:It o 0|
Lo t o_]
. For B, we check directly that
r cos6 sinďllcos0 -sin0lBrB - L-*i,,B oo*á]Lil; il;]
:t cos29+sin26 *cOOsin6+sin6cosál
[*sin0cos6+cos$sin0 sinzO+cosz j
ecticn 5.l" Orthogonality in K*
[t 0l:l 1-I
L0 IJ
Therefcre, . is orthogonal, by Theorern 5.5, and
B-1 *fiT:
n0mlíl Matrix Á in Example 5.7 is an example of a permutation matrix, a matrix
obtained by permuťng the columns of an identity matrix. In general, arty n X n permutation
matrix is orthogonal (see Exercise 25). Matrix B is the matrix of a rotation
through the angle 0 in R2. Any rotation has the property that it is a length-preserving
transformation (known as art"isometry in geometry). The next theorem shows tlrat
every orthogonal matríx transformation is an isometry, Orthogonal matrices also
preserve dot products. ln fact, orthogonal matrices are characterized by either one of
these properties.
r co 0 sinÉl
L-sin 0 cos 0 j
a
PíOOíWe will prove that (a) .+ (c) =+ (b) + (a). To do so, we will need to make use
of the fact that if x and y are (column) vectors in Rn, then x .y = *'y,
(a) ,+ (c) Assume that Q is orthogonal. Then Q?Q : 1, and we have
Qx. Qy: (Qx)rQy = *'Q'Qy : *? : xŤ = x.y
(c) =+ (b) Assume that Qx. Qy : x. y for every x and y in Rn. Then, taking y : x,
wehave Qx.Qx: x.xl so l|Qxl| _ Qx. qx : \Áli : ll"ll.
(b) + (a) Assume that property (b) holds and let q; denote the lth column of Q.
Using Exercise 63 in Secťon 1.2 and property (b), we have
x.y: i(||x + yll' - ll* - yll')
: i(llo(* + y)ll' - llqr* - v)ll')
= i(||qx + Qyll'- llq* - qyll')
: Qx,Qy
for allxandyin R". [This showsthat (b) + (c).]
Now if e; is the ith standard basis vector, then q : Qe;. Consequently,
{i' {li :
Thus, the columns of Q form an orthonormal set, so Q is an orthogonal matrix.
, ,]:#
Qei.Q j : l-ej:
{l
rrrr',!',
..
Chapter 5 ůrthogonalíty
In ,xcrclses I*6, determins which sats af vector fr,re,
orthogonal.
1L;][1]Li]2,
3[l][i][1]4,
Irooking at the orthogonal matrices Á and B in Example 5.7 , you may notice that
not only do their columns form orthonormal sets-so do their rows. In fact, every
orthogonal matrix has this property, as the next theorem shows.
liíť}tiiFrom Theorem 5.5, we know that Q-t : QT. Therefore,
l (Q)-1 : (Q*')*' : Q: (Q')'
so Q7 is an orthogonal matrix. Thus, the columns of QT*which are just the rows of
Q-form an orthonormal set. ,:;l]*#
The final theorem in this section lists some other properties of orthogonal
matrices,
PlO0í We will prove property (c) and leave the proofs of the renraining properties
as exercises.
(c) Let ,\ be an eigenvalue of Q with correspond.ing eigenvector v. Then Qv = ,\v,
and, using Theorem 5.6(b), we have
ll"ll Ť llq"ll : 11,1oll
*
Since ||v|| + 0, this implies that lAl
: 1.
i: j;lí!::ii property (c) holds even for complex eigenvalues. rrr. *utri*
|!
-;]
is orthogonal with eigenvalues i and - i, both of which have absolute value 1.
':,
[ -4l
| -0 l
I zl
L 7)
L:]Ll]
-21
-iI
0]
;]
;]
1]
il
l]
5.
i],[-i]Ll]
l]Li]L1]
In Exercises 7_10, show that the given vectors form an orth,ogonal
basis for R2 0r R3. Then use TLteoťem 5.2 to exprťs*w
ňs a linear combinatiort of these basis vectors, Give the
coordinate vecfor [w] n oíw with respect to the basis
8 -- {ur, vz. oíR2 or ffi ; y1l y2l v3 o/R3,
t 41 ltl t 1l
7.vt:
L_;],o,: L;],w
- L_;]
t-rl t--ol trl8.v,:
Li],": L i1'*- L,J t
[1 0 0 Ll\/61
lo 2/3 I/\n :l\re l
2|.l ', ' ,*lbll
Io -2l3 ll\fr *Il\/6|
Lo Ll3 0 Ll\fr )
22. Prove Theorem 5.8(a).
23. Prove Theorem 5.8(b).
24. Prove The rem .8(d).
Section 5.1 Orthogonality in R" i iii
2 . Prove that evsry permutatinn mntrix is orthogonal.
26,IrQ is an orthogonal rnatrix, prov* that arry matrix
obtained by rearran$ins the rowg oť Q is alsn
orthogon&l"
27 . Let Q be m orthngonal 2 Y, 2 matrix and let x and y
be vectors in R2" If 0 is the angle between x and y,
pr ve that the angle betw*en Qx and Qy is also S.
(This proves that the linear transťormations defined by
orthogonal matric frr angJe-pres*rving in R2, e ťact
that is true in geneťel)_
28. (a} Prove that an orthogonal žX 2 matríx must have
\
the form
la *b1 la b1
' o) CIr
l, *-)Lb
whe
l a1
.nit vector.
'* Lr] 'a
unit vector.
(b) t}sing part (a), show that very orthogonal
. žx 2 matrix is of the fnrm
where 0 *0ť*.žn,
(c) how that v ry orthogonal 2 >< 2 matrix corre sponds
to eitheť a rotation r a reflection in R2,
(d) how that an orthogonal 2 Y, 2 matrix Q corre
půndsto a rotation in R2 if ďct Q - 1 and a
reťlection in R2 if det Q : * 1.
In Exerct. ť29-32, use Exercise 28 tu determine whether
the given orthogonal matrix repr§s nts u rotation or fr,
reflection. If it is a ratation, giv* the ang}u af rotation; if it is
a r*flectiorc, give ths line af rrflectian,
ltt\/1 -Ll\/1129,1 ,, ,*i-z' flv2 tlr/zJ
t- 1l
9.v,: | 0l,r,
L-r]
Itl
10.v,: I r 1,or-
Lrl
-Li]",=[l];W*[1]
Il]--Li];w:Ll]
*21, determina w\tettter the giuen matrix is
t is, find its inverse.
l7,t :l t'l 11Ll l
L* r l\/í L/\/í)
i"-I/2 ÝÍl21
30. l ,*,JVr
L*v5tz -Ll2]l
r 3 4l| _*,
32. l ; ilL.-5 íj
| -tlz v1 /21
31.1 ._, , I
LV3l2 ll2J
l 11
1 5ll
1 il l
2 2l
l 1l
) 1l
; ilI
11l
lcos0 *sin0l lcos6 sin6l
Lsin0 .oro]
or
Isin0 -co*o]
In Exgrcises 1I*15, d#ermine whetLter the given orthogorcal
set af vectors is orthonormAl. Iíit is nat, nzrmalize the
vectors to form an orthonormal set.
ril r _{l 1-1l r ll
,,Lí],L-í]!,.Li],Ii]
ti
"Li] L í]Lj] L.,L
1] L?] L:i]
ttr [
,l,:1[
#,,_] | Wll[ ; ]
'u, l
*ttz|,
1 ,l.6 1, 1 \/516l'Ivq|
L L/2) L
*Ll\6 ] L*tll6) LrtÝž)
In Hxerrises 16
artílnganal, If i
[o -1l16.i l
Ll 0]
111
lT ž
l8.I i *b
| *l 0L3
I cos 0 sin
l
19. I cos'0
l
L sin0
111
| 1-1
lt1
za,l ? ',
|-l 1
li1
L1 1
*cas #
sin 6
ů
*sinz á l
-c0 0 sinll
cos0 l
33" Let Á and .B be ru X rr orthogonal matrices,
(a) Prove that Á(Ár + _n').g - 1 + B.
(b} tJse part (a) to prCIye that, if det á + det .
then Á + B is not invertible.
34, Letx be a unit vector in Rn. Partition x as
lx, l
x
[1]
L;]
Let
LXn J &
:.:
Chapter 5 Orthogonality
f *,l
a :- |--,-i------
ll
[r i r _
, w'ith a prescribeď first vector x, fl construction that is
. frequently useťul in applications.}
: 0, 35. Prove that if an upper triangular matrix is orthogonal,
then it must be a diagonal matrix.
36. Prove that ífn > m, thenthere is no m X ru matrix Á
such that lla*ll : ||x|| ror all x in Rn.
37. Let -- {yo, , . . , yn} be an orthonormal basís ťor Rn
(a) Prave that, for any x and y in Rn,
N.y * (x"v1)(y.or) + (x.v2){y.vr) *P,. "
+ (x. v,,}{y. vn)
(This identity is ca}le d, Parseval's ídentity.)
(b) What d.oes Parseval's ldentity imply about the
relationship between the dot proďucts x. y and
[x]r . |ylr?
yr
j
,r
l
llomplemsnt tnil
ProiGGtiOn$
]Igutt .
t: WL and W * 8L
(#-)*
Prove that Q is orthogona}. (This procedure gives a
quick method for finding an orthonormal basis for Rn
ílrtnOsítnaI
ílthsstlntl
In this section, we generalize t\.vo concepts t}rat we encountered in Chapter 1, The notion
of a normal vector to a plane will be extended to orthogonal complements, and
the projection of one vector onto another will give rise to the concept of orthogonal
projection onto a subspace,
0rtfio0onal GOmpIGmGnt$
A normalvector n to a plane is orthogonal to every vector in that plane. If the plane
passes through the origin, then it is a subspace lť'of R3, as is span(n), Hence, we have
two subspaces of R3 with the property that every vector of one is orthogonal to every
vector of the other. This is the idea behind the folIowing definition.
If ilris a plane through the origin in R3 and / is the line through the origin perpendicular
to W (i.e., parallel to the normal vector to W), then every vector v on { is
orthogonal to every vector w in M hence, (, : Wr , Moreover, W consists precisely
of those vector w that are orthogonal to every v on .ť; hence, we also have W = 4L .
Figure 5.5 illustrates this situation.
Section 5.2 Orthagonal Complements and Orthogonal Frojectians
In Example 5.8, the orthogonal complement of a subspace turned out to be another
subspace. Also, the complement of the complement of a subspace was the original
subspace. These properties are true in general and are proved as properties (a) and
(b) of Theorem 5.9. Properties (c) and (d) will also be useful. (Recall that the intersecilon
A n B of sets Á and B consists of their common elements, See Appendix A.)
a.
b.
c.
d.
WnWr*{0}
Prmm (a) Since 0 , lry
c be a calar. Then
Therefore,
(u + Ý),w -- u,w * v*lry,: fi + ůff 0
ou*visinlť.L.
We also have
(cu),w* c(u,w)* c(0) * S
from whích we ee that cu is in Tťr. It follows ttrat WL is a subspace of R'?
(b) We will prove this property as Corollary 5.12.
(c) You are a ked to prove this property in Exercise 23.
(d) }fou are a ked to proye this property in Exercis ť, 24.
: S for alI w in W, 0 is in WL. Let u and v be in IťJ and let
u, w - rr,lry : 0 for all w in W
_ jiĚ$ffi
m 5.í0
Wb can now express some fundamental relationships involving the subspaces
associated with an m x nmatrix.
(row (Á))PlO0í
If x is a vector in R', then x is in (row(Á))a ir ana only if x is orthogonal to
every row of Á. But this is true if and only if Áx : 0, which is equivalent to x being
in null(Á), so we have established the first identity. To prove the second identity, we
simply replace Á by Á? and u the fact that ťo\,ť(Á') * col(Á).
Thus, an m X n matrlxhas four subspaces: row(Á), null(Á), col(Á), and null(Ár).
The first two are orthogonal complements in R', and the last two are orthogonal
].
Chapter 5 Orthogonality
nultr(Ár)
TA
mfi míí?
rlsurg ,B
The forrr fundamental srrbspaces
complements in R"'. The m X n maťix Á defines a linear transformation from R'
into R'whose range is col(Á). Moreovero this transformation sends null(Á) to 0 in
R*. Figure 5.6 illustrates these ideas sc.hematically. These four subspaces are called
thefunilanental subspaces of the m X r matrix Á.
Find bases for the four ťundamental subspace §f
andveri$Theorem 5.10.
.:ii;;iii i] In Examples 3.45,3.47, and 3.48, we computed bases for the row space,
column space, and null space of Á. We found that row(á) : span(u1, ll2, u3), where
ut: [1 0 1 0 -1], rr2: [0 I 2 0 3], u:: [0 0 0 L 41
Also, null(Á) : span(x1, xr), where
xl : ,X2
To show that (row(Á))a : nrrll(Á), it is enough to show that everf u; is orthogonal to
ŘF-"& each x;, which is an easy exercise. (\{hy ir this suf&cient?)
L,i]
:l]
1l
0l
0J
1 1 3 1 6l
? *1 ů 1 *1
l
-3 21 -2 1I
4 1 6 1 3J
B81
-3y^.//a
is the &rnc as the column
l]w:L,l]
basis ťar W* .
ii The subspace W spanned by w1, Iť2l &ítd w,
,f
r 1 -1 ol
|-, 1 -1 l
^:|
s 2 4l
l o -2 *1
l
L 5 3 sl
Let W be the subspace of R5 spanned by
t_ll
wi_|
'l,w2--I ol
L sJ
Fínd
nlut
pac
Section 5.2 orthogonal Camplements and Orthogonal Projections
The column space of A is col(Á) : span {a1, &2, er), where
!-ll rtl tllít1 :i ;l l*, l
lAIe stil1 need to compute the null space of ÁT, Row reduction Produce s
[l _i -;
il:] [; ? ; ;l;]
Á,i0 :ll ? _: il:l--l; 3;il:l
[o --1 1 :loJ [0 0 0 0l0]
o, if y is in the null pace of ÁT, then y, * *!a,
lz* -6yn, and y3 It
ťollows that
null Á :
{ L,ť|:
span[:i]
)
and it is eesy to check that this vector is orthogonal to a1 , &2, and a3.
The method of Example 5.9 is easily adapted to other situatÍons.
Chapter 5 Orthogonality
perpu(v)
ílsure s.l
v : proj,,(v) * perp,r(v)
r 1 -3 5 0 slol [r
tÁ'lo] :|-, 1 2 *2 r|o|-*|*
L 0 -1 4 -1 sloj Lo
Hence, y is in Iťr if and only if yr: *hyn
- 4yr, lz: -?q
follows that
LL Is
Therefore, by Theorem 5.10, Wr : (col(Á))a = null(ÁT), and we may proceed as in
the previous example. We compute
0 0 3 4l0l
1 0 1 3l0l
0 1 0 zlo]
- 3yr, and ll : -2y5,It
|-i])
and these two vectcrs form a basis for Tťl
0tthOsonil Proiestion$
Furthermore} the vector perp"(v)
candecompo eya
prajo(v) is orthogonal to projo(v), and we
v: projo(v) * p.rpo(v)
as shown in Figure 5.7.
If we let I4r: span(u), then w : projo(v) is in Iť and wl : perpo(v) is in I,V'l. We
therefore have a way of ?ecomposing" v into the sum of two vector , one from l,l/ and
the other orthogonal to lť-namely, v : w * wJ. We now generalize this idea to Rn.
Recall that, in R2, the prajection of a vector v onto a nonzero vector u is given by
projo(v)-(X)"
lIsflnlfi
basis for W.
defined as
1l "\u,. /
proj"{v)
Section 5.2 orthogonal Cornplements and Orthogorra1 Prorections Ess
ílgu]o 5.8
P:Pr*Pe
Since the vectoíS tl;8íe orthogonď, the orthogonal projection of vonto-W'is the sum
of its projections onto one-dimensional subspaces that are mutually orthogonal, Figure
s.-8 iťustrates this situation with I4l = span(u1, 2), P : Projry(v), P, : Projo,(v),
andpz : projo,(v).
Á,
"
rp.á.ase of the definiťon of projry(v), we now also have a nice geometric
interpretation of Theorem 5.2. In terms of our pre nt notation and"terminology,
that theorem states that if w is in the subspace W of R', whió has orthogonal basis
{vl,Y2,,..,v/,},then
w:(H",+
* proju,(w) +
+( *)"*
projou(w)
Thus, w is decomposed into a sum of orthogonal projeďions onto mutually orthogonal
one-dimensional subspaces of Iť.
The definition above J..*, to depend on the choice of orthogonal basis; that is,
a dtíferent basis {u|, . . . , ui} for I4l would appear to give a "different" projyy(v) and
perpry(v), Fortunately, this is not the case, as we will soon prove. For noq let's be
content with an example.
Let lďbe the plane in R3 }yith equation x * y + 2z :0, and let v :
§ ffi In Example 5.3, we íoundan orthogonal basis ťor 1ť.Taking
U1 x and 82 =
orthogsnel prcjection of v ontů W and the component of v orthogo
r 3l
| -' |.
r'r,a,r,*
L ?,)
nal to W.
Ll]
Itl
Ll]
wehave ll1 ,y:2
ll1 'lli:2
llž'Y : *7
U2' tl2 :
perpry(v)
projry(v): (# )",
* (ffi.,
{i]{i][;]r J L -3J
and n;rnw(v) =v- proj,"*, =
[ il I l]
=
[ i]L2J
. It is easy to see that projlv(v) is in I44 since it satisfies the equation of the plane. It
is equally easy to see that perpy(v) is orthogonal to W, since it is a scalar multiple of
l- rlll
the normal vector | - t I
to },lz. (See Figure 5.9.)
l l \ 9-,-- --- ,,
LzJ
Thereťore,
Hsu B .g
y l* projp{v} + perpry{v)
The next thearem shows that Ýye can always find a decomposition of a yector with
re pect to a subspace and its orthosonal complement.
, i. Chapter Orthogonality
ísm.ll
Pt00l We need to show two things: that such a decomposition exists and that it i
unique,
To show existence, we choose an *rthogonal basis {ur,
w : projw{v) end let wr * perpnr(v). Then
ut} for W Let
w * wr - projir(v) * perpw(v) * projfir{v) + (v * projw(v}) : y
Clearly, w = projry(v) is in 14| since it is a linear combination of the basis vectors
u1, . . . , u6. ro snow *r,at wr is in Ýý'r, it is enough to show that wJ is orthogonal to
eaih of the basis vectors q, by Theorem 5.9(d). 14Ie compute
u;.1111 = u,. perp1a,(v)
= út,(v - projry(v))
l (ul.v\ /uo,v\ \
- o,,(o - (_ř*l"1-, , ,- (ua/*,l
. =lli,v*(#*)*,,ur)-, (ffi)(u;,u)-",
_/ o*'o \r,, .,,-,,
\o*,
lto)\"i "r'
- 0 -...- í:r:)(q.u) _..._ o=ui.y-U-...-\"ťu,i'
= Ui'V - Q'V_= 0
since u1. q = 0 for j * i. This proves that wJ is in 1ťr and completes the existence
part ofthe'proof.
' To show the uniqueness of this decomposition, leťs suppose we hlve another de_
compositionv= y, i1,|,wherewlisin lťandwiisin w].Tlrenw*wJ = w1 twf,so
\ť-'Wt=wť-w'
But since Iť - I{t is in lť and wf _ wJ is in WJ (because these are subspaces), we
know that this common vector is in W í''l l,tl'J = {0} [using Theorem 5.9(c)]. ThuS,
Section 5.2 Orthclgonal Complements and Orthogonal Projections ::
\f*wt*wi*w-L:0
o1ť1 = Wandwi = wt.
If W is a subspace of R"
,:lil
Example 5.11 illustrated the ort}iogonal Decomposition Theorem. Ywhen W is
the subspace of R3 given by the plane with equation x _ y * 2z : 0,the orthogonal
[ ,l
decompositionofv = 1 -r | *itrrrespectto Wisv: w * wJ,where
- L ,)
ril [3l
y = pro;*(v) = | i l and wr = pe{pw(") = | -íl
L-t] L 3l
The uniqueness of the orthogonal decomposition guarantees that the definitions
of projry(v) and perpry(v) do not depend on the choice of orthogonal basis. The
Oritrogóna Decomposition Theorem also allows u to prove property (b) of Theorem
5.i. We state that property here as a corollary to the Orthogonal Decomposition
Theorem.
(w')'
Chapt*r 5 Orthogonality
P1O0í If w is in W and,x is in lť'1, tlren w. x : 0. But this now implies that w is in
(I4/')'. Helce, tď G ( MÉ)a.Now let vbe in ( Iťa)a. By Theorem s. t 1, *. can write
y = w * wr for (unique) vectors w in 1ť and w' in Wi. But now
0 : v.wr = (w + wl).wJ : w.wt + wJ.w] = 0 + wf.wr = lťl.w.
so wJ : 0. Therefore, v = w i wJ = W and thusvis in }ť. This shows that (Wa). e W
and, since the reverse incluďon is also true, we conclude that (I4lr)' : Iť, as required.
,]il
There is also a nice relationship between the dimensions of 14/ and lťr, expressed in
Theorem 5.13.
Tlt$OťGm S.lff ífW is a subspace ůfR"
P]OOí Let{u1,....,ut}beanorthogonalbaďsfor14/andlet{v,,,,.,vl}beanorthogonalbasisfor
Wa.Thendim t4l: kanddim Wt : LLetB = {ú,,..,',upv1,,..,rn}.
We claim that 6 is an orthogonal basis for Rn,
lt/e first note that, since each u1 is in W and each v; is in 1ť'l,
, u;.yj=0 forí= l,,,.,kand7: l,,,.,l
Thus, B is an orthogonal set and, hence, is linearly independent, by Theorem 5,1,
Next, if v is a vector in Rn, the orthogonal Decomposition Theorem iells us that v =
w * wa for some w in Iť'and wa in wl. since w can be written as a linear combinaťon
of the vectors u; and wl can be written as a linear combination of tlre vectors v;, v
can be written as a llnear combination of the vectors in B. Therefore, 8 spans Rn aiso
and so is a basis for Rn, It follows that lc * l + dim Rn, or
dím W + dim 1g'' -Ť n
As a lovely bonus, when we apply this result to the fundamental subspaces of a
matrix, we get a quick proof of the Rank Theorem (Theorem 3.26), restatid here as
Corollary 5.14.
SorolltíU S.I4 The Rank Theorefil
If Á is an rn X ru matrix, then
rank(Á) + nultity(Á) : ?x
i,!i,:i,li.ii In Theorem 5.13, take I4l : row(Á). Then Iťa : null(Á), by Theorem 5.10,
so dim W: rank(Á) and dim L4za : nullity(Á). The result follows. :.;-ff
Note that we get a counterpart identity by taking I4l : col(Á) [and therefore
Wa : null(ÁT)]l
rank(Á) + nullity(e')
Section 5.2 orthogonal Complements and Orthogonal Projections
Sections 5.1 and 5.2 have illustrated some of the advantages of working with
orthogonal bases. However, we have not established that every subspace has art or,
thogo;al basis, nor have we given a method for constructing suc;h a basis (except in
pur[.lrlar examples, such as Example 5.3). These issues are the subject of the next
section.
xnrsi 8 s,a
Wr afw
In Exercťses 9 and ffi, fi.nd bases for the column space af
A and the null spnce af Ar for t?te given xeťcise. Yerily
that every ve tor in caí(A) is orthogzrral to every vectoť
in null@r).
11. \ťt :
12.'lťt :
13. \ťt :
14. w, :
In ,xercises 15_18, find the orthogúnal projection alv ontz
the subspace W spanned by the vectors u;. {}fru m#y §, síÁrrre
that the vectors |l;6.f orthogofial.)
15"y:
16.Y:
17.v:
18"Y:
e 11*14, Iet W be the subspece spann á by thr
ors. Find a basisfor WL.
L1],*,:Ll]
L-l]-:Ll]
Il]-,:L-i],ť:Li]
r4l [.1l tzl
Iol -l ;l,*,:\:l
1-1l,*,
L_,]L-l]Llj
In ,xercis
given vect,
In xercises 7*6, frnd the orthogonal comPlernent
and give a basis ío,WL.
1. Iť:
{L;]
:2x - y - -}
2.w:{i-'l :3x*4y:o}
LLIJ / )
(fxl )
3.w:{l;l,, +y-z:oIJ,
tl;l
- ,/ )
íl."lrr l4.w:
tL;]
,r- * y + 3z: oí
í[-l 1
5. Tť : { l yl,* : t,! : *t,7 :rr|
|, L;] )
6.W:{
L;]
:x:l,,, -it,Z:,,}
L -i], ",
:
[1]
Ll],",:[l],1l2:Li]
L_1],",
:
Ll]
",
:
L-l]
",
:
Ll]
In Exercises 7 and 8, find bases for the row space and null
sp1ce af A"Vertfy that avery vectar in row(A) is orthogonal
ta every vectar in null(A).
t 1 *1 3l
l s 2 1l
7.A_l o 1 *2l
L-r *t rj
t- 1 1 -1 0 2i
|-z 0 2 4 4l
8,Á_| , 2 *2 0 1l
L*s *1 3 4 s_l
10. Exercise B9. xercise 7
Chapter 5 Orthogonality
In Exercises I9* ,2, rtnd thg orthogonal decompasitian af
v witLt respect to W.
19.v: r _il, *,: pan([i])
L*
20v:Ii],,:pan{Li])
'l.v:[ 1],,:spanfii] L i])
22,v:L;]|fipa{Li]Li])
23. Frave Theorem 5.9{c},
24, Prove Theorem 5.9(d).
2 . Let lťbe * snbspacc cť ffin and v a ys tcr in ffin. Suppn
that w a-nd $, g"r *rthogonal vectors with w in W anď
that v ffi w * w'. r ít necensarily true that w' is in WL?
fiither provť that it is true or find a counterexflmple,
26, Let {vr, . ) . ,vr! be an orthogonal basis for ffin and let
|rtí:- span(v!).,,,y*}. Isitn ťearilytruethat WL *
Pan(Yt*t, , * l } Yn}? ithcr proye that it ístrue or find a
cnunterexarrrple.
In frxerrises 2,7*29, Iet W be a subspace af Wn, and let xbe
a vector ln ffi'.
27. Prcve that x ísin W iť*nd anly if proj w{x} ffi x.
28, Prove that x is orthogonal t* W if and only íť
projw(x) : S.
ž9. Frove that projvrr(prnjn (x)) * pťojw(x).
30. Let n {or, r , r ! v1} be an orthonsrm*l set in Rn, and
let x be a vector in ffin.
(a) Prove that
ll"ll' x l*.u,l' + l*"orln í*, o*l'
(This inequaliťy is called Bessgl's In*quality,}
{b) Prove that Bessel's Inequelíťy í* an equality if and
only iťx is in span(, ).
Tlts firnm- §hffiidil Prgt$
anfi tn$ ť#rtst0rnfitlún
ln this section, we presenť'a simple method for constructing an orthogonal (or orthonormal)
basis for any subspace of R'. This method will then lead usio one of the
most usefirl of all matrix factorizations,
T[e Gram- Ghmidt Procots
We would like to be able to find an ortlrogonal basis for a subspace I4rof Rr. The idea
is to begin with an arbitrary basis {x1, . . , , xti for W aln,d to "orthogonalize" it one
veďor at a time, We will illustrate the basic construction with the subspace W'from
Example 5.3.
Let Tť* p&n{xr, xr), where
Xl ffi
Construct an orthogonal basis for M
{ll6í{iť3Starting with x,, \rye get a second vector that is orthogonal to it by taking
the component ofx. orthogonal to x1 (Figure 5.10).
Li]Ii]
ection 5.3 Thc firam- chmidt Process and the fift Sactarization
ls s .l;}
Constructing v2 orthogonal t* x1
Etí?iťf|Observe that this method depends on the order oíthe original basis
| -zf l-tl
vectors.InExample5.12,ifwehadtakenx,: I O |*a*r: l
' |,wewouldhave
L , _] Lo_]
,dť;* obtained a different orthogonal basis for W. (VeriŤthis.)
The generalizatian gf this method to more than two y ctor begins as in
Example 5.12. Then the process is to iteratively construct the components of subsequent
vectors orthogonal to all ofthe vectors that have alreadybeen constructed. The
method is known asthe Gram-schmidt process.
.l The Gram-Schmidt Process
*1l
1l
1l
{v,,v2} is a linearly indepenÁlgebraically,
w,e ct yl * xlr o
y2 * perp*,{x2) : x2 * proj*,{x,
{*r,xz\: X2 \,*ol-'
|-*zl ttl r
l ; l - r-:1)l ; l : I
L;] \2/L;] L
Then {v,,vr} is an orthogonal set of vectors inW. Hence,
dent set and therefare abasis for w, since dim w :2.
1
Vl : Xl,
Y2 x X2
Y3*X3
Chapter 5 Orthogonality
XF-b
Stated succinctly, Theorem 5;15 says that every subspace ofR'has an orthogonal
basis, and it gives an algorithm for constructing such a basis.
Pío0l Wewillprovebyinducťonthat, for each i : 1,.. ., k, {vr,...,q} isan orthogonal
basis for I4|.
Since v1 = xr, clearly {v1} is an (orthogonal) basis for W'1 : span(x1). Now assume
that, for some i d ft, {vr, . . . ,Ý} is an orthogonal basis for fi|. Then
/", ''q*,\ 1"r,,q*,\ /t'q*,\yi+t : ,.'+r - (ffiJ" - (ffi')y2 - ", - (ffil",a
Bytheinductionhypothesis,{v1,...,vi}isanorthogonalbasisforspan(x1,.,.,&)=
Iť;. Hence,
vl+t : 4+i - projr,,r(x;a1) : perpa(ry*r)
o, by the Orthogonal Decomposition Theorem, y,+t ísorthogonal to YV;.
definitíon, y1, t t - , yi are line&r combination of X1, , . . , Xi efid, hence, are in
Therefore, {\ 1, r . . } vi+r} is an orthogonal set of v tor in }ť;a1.
Moreover, v;+t * 0, since otherwise 4+r : projry,(4+r), which in turn implies
thatx;al is in I4{. Butthis isimpossible, since Iď'; : span(x1,. . .,xi) and {x1,. . .,xl+t}
is linearly independent. (Why?) We conclude that {vr, . . . , vl+ t} is a set of i * 1 linearly
independent vectors in l4li*r. Consequently, {vr, , . . , vl+r} is a basis for Wi+p
since dim Wo*l * i + 1. This completes the proCIf,. _ *..iifiW
If we require an orthonormal basis íot W, we simply need to normalize the
orthogonal vectors produced by the Gram-Schmidt Process. That is, for each i, we
replace v; by the unit vector qr = (1/ ||v,|l)v.
Apply the Gram-Schmidt Process to construct an orthonormal basis for the subspace
|l{ : pan(xlt x2l x3} of ffia, where
xl : , X2: ,X3
L1]
[1]
Ltl
[-l]
L 1l
liiii::l];;}* First we note that ix1, x2, x3} is a linearly independent set, so it forms a basis
for W. We begin by setting v1 : x1. Next, we compute the component of x, orthogo/v,
, x.\
nalto I4l': span(v1):
v, : perply,(x2) : x2 - (;*/o
[rl t 1l
Ll]$L-i]
t1l
lillil|-l
l-]J
Section 5.3 The Ciram- chmidt Process and the Qfi Factorization
Fot hand calculations, it is a good idea to "scalď'v2 at this point to eliminate fracťons.
When we are fi,niďred, we can rescale the orthogonal set we are constructing to obtain
an orthonormal set thus, we can replace each v; by ary convenient scďar multiple
without ďfecting the final result. Accordingly, we replace v2 by
[,1lgl
v'r= 2v2 =
l , l
L,]
Wg now find the component of x3 orthogonal to
W', : span(x1, x2) = span(v1, v2) : span(v1, vj)
using the orthogonal basis {v1, vj}:
v,: perpiy,(x3): x3 - (HŤ)",- (",.r)"'
:ii]_,},[_i],*,Ii]
Lll L,] L,j
:til
Li]
Again, we rescale and use v| : 2v, :
l il
L;]
B**Ť' We now have an orthogonal basis {v1, v'2,v!r} for W. (Check to make sure that
these vectors are orthogonal.) To obtain an orthonormal basis, we normalize each
vector:
L 1] L LllJ
r3l r3l2V5 l
Q: : (lMl|)", : (#)| i |
:
|:,:#l
Lr] !lzv6j
t-1l [-1lv6
{l:-:(r,rlr)",Ť(*)
| l|:i ,i*
L 2) L 2l\/6
Then {q,, {b, Qgi is an orthonorrnal basis for 1ť.
Irql10l
I lÝs /t0 l
I rrslro l
L ÝshOJ
;*Ýsla
|0
| Ýata
L Ýal3
1
],:
Chapter 5 Orthogonaliťy
One ofthe important uses ofthe Gram-Schmidt Process i to con truct an orthogonal
basis that contains a specified veďor. The next example illustrates this application.
váx
Find an orthogonal basis for R3 that contains the vector
[,l
v,:l2l
Ll]
,sťj l,
"ťt$s W. nrrt frnd anybasisfor R3 containingv1. If we take
|-o-1 1-ol
*,:|1| u.,a *,:IOI
Lo_] L, _]
then {v1, x2, x3} is clearly a basis for R3. (\ťhy?) We now apply the Gram-Schmidt
process to this basis to obtain
Vr : x2 (\' x'\
,
\v,í)"'
and finally
[:i]
Y3:
v{x
X3
[ *l
Ll
Y1, Y|,
Il],-,Li]
R3 that cantaínsThen {
Similarly, given a unit vector, we can find an orthonormal baois that contains it by
using the preceding method and then normalizing the resulting orthogonal vectors.
Remilt When the Cram-Schmidt Process is implemented on a computer, there
is ďmost always some roundoff error, leading to a loss of orthogonality in the vectors
q;. To avoid this loss of orthogonality, some modifications are usually made. The
vectors v; are normalized as soon as they are computed, rather than at the enó to
give the vector 9i, and as each q; is computed, tlre remaining vectors ry are modified
to be orthogonal to q;. This procedure is known as +he Moiltfieil Gram-Schmiilt
Procus. In practice, however, a version of the QR factorization is used to compute
orthonormal bases.
Tns ofr índorlratlon
If Á is an m X nmatrix with linearly independent columns (requiring that m > n),
then applying the Gram-Schmidt Process to these columns yields a very usefirl factorization
of Á into the produď of a matrix Q with orthonorrnal columns and an
ection 5.3 The ram- chrnidt Process and the QR actorization ". .
upper triangular matrix R. This is fhe QRfactoriaation, and it has applications to the
nrrmerical ipproximaťon of eigenvďues, whió we expl9re atthe end of this section,
and to the p.out.* oťleast squares approximation, which we discuss in Chapter 7.
To see tow the QR factorirxionarises, let a1, . . . , an be the (linearly independent)
columns of Á and let ql, . . . , ílnbe the orthonormal vectors obtained by applying the
Gram_Schmidt ProcesstoÁ with normalizations, FromTheorem 5.15, we knowthat,
foreachi=I,..,,n,
||/, .
Theqefore, there are scalars
3i * ťri{lr
Pan(*|) . .,, fir) ffi Pen(qtt . . ., {i)
Tli. ?2p l r, rr11 SUťh thet
+ Tziaz+,,, *l, riiryl for i * 1, .,,, ťl
That is,
a1 :' rr tílt
il|: ťnqrt + ťzz\z
a
:
, tn x ťyo{1t + ťznftz + ", ,T ťno{,n
which can be wrítten in matrix fórrn as
A : [a, il,2 an] : [q, í[z
Tu,
ť2n
')r,,
['"
Tn
, n,,l ? 'i,
L0 0
QR
Clearly, the matrix Q has orthonormal columns. It is also the case that the diagonal
entries of R are all nonzero. To see this, observe that if 1; = 0, then q is a linear
combinationofql,..;,;-1aild' hence,isinWi_1,Butthena;wouldbealinearcombinationof
a1,..., ai_r,ivhich isimpossible, incea1,..., alarelin arlyindependent,
Weconclude thatrii4 0fori = 1,...,1,1.SinceRi uppertťiangular,itfollowsthatit
must be invertible. (See Exercise 23.)
We have proved the following theorem.
Tltoorcm .16 The QR Factorization
Let Á be an m x n matrix with linearly independent columns. Then Á can be factored
as Á = QR, where Q is an rn X rrmatrix with orthonormal columns and R is
an invertible upper triangular matrix.
nOmulE
. We can also arrange for the diagonal entries of R to be positive.If any 1; { 0,
simply replace qi by -ql andriiby -rii.
l '
n
"
requirement that Á have linearly independent columns is a necessary one.
To prove this, s'uppose that Á is an rn x n mattkthat has a QR factorization, as in The,
or.* S.re. Then, since R is invertible, we have Q = ÁR-1. Hence, rank(Q) = rank(Á),
by Exercise 61 in Seďion 3.5. But rank(Q) = n, since its columns are orthonormal and,
therefore, 1inearly independent. So rank(Á) = ntoa)and consequently the columns of
Á are linearly independent, by the Fundamental Theorem,
Chapter 5 Orthogonality
. The QR factolization can be extended to arbitrary matrices in a slightly
modified form. If Á is m X n, it is possible to find a equence of orthogonal matrices
Q1, . . ., Qr-1 such tlrat Q,n_r ", QzQrÁ is an upper triangular m X nmatrix R. Then
Á = Qi, where Q : (Q,z-r ",QrQr)-l is an orthogonal matrix. We will examine this
approlh in Exploration: The Modified QR Factorizaťon,
Find a QR ťactorization of
112I
| *1 1
Al
l
*1 0
Ll1
{lz
Ť:*
{ls
*
21
2l
1l
2J
just the vectors from xample 5. t3. The orthonorrnal
Gram-Schmidt Proces wa
Ss Ě &nt The columns oťÁ are
basis for col{Á} produced by the
r Ll21
| *ttz l
Qr
:l*rnl'
L I/2 )
* \/6/61
0l
Ýsla l
,laft l
3\/Íl10l
3Ý5 /La l
I
r/s/rc l'
\/5/ 10 J
l t/z zxfslto -Ýs/e1
Q: [q, q2
",:| _i',i 'W:i }r,|
l_ t/z l/5/to {a/z)
From Theorem 5,16, Á = QR for some upper triangular matrix R. To find R, we
use the fact that Q has orthonormal columns and, hence, Q'Q : I Therefore,
Q% : Q QR: IR: R
y? lt-l ::1
#;:]
L li I
}1/e compute
R*Q%.:
*L/2 *!/2
lr6/tn r/s/to
0 Ý6/6
r I/2
l
I tVsltol*
L - ala
|z 1 Il2 l1"*1
: | 0 Vs ll,/s/z l
Lo 0 Ý6/? J
t- 3"l t-:l
_
L*r],*,-1 L,]
-[-ll,- [ol [:l
L_,]
J:
L;.J,":-- L;]
2. x,
3" x,
In Exercises 1*4, the given vectors farm a basis for W2 or R3.
Apply the Gram-Schrnidt Proce to obtain an orthagonal
basis. Then ncrmalize thtis basis to obtairu an arthon*rmal
basis.
I-rl l-rl
1.x,:Lr],*r:Lr]
]1 .', 'll,
':] ,', .: i: , , ]-,
Section 5.3 The Gram-Schmidt Process and the Q"& Factorization
4. Xt
TJs* the Gram-Schrnidt Proťes to nd an orthogonal basis
ío,the columft spa.ces of tha matrices in Exercťses 9 and La.
In Exgrcises 15 and 16, find a QYfactorization of the
matrix in the given gxercise.
15. Exercise 9 16. Exercise 10
18.Á:
1
?
*1
0
19. If Á is an orthogonal matrix, find a QR factorization
of Á.
20" Prove that Á is invertible if and snly if Á : QR, where
Q is orthogonď and R is uppťr triansular with nonuero
entries on its diagonal.
In Exercťses 21 anď 22, use t|te methad suggested by
Exercise 20 to cornpute A*' ío, the matrix A in the given
gX rcasg.
ž1. Exercise 9 22, x*rcis e 17
?3. Let Á be an m X n matrix with linearly independent
columns. Give an alternative proof thnt the upper
triangular matrix R in a QR factorization of Á must
be invertible, using property (c) oť the Fundamental
Theorem.
24. L*tÁ be an m X n matrix wíth linearly independent
calumns and let Á * QR be a QR factorization oťÁ.
how that á and Q have the same column pace.
In Exercťses 13 and 14, frU in the rnissing entries af Q
to make Q an orthogonal matrix
r I/\/1 Ll\/íxl
13.Q:I o tlrtr *l
1_1
L-t/\/í Ll\/5 *J
[l]",š[l]-:Li]
2/ Á E
Il\h* #
0s
-3/\h4 *.
;
form a basis for a
ram-schmidt process
a
tors
eGl
rcro
hel
,14f.
|,8
glven ve
l*.Ápply tl
.l basis for
ql
I
4l
2)
d 6, tílr,
13 or Ra
wgonal
|-l
|,lÁ
,1 l':kl
L;
,'
,X2
l]",Li]
X2
In Exercises 7 and B, find the arth*gonaí decompositian oív
with respect to the subspace W
r 4l
7.y : | -n |,
ras in Exercise 5
L 3j
til8. v :
l o l,
Was in Exercise 6
L;]
l1. Find an orthogonal basis ťor K3 that contains the
[,l
vector
I
t
l.
Lsl
12. Find an orthogonal basis for Ra that contains the
vectors
fr"ťlt
rR
rth
5a
rf
or
l
,]'
2
*1
1
2
ig.
W,
an
Itl
|1
lo
íse
,il
lal
r
L
In Exerci
subspace
to obťain
6. Xt
;]
-]
|'lz
l t/z:
l,n
Ltlz
In xercíses l 7 and 18, the columns af Q were obtained by
applying the Gram-Schmidt Proc s tg the colurrlns al A.
Find the upper triangular maťrix R such that ! : QR.
|- 2 8 21 [i } 1l
17,A:I r 7 *1
|,c:I l ? -*l
L--z -2 1] L-i ! }J
tlxG tlu"5)
ll\/a 0 l
-Ll\/6 tlrtl l
0 LlÝ1 )
3]
4l
1l,Q:
1]
14.Q
r 1 1 1]
10. l , -1 2l
L-ll:]
[o 1 1l
9.|t 0 1l
Lt 1 0l
i]andLl]
-1
[xpl ratl0
tlrnrn s-!l
The Modified Q^R Factorir,ation
When the matrixÁ does not have linearlyindependent column , the Gram-Schmidt
Process as we have stated it does not work and o cannot be used to develop a generahzed
QR factorization of Á. Ťheie is a morlification of the Gram-Schmidt Process
that can be used, but instead we will explore a method that converts Á into upper
triangular form one column at a time, using a sequence of orthogonal matrices. The
method is analogous to that af LU íactorization, in which the matrix t is formed
using a equence of elementary matrices.
The first thing we need is tlre "orthogonal analoguď' of an elementary matrix;
tlrat is, we need to know how to.construct an orthogonal matrix Q that will transform
a given column of Á-call it x*into the corresponding column of R-call it y.
By Theorem 5.6, it will be necessary that ||x|| = llq*ll = llyll. Figure 5.11 suggests a
way to proceed: We can reflect x in a line perpendicular to x - y. If
":(r5)r--rl :l';,]
is the unit vector in the díreďion of x - y, then ul :
|-';)is
orthogonal to u, and
we can use Exercise 26 in Section 3.6 to find the standard matrix Q of the reílection
in the line through the origin in the direcťon of ul.
ffi I * }rttlT.
?. Compute Q for
{b) x
We can generalize the definition of Q as follows. If u is any unit vector in Rn, we
defineann X n matrixQas
Q * I* Zuur
1" Show that o -:* [
t 2d? *2d,d",1
L*zd,d, l * ,o-i|
L;],
T :
Li]
r3l
L5J
-ťf,
-J
C
"e
Z-
í)
i:ť
_ll
(f
u2
qJ
(]
L)
(a) Q is symrnetric. (b) Q is orthogonal. (c) Q2 = l
4. Prove that if Q is a Householder matrix corresponding to the unit vector
u, then
^ ( -v if v is in span(u)
Qv= {
L Y trv,u=0
5. Compute Q for u : arrd veri{T Problern 3 and 4.
6, Letx+ywith ll*ll = llyll andsetu = (t/llx*yllX*- y).Provethatthe
corresponding Householder matrix Q satisfies Qx = y. |Hint: Apply Exercise 57 in
Section 1.2 to the result in Problem 4,]
7, Find Q and veri$, Problem 6 for
We are now ready to perform the triangularization of an tn X n matr*Á, column
bycolumn,
8. Letxbethe firstcolumn ofÁ andlet
y:
Show that if Q1 is the Householder matrix given by Problem 6, then QlÁ is a matrix
with the block form
*l
Á,_]
whereÁ1 iš(m * l)X(r - 1).
ťwerepeatProblem 8 on the matrixÁ1, w u aHouseholderm atrkPrsuch that
Li]\ďas one of the pioneťrs in the field
of numerical lirrear algebra. He
wa the first to present a systematic
treatment oť algorithms for solving
problems involving linear systems.
In addition to introducing the
widely u ťdHousehcllder transforrnations
that bear his name, he
wa on of t}re first to advocate the
ystematic use of norms in linear
algebra, His 1964 book The Theory
af Matrices in Numerical Analysis is
collsidered a classic.
x*[i]andy:Li]
9, et Q, Ll
Q,Á
Pň,
QzQrA_
Ix
I
Lo
Ixl
I
L0
: l, Show that Qe is an orthosonal matrix end that
Prl
\4 l)
[;;;]
Lo 0 Ar)
-l
Ar)
where A2ia {m
* Z)X{n * ž},
/
]
Q,,n* t
11. Deduce that ! - QR with Q : QtQz, " Q*-1 orthogonal,
L2. t]se the method of this exploration to find a QR factorization of
t- 3 g 1l [t 3
L*+ J l)
L2 -5
321
1 1l
-1 -2J
&Approxim
ati*g Eigeíryalue
with the QR Algorithm
one of the best (and most widely used) methods for numerically approximating the
eigenvalues of a matrix makes use of the QR factorization. The purpose of this exploration
is to introduce tlris method, the QR algoňthm, and to show it at work in a
few examples. For a more complete treatment of this topic, consult any good text on
numerical linear algebra. (you will find it hebful to use a cAs to perform the calculations
in the problems below.)
Given a square matrix á, the fust step is to factor it as Á : QR (using whichever
method is appropriate). Then we define Á, = ftQ.
1, First prove that Á1 is similar to Á. Then prove that Á, has the same eigenvalues
as Á,
2, Oo : l1 |j, n"aÁ, andverifythatithasthe same eigenvaluesasÁ.
Ll 3), ,--_/ ----_
Continuing the algorithm, we factor Á1 as Á1 = QrR, and set Á2 = RlQ1. Then we
faďor A2 = QzRz and set Á3 = RzQz, and so on. That is, for k ž1, we compute Á& =
QlR1 and then set Ála1 : RtQr.
3. Prove that Áp is similar to Á for all k > 1,
4, Continuing Problem 2, compute Á 2, A3, A4,and Á5, using two-decimal-place
accuracy, What do you notice?
It can be shown that ifthe eigenvalues ofá are all real and have distinct absolute
values, then the matrices Á1 approach an upper triangular matrix U.
5. What will be true of the diagonal entries of this matrix U?
6. Approximate the eigenvalues of the following matrices by applying the QR
algorithm. Use two-decimal-place accuracy and perform at least five iterations.
(a) |z 3l
L2 lJ
t 1 0 *1l
(c)l r 2 1l
L*n 0 1]
QR algorithm to the
(b)
L; i]
t t l *1l
(d)
i o 2 0l
|.-z 4 2)
7. Apply the
1ď'},y?
t 2
:l what happens?:
L-r -2J
matrix A
ťr*m the (approximat*)
, *-.)*---,,
eigenvalues rrf S. Verify that this method approximates the
eigenvalues of Á.
g, Let Q6 = Q and Ro : R. First show that
QoQr''' Qt-,Át : ÁQoQr''' Qr-,
for all k > 1. Then showthat
(QoQ,. . .QJ(no. , .RlRo) : Á(QoQr, , ,Q*-,XRo-,, , ,RlRo)
[HlnŤ: Repeatedly use the same approach used for the first equation, working from
the "inside outl'] Finally, deduce that (QoQr ...
QrX&, , ,RrRo) is the QR factorization
oíAk+l.
Chapter Orthogonality
0ítnogonal lllagonallration oíymmgtílo I[at]lccs
l{e saw in Chapter 4 that a guáre matrix with real entries will not necessarily have real
eigenvalues. Indeed, m" *ut i* ||
-1l n* complex eigenvalues l and -i. We also
Ll 0]
discovered that not all square matrices are diagonalizable. The situation changes
dramatically ťwe restriď our attention to real symmetric matrices. As we will show
in this section, all of the eigerrvalues of a reď symmetric matrix are real, and such a
matrix is always diagonalizable.
Recall that a symmetric matrix is one that equals its own transpose. Let's begin by
studying the diagonalization proce for a symmetric2 X 2 matrix.
If possible, diagonalize the matrix
^
:
l: _X
*crinl: ThecharacteristicpolynomialofÁis,\2 +,\ - 6: (^ + 3X^ - 2),from
which we see that Á has eigenvalues ,\, : - 3 and lz: 2. Solving for the corresponding
eigerwectors, we find
y1 : t_ll and y2:
L LlJ
r p*ctively- o Á is diagonalízable, and iť we et P
r-3 0lp-LAp*l r:*:D,
L 0 2J
[i]
* [v, ve ] , then we knnw that
However, we can do better. Observe that v, and v2 are orthogonal. So, ťwe normalize
them to get the unit eigenvectors
t l/\fr1 ,
Ui :
l_rt\6]
and r2:--
and then take
Q * [u, uz] :
|_y,s y, 1
we have Q-'ÁQ = D also. But now Q is an orthogonal matrix, since {u1, u2} is an
orthonormal set ofvectors. Therefore, Q-' = QT, and we have QlÁQ = D. (Note that
checking ísa y} since cornputins Q*l only involves takins a tran po e!)
The situation in Example 5.1ó is the one that interests us. It is important enough
to warrant a new definiťon.
We are interested in finding conditions under which a matrix is orthogonally
diagonďizable. Theorem 5,17 shows us where to look.
lztvsl
Lrtvs]
ection 5.4 Orthogonal Diagonalization of Symmetric h{atrices .]...
Thno]0m
Pí00l If Á is orthogonally diagonalizable, then tlrere exists an orthogonal ma,
trix Q and a diagonal ňratrix D suČhthat QTÁQ = D. Since Q-l = Qr, we have QrQ =
1= qqr, so
QDQT=QQ%QQT=IAI=A
But then
, Ar = (QDQr)? = (QT)TDTQT = QDQT = /
since every diagonal matrix is symmetric. Hence, Á is symmďric, ,
n6nryt Theorem 5-17 shows that the orthogonally diagonalizable matrices are
all to be faund amongthe symmetric matrices. It does not say that every symmetric
matrix must be orthogonall/ diagonalizable. Howeve& it is a remarkable fact that this
indeed.is truel Finding a proof for this amazing result will occupy u for much of the
rest ofthis section.
We next prove that we dont need to worry about c omplex eigenvďues when working
with symmetric matrices with real entries.
.18
Recall that the complex conjugate of a complex number z = A * bl is the number
ž= a _ bi (see Appendix C). To show that a is real, we need to show that b = 0. One
way to do tLis is io show that z = ž,for then bi = _bi (or 2bi = 0), from which it
followsthatb:0.
We can also extend the notion of complex conjugate to veďor and matrices by,
for example, defining Á to be the íňatrix \,trhose entries are the complex conjugates of
the entries of Á; that is, f A = |ai),then Á = [áu]. The rules for complex conjugation
extend easily to matrices; in particular, we have AB = AB for compaťble matrices
Á and 8.
P]oot uppose that
^
is an eigenvalue ofÁ with corresponding eigenvector v. Then
Áv = z\v, and, taking complex conjugates, we have Áv : ,\v. But then
aT:aT =Ň:Ň:rrv
since Á is real, Taking tran po es and using the fact that Á is symmetric, we have
ťA = írAT = (aÝ)r - Ň)'= I7r
Therefore,
(v14)v*(,Wlv:l(V\)
u?,,l * 0
bril
: l, o
b-i )
.+ k?, +
Chapter 5 Orthog*nality
-ň-OrorA-.&.]
;í
sÍnce v * 0 (becau e it is an eigenvector). We conclude that ,i
Hence, ,,\ is real.
jheorem 4.20,showed_that,,for any square matrix, eigenvectors correqponding
to distinct eigenvalues are linearly independent. For symmitric matrices, somethin!
stronger is true: Such eigenvectors are orthogonal,
&:í *i Let v, and v2 be eigenvectors coíre pon{ing to the distinct eigenvalues
)q * lzso that Áv, = r\rv, and Áy2 : i2y2. Using Ár : Á and the fact that i. y : xry
for any two vectors x and y in R', we lrave
^,(vr.vz):
(^,v,).v2: Áy' .y2 - (Áv,)Tv,
: (vIÁlv, : (vf,4)v, : v1(avJ
: vl(lrvJ : ,\r(vfur) : ,\2(v,.v2)
Hence, (,\, * ,\z)(vr.vr) : 0, But,\r - ,\z * 0, soyt.yz : 0, aswewishedto show
1" Thus, eigenvectors c*rrc ponding to the
Veri$l the result of Theorem 5.19 for
o:l?; ll
L; ;;)
0luilOn The characteristicpolynomial ofÁ is -.,\3 + 6^2 - g^ + 4: -(^ - 4).
(^ - 1)2, from which it follows tirat tt e eigenvalues of Á are ,\, ] + and ,\2 : 1. The
corresponding eigenspaces are
- :.,-(li]) and u, :,n*(l
i] t ;])
(Check this.) We easily veri that
rl]|i]Oand|i]|-;]:,
L,_
from which it follow that every vector in Ea is orthogonal to every vector in č,,
$ryhy?) ^
W@"
[-,l [-,l
u ne d not be orthosonalsfrme eigenval
ection 5.4 Orthogonal Diagonalization of ymmetric Matrices
We can now proye the main result of this section. It is called the Spectral Theorem,
since the set of eigenvalues of a matrix is sometimes called the qrcctrum of tJae
matrix. (Technically, we should call Theorem 5.20 the Real Spectral Theorem, since
there is a corresponding result for matrices with complex entries.)
TilaOrsm .20 The Spectral Theorem
Let A be an n X rr real matrix, Then Á is symmetric if and only if it is orthogonally
diagonalizable.
Plo0t We have already proved the "if" part as Theorem 5,17. To prove the "only if'
implication, we proceed by induction on n. For r, : 1, there is nothing to do, since a
1 i 1 matrix is already in di4gonal form. Now assume that every ft X k real s}rrnmet_
ric matrix with real eigenvalues is orthogonally diagonalizable . Let n = k * 1 and let
Á be an n X n rea|symmetric matrix with real eigenvalues.
Let i1 be one ofthe eigenvalues ofÁ and let v1 be a corresponding eigenvector,
Then v1 is a real vector (why?) and we can a ume that v, is a unit vector, since
otherwise we can normalize it and we will still have an eigenvector corresponding
to ,\r. Using the Gram-Schmidt Process, lve can extend vl to an orthonormal basis
{v1,
y2, . . , , vn} of R'. Now we form the matrix
Qr [v, y2" , yn]
Then Qr is orthogonal,, and
QLAQl Y2'" Yr,] * [Áv, A:or" " Ar,)
[irvr Avrt,iÁ*n]
:B
In a lecture he delivered at t}re University of Góttingen in 1905, the German mathematician
David Hilbert (1862_1943) considered linear opeťator acting on certain infinite-dimensiona]
yector paces. Out ofthis lecture arose the notion ofa quadratic form in infinitely many
variables, and it was in this context that Hilbert first used the term spe ctrun to mean a
complete set of eigenvalues, The spaces in question are now called Hilbert spaces.
Hilbert made major contributions to many areas of mathematics, among them integral
equatiorrs, number theory geometry, and the foundations of mathematics. In 1900, at the
Second International Congress of Mathematicians in Paris, Hilbert gave an address entitled
"The Problems of Mathematics]'In it, he challenged mathematicians to solve 23 problems
of fundamental impoťtance during t}re coming century. Many of the problems have been
solved_some were proved true, others false-and some may never be solved. Nevertheless,
Hilbert's speech energized the mathematical community and is often regarded as the most
influential speech ever given about mathematics.
,_],. , , , ;: ,, is a Latin word meaning
"irnagei' When atom i vibrate, they
mit light. And when light passe
through a prism, it spreads out i!., ,.,,,,.
into a spectruffl*a band of
rainbow colors. vibratiorr
frequencies correspond to the
eigenvalues of a certain operatCIr
and are visible as bright lines in the
spectrum rlf light t}rat is emitterJ
from a prism. Thus, we can literally
see the eigernralues of the atonr
in its spectrum, and for this reasoíl,
it is appropriate that the word
spectrum has come to be applie d
to the set cf all eigerrvalues of a
matrix (or operator).
Lí]
|'']Á
v,
Lq]
:
|$]
L
"íJ
[A, i *lt *-----*_-*--* l
Lo i Á,]
{-l)
=lJ
č
cJ
(
š
L)
c
>!-
-f,
C
5Z
o<-)
-(:)
rJ
3
:tr
L
E
ELU
cn
Chapt*r Srthag*nality
fnce
v|(r\,v,) : ,\(vfo,) : ,\t(vt.vt) : ,\r andv|(r\rv,) : ,\,(v,fo,) : ,\r(vl.vr) = 0
íori * 1, because {vr, vr, . . . , vri is an orthonormal set.
But
BT : (QTAQ,)T : Q| T(QI)T:
QhQ, : r
so B is symmetric. Therefore, B has tlre block form
and Á1 is syrqmetric. Furthermore, 8 is sirnilar to Á (why?), so the characteristic polynomial
of 8 is equal to the characteristic polynomial of A, by Theorem 4.22, By
Exercise 39 in Section 4.3, the characteristic polynomial ofÁ1 divides the characteristic
polynomial of Á, It follows that the eigenvalues of Á1 are also eigenvalues of Á
and, hence, are real. We also see tlrat Á1 has real entries. (Why?) Thus, Ár is a k X &
real slrrnmetric matriíwith real eigenvalues, so the induaion hlpothesis applies to it.
Hence, there is an orthogonal matrix P2 such that ďALPzis a diagonal matrix*say,
D1. Nowlet
}"n a, is an orthogonat (/c + 1)X(& + 1) matrix, and therefore so is Q = QrQz.
Consequently,
'a%a
= (Q,QJ%(Q,QJ = (QTaI)Á(a,QJ = aT(aÍ,{QrEz: QTBQ,
which is a diagonal matrix, This completes the induction step, and.we conclude that,
íor ďl n > 1, trl n X n rea7 symmetric matrix with rea] eigěnvalues is orthogonally
dingon*lír*ble,
§rthrgonally diagan a\ine the matrix
B_ [,l, 0
l
L 0 Á,]
[t 0l
Qz :
Lo i,)
[,t,i0ll -----"l------ |
L0 iD,J
A*
1 1l
21l
1 2)
olullon This is t}re matrix from Example 5,I7. We have already found that the
eigenspaces ofÁ are
E4:* and 1
lz
Ll
:pen{Li]Li])pan([i])
ffi-e
ffii-*#b
Section 5.4 Orth*g*na} Diagona|izatian rrf Symmetric Matrices
Ure ne d three ortJronormal eig lythe fir*m- chmiďt Frocess to
to obtain
, [-tl
: The newvector, which has been con tructed to be orthogonal to l O
l,
is still in E,
L ,_]
|-tl
tt+ (why?) and so is orthogonal * l
'
|. Thus, we have three mutually orthogonal
Lr]
vectors, and all we need to do is normďize them and construct a matrix Q with the e
- Ýectors as its columns. Wb find that
;cí}\r$ t#r . First, we flpp
Li]andLi]
Li]*ndL:i]
l tt\,5 *I/\/1 -Il }Gl
a *lrt 5 0 2lÝ6i
L,rv3 L/\/T -Ilvtsl
and it is straightforward to veri r that
a?Áa
[+
0 0l
:|0 1 0l
L0 0 lJ
The Spectral Theorem allows us to write a real symmetric matrix Á in the form
n: QDQT, where Q is orthogonal and D is diagonal, The diagonal entries of D
are just the eigenvalues of Á, and if the columns of Q are the orthonormal vectors
91, . . . , qn, then, using the column-row representation of the product, we have
[^, o llqll
tr: qpqr: 1q, ... q,ll , " : ll : l
v< t:| *'L
; i,]L*]
IqIl
: [,\rgr ", ,r"q,tl , I
LqIJ
= r\rqr{I + }rzjz9tr + ", + A,q,qT,
This is called the spectral decomposition of A, Each of the terms );qq| is a rank 1
matrix, by Exercise 62 in Section 3,5, and q,ql i, actuaíly the matrix of the projection
onto the subspace spanned by q;, (See Exercise 25,) For this reason, the spectral
decomposition
Á :,\r9,ql + hz1zjT +,,, + }r,qnq|,
is sometimes referred to as the projection form of the Spectral Theorern.
Chapter 5 Orthogonality
Find the spectral decomposition of the matrix Á from ,xample 5.18.
Sg nxt mxg From Example 5.18, we have:
'1,1 4,
^2
= 1, ,A1 :
Q:*
Thereťore,
{r{I :
{z{lí *
qlqí:
[_1-i-l]
The rank 2 mattlxq2$ + qrq| is the matrix of a projecťon onto the two-dimensional
subspace (i.e., the plane} spanned by q, and {b. ( ee Exercise 26.)
Observe that the spectral decomposition expresse a slmrmetric matrix Á explicitly
in terms of its eigenvalues and eigenvectors, This gives us a way of constructing a
matrix with given eigenvalues and (orthonormal) eigenvectors.
Finda2X2matrixwitheigenvaluesr\1 =3andr\2=*2andcorrespondingeigenvectors
-Il\C)
zlvE l
l
-Ilxta)
9:|Il#] ,az:
L :;:1
I rt\^l I rts Ll3 I/31
lrt\/i_I ttlÝ3 Il\/3 I/\/3] - lrtu L/3 L/3
l
Lth6] Lt/l L/3 tll )
[-t lÝŽ1 t L/2 0 *Ll27
I o |t-L/\/1 0 L/\/íl _I o 0 0 l
L L/\/1 J L-Uz 0 I/2)
[-r/Ý61 r L/6 -L/3 L/61
l ,tÝ6 I
t- L/\/6 2/\/6 -L1\/6l - l -rtu 2/3 -Ll3l
L-t/\/6) L I/6 -I/3 L/6)
r á -i il
l r , ?l
|*i 3 -3 l
l r -L tlL 6 3 6J
last two terms ňzqz{Í + Al{le í
A* ň,qrql + ilqzqí + il{llílí
t-i i il t * 0 -il*4l l i il+| o 0 0|+
Ll i il L-l 0 }j
which c&n be easily verified.
In this example, ň2 * i3, so we could combine the
to get
so
[-+l
L 3]
and
§{tit(ři$,,,,,,,,,,,,,,* We begin by normalizing the vectors to obtain an orthonormal basis
{q1, q2i, with
t]l : 1-1]o,: Li]
and 9z L i]
NoW we compute the matrix A whose spectral decomposiťon is
A: ň,q,ql + izQr{lí"
r1] r *{l
:3i;lti *i z|;lr*g it
LTl l- EJ
19 12a r i6 l2llal
25 25| ri 25 25 l
L25 25l L -25 rSJ
|-_! i2l
l 5 5l:-i i: ól
L 5 5_]
It is easy to check that Á has the desired properties. (Do this,)
Sectiorr5.4or'ťhogona1Diagona1izationofSyrnmetrich{atrices
1l, If b *0,orthogonallydiagonalizeÁ - |; :1
|a 0 b1
lL.If b * 0, orthogonally diagonalize Á _ | n a 0 l.
|,u 0 o)
13. Let Á and B be orthogonally diagonalizabl e n X n
matrices and let c be a scalar. TJse the Spectral
Theorern to proye that the following matrice are
ortho gonally diaganalizable :
(a} Á+ (b) cA (c) A?
14. If Á is an irnrertible matrix that is orthogonally ďiagonalizable,
show that Á*1 is orthogonally diagcnalízable.
1 . If Á and . are orthagcnally diagonalizab}e and Á"B :
BA, show that ÁB is orthogonally diagonalizable-
16- If Á is a ymmetric matrix, show that eyery eig*rwalue
of Á is nannegative if and only if Á : B2 ťor some
}rínm tric matrix - "
Irc Exertises }7-20, frnd a spectral dgcompositiun af th*
watrix in the given exercise.
Ort?togonally diagonalixe the matrices in Exarcíses 1*10
by nding an orthogonaí matrix Q and a diagonal
matrix D such that a?Áa - D.
f+ 1l [*t 3l
l,Á:L, 4.] 2,A:L 3 -'j
l- 1 11 rg *21
3, A:
l* 0] 4, AL_,
6]
[s 0 0l |z 3 0l
5.A:Io 1 3l o.Á-|l 2 4l
Lo 3 1l Lo 4 2)
t 1 0 -1l [t 2 2l
7,A--Ť| 0 1 0l *.o-|2 1 2l
L-r 0 1l Lz 2 1]
[t 1 0 0] r2 0 0 1l
l, 1 0 0l l0 1 0 0lg,A:Io 0 1 1l'o,A:Io 0 1 0l
Lo 0 1 lJ [r 0 0 2J
Chapt*r 5 Orthogonality
17- Hxercise 1
19. Hxercise 5
18. xercise 2
2§, xercine 8
In Exercises 21 and 22, find il syrnmstric e Y. žmntrix wit|t
eigenvalues h7 and h2 anď corťťspondingorth,ogonal
eigenvtctors y7 áfiď y2-
21. ,l1 ffi *1, Áz LrYt
22,
^I
ffi 3r he tr *3, yt
In ,xercises 23 and 24, find a ymmetric 3 X 3 matrix with
eigenvalues.o\1, A2, and \ and cCIrre punding arthoganal
eigenvectors y1l y2, Andy3.
23.ňi ffi l,iz * 2,r\l * 3,Yt fi
Y*
24.
^1
H 1, ie ffi *4, A: ffi *4, Yt
Y3 Ť:
25. Let q be a unit vector in R' and Let W be the subspace
pa,nned by q, haw that the orthogonal projection oťa
vector y onto Iť (as defined in Secťon l,žand 5"2) is
given by
.
projw(v) (qqr}v
and that the matrix oť this prajeetian is thus qqr.
|Hint: Remember that, ťcr x and y in R', x. y; x?.]
26. Let {q,, r l r , &} be an orthcnormal set of vectors in Kn
and let 14/ be the subspa e spanneď by this set.
(a) how that the matríx of the orthagonal projection
onto 1ť is given by
p ff qrql + .*. *|*
qJrď
(b) how thnt the projectínn rnatrix in pert (a) is
ymmetric and satisfies FŽ x P.
(c} Let Q * [q, { l . *h] be the rr Y" k matrix whose
column ar the orthffnormal basis vectors oť w.
how that P : QQŤ and dgďuce that rank(P) * ft.
27 , LetÁ be an n X n rea|rnatrix, ell of whose eigenvalues
are real. Prův that there exigt nn orthogonal matríx Q
and an upper triangular matrix f sueh that a?Áa - T,
This very useful resnlt is known as Sehur's Triangular*
ixation Thearem, |Hirrť; Ádapt the prooť of the pectraI
Theorem.]
28. Let Á be a nilpotent matrix (see Exercise 56 in Section
4.2). Prove that there ísan orthngonal matrix Q
such that Qr eQ is upper triangular with zerff on its
diagon ď,, |Hiní; IJse xereis e 37 ,}
Ii]
[l],",:L1]
L;],",:Ii] t
Ll],",=Li]
ffi[1],,,ffi[l]
[1]
lluailratic íOtms
An expression of the form
axz+b72+cxy
is ca]led a quadratic form in x and y, Similarly,
ax2 + by2 * cz2 * dxy * exz * fyz
is a quadratic formin x, y, and z. In words, a quadratic form is a sum of terms, each of
which has total degree íwo in the variables, Therefore, 5fi2 - 3y2 * 2xy is a quadratic
form,butx2 + y'+ xisnot.
We can represent quadratic forms using matrices as follows:
ax'+ by'+ cxy * |x ,,[ ,i, '','1|;]
ectinn . Applications
and
I a d/2 ,/zfl x1
ax2 + byz * czz * dxy * exz + fyz : |x y "1|alz
U rtr|| yl
Le/2 í/Z , )LrJ
.}.ii*e (Verify these.) Each has the form xTÁx, where the matrix Á is symmetric. This observation
leads us to the following general definition.
W §ffi "W
form
where Á is a symm etric n
frs ociated with f,
ffiffie&#ffi lfx:
wiih ils cíated matrix ÁWhat is the quadr
-;]L;;] ::
,atic fo
L;;],
,
3rm
theren
2*? + 5*3 6xrx2
Observe that the off-diagonal entries dp = a2t * -3 of A ue combined to give
the coefficient *6 of xlx2. This is true generally, We can expand a quadratic form in
rr variables xTÁx as follows:
Thus, if i * j, the coefficient of xixiís 2ni3,
Find the matrix associated with the quadratic form
f(xl,x2,4):2x| - r} + 5x! * 6xrxr- 3xrx3
i}liixii,,] The coefficients of the squared terms xf go on the diagonal as a;;, and the
coeffrcients of the cross-product terms .Ti.tj are split between ailanď alt. This gives
t-2 3*+l
L-i 0 5J
/(x)_ xrAx- [x, *r][ :
L-*J
Chapter 5 Orthogonality
.,ť
(*,, Jť2, X3) :-:
I x, xl
e you can easily check.
In the case of a qu.adratic formf(x, y) in two variables, the graph of z = f(x, y) is
a surface in R3. Some examples are shown in Figure 5.12.
Observe.that the effect of holding x or / constant is to take a cro s section of
th. gruph parallel tothe yz or xz planes, respectively. For the graphs in Figure 5.12,
all of these cross sections are easy to identift. For example, in Figure 5.12(a), the
cross sections we get by holding íor / constant are all parabolas opening upward,
so/(x, f) >g for all values ofx and y. ln Figure 5.12(c), holding x constant gives
parabolas opening downward and holdingy constant gives parabolas opening upward,
producing a saddle point.
(b) í: -2x2 *3y?
(c) z:žxZ*3yz
]lgll 8 .12
Graphs of quad raticformsfl x, y)
(d) z : }xZ
t 2 3 -il[,, l
L--i 0 5_] Lx:l
(a) r, : 2x2 * 3y2
ection 5. Applications
What makes this type of analysis quite easy is the fact that these quadratic ťorms
have no cross-product terms. The matrix associated with suclr a quadratic form is a
diagonal matrix, For example,
2x2 3y2 :-/
In general, the matrix of a quadratic form is a symmetric matrix, and we saw in Section
5.4 that such matrices can always be diagonalized. We will now use this fact to
show that, íorevery quadratic form, we can eliminate the cross-product terms by
means of a suitable change of variable.
Let/(x) = xlÁx be Jquadratic form in n variables, wittr A a s}rrrrmetric n X n
- matrix, Bythe Spectral Theorem, there is an orthogonal matrix Q that diagonalizesÁ;
that is, QrÁQ = D, where D is a diagonal matrix displaying the eigenvalues of Á. We
now set
1 = Qy or, equivalently, y : Q-lx = Qt
Substitution into the quadratic form yields
x%x: (oy)%(ay)
= ťQkQy
: yTDy
which is a quadratic form without cross-product terms, since D is diagonal. Furthermore,
if the eigenvalues of Á are r\1, í.. , ,\r, then Q can be chosen so that
Iíy= íy,. , , /,fr, then, with re pect to these new variables, the quadratic form
becomes
y'Dy: try? + ,.. +
^,,y1
This process is called diagonalizing a quadratic form. We have just proved the following
theorem, known as the Pňncipal Axes Theorem, (The reason for this name
will become clear in the next subsection.)
[x ,,
[; -;] L;]
[l, ", 0l
D:
L; ... o,]
§.21
Chapter 5 Orthogonality
&*e"ffiť-,",
,
Find a clrange of variable that transforms the quadratic ťorm
f {*r, xz} : 5*? + 4xrx2 + Z*i
into one with no cro s-product term .
;,]ii.lii;i,ir.,ii The matrix of/is
[s 21
with eigenvalues At : 6 and hz:r,:*r[r:jing unit eigenvectors are
'
nl : |ri\6l t LlÝi1
(Check this.) If we set
LrtÝi -2/V5]
then a?Áa ":=
P, Tlre change of variable x :
x -T [-'l and
LxrJ
conyertr Íinto
r
ftv}:f{yr,?r): {y,
';l
and =
Qy, where
y: Lí;]
íl]L;;]
[a 0l
L0 1l
ayť+yi
The original quadratic form xŤÁx and the new one y,lDy 1r.ftrred to in the PrincipalAxesTheorem)areequalintlrefollowingsense.InExample5.23,suppo
ewewant
* l-r ''i
toevaluate/(x) = ar4*atx = | l l. W.lr"ue
L3j
í(-t,s) = 5(-t)2 + a(-tXr) + z(3)2 = 11
In terms of the new variables,
|:;,1:y:
Q,x:
|Y,# -ť,*] [-1]
so
í{yr,yr}: ay? + ytr= s{t l Ýil'+ ?r lvts)'
exactly as bef,ore.
The Principal Axes Theorem has some interesting and important con equences.
We will consider two of these. The first relates to the possible values that a quadratic
form caírtake on.
t
L
LlÝš1
*7
1V5J
55/5 Ť:: 11
Section5.5 Applications
A symmetric matrix Á is called positive definite, positive semidefinite, negative
definite, negative semiilefinite, or indefi,nite if the associated quadratic form
í(x) = x?Áx has the corresponding property.
The quadratic forms in parts (a), (b), (c), and (d) of Figure 5,12 are positive definite,
negative definite, indefinite, andpositive semidefinite, respectively. The PrincipalAxes
Theorem make it ea yto tel if a quadratic form has one of these properties,
TltOOtGm .22 Let Ábe an n X rr symmetric matrix. The quadratic form/(x) = xrAx is
a. positive definite if and only if all of the eigenvalues of Á are positive.
b. positive semidefinite if and only if all of the eigenvalues of A are nonnegative.
c, negative definite ifand onlyifall ofthe eigenvalues ofÁ are negative.
d. negative semidefinite if and orrly if all of the eigenvďues of Á are nonpositive.
e. indefinite if and only if Á has both positive aná negative eigenvalues.
)fuu nre a ked to prove Theorem 5,22 in Exercise 27.
Classi /(x,1, z) = 3i +
'}
+ 3} - 2xy - Zxz - 2yz as positive definite, negative
definite, indefinite, oť none of tlrese.
*1l
*1
l
3J
If a quadratic form /(x) = xlÁx is positive definite, then, since /(0) = 0, tlre
minimum value ofl(x) is 0 and it occurs at the origin. Similarly, a negative definite
guadratic form has a maximum at the origin. Thus, Theorem 5.22 allows us to solve
certain tlpes oťma,tima/minima problems easily, without resorting to calculus. A ty?e
of problem that falls into this category isthe constraineil optimization problem.
It is often important to know the maximum or minimum vďues of a guadratic
form subject to certain constraints. ( uch problems arise not orrly in mathematics
but also in statistics, physics, engineering, and economics.) We will be interested in
finding t}re extreme values of/(x) = x?Áx subject to the constraint that ll*|l = t.
In the case of a quadratic form in two variables, we can visualize what the problem
mearr . The graph of z = í(x,y) is a surface in R3, and the constraint ll*ll = t restricts
the point (x, y) to the unit circle in the xy-plane. Thus, we aíe considering those
points that lie simultaneously on the surface and on the unit rylinder perpendicular
to the 17 plane. These points form a curve lying on the surface, and we want the highest
anó lowest points on this cuťve. Figure 5.13 shows this situation for the quadratic
form and corresponding surface in Figure 5.I2(c).
n.) Sinc ell oťthese eígenvalue are p i-
*1
3
*1
thi
1.
trix
lues
defi
mat
tvalr
vť(
:ffi
nYi
rivt
The
eigei
pO it
,ffiF-4p urhich has
twe, f is ft
Chapter 5 Orthogonality
-2s
]lsure 5.18
The intersection of z * 2x2 * 3 with the
In this cí[el the maximum and minimum values af f{x, y)
est and lowest points on the curve of intersection} are 2 and
are just the eigenvalues of the associated matrix. Theorem
always the case.
: 2x2 * 3y2 1the high*
3, respectively, whic}r
5.23 shows that this is
ňr í{x}= io
The maximuíítvalue
corre ponding to r\r.
The minimum value of/(x)
ťorre ponding to .trr,.
P1O0í As usual, we begin by othogonally diagonalizing Á. Accordingly let Q be an
orthogonal matrix such that Q%Q is the diagonal matrix
[l, "t 0l
D-|, '.. :l
Lo ...
^,l
f|ren, by the Principal Axes Theorem, the change of variable x : Qy gives xláx =
yrDy. Now note that y - Qrx implies that
y'y : (Q'*)'(Q'*) : *'(Ql'Q'* : x'QQ'x : x'x
since Qr : Q-', Hence, using x. x : xTx, we see trrat ||y||
: r/ťy : {*:
ll*ll :_ t, Thus, if x is a unit vector, so is the correspondins i *a the values of xTÁx
and yrDy are the same.
Section 5.5 Ápplications
(a) To prove property (a), wc observe that if y -- ly, ?nlT, then
Thus, /{x) ň1 for
{ ec xercise 37.)
/(x) Ť xl{x : yT) y
: hrlr? + xryš +.,.+ h,,y}
hr!ť + h;1i +.-.+ hrtí
-Ť Á,(yi + yi +.,.+ yí,)
-:: n, llYll'
:^1
all x such that il*ll : ].. The prooť that /(*) ry An is similar.
(b) If qr is a unit eigenvector corresponding to i1, then Áq1 = ,\1q1 and
' í(q,) : qiqq, = ql,\,q, : ,\,(qlq,) : ,\,
This shows that the quadraťc form actually takes on the value A1, and so, by property
(a), it is the ma;cimum value ofl(x) and it occurs when x = qr.
(c) You are asked to prove this property in Exercise 38. ,
Find the maximum and minimum values of the quadratic fotm f(xt, x2) : 5x! +
4xp2 * 2x22 subject to the constraint x', + x?, = 1, and determine values of x1 and x2
for which each of these occur .
olullon In Example 5.23, we found that/has the associated eigenvalues ii : 6 and
ia = 1, with corresponding unit eigenvectors
lzl\/ii , I L/\/i1
Qi --
lrt\6j
ano {z :
L*rtVE]
Therefore, the maximum value of/is 6 when x, = Z/ Ýi and x, = l / x/E, The minimum
value of/ is 1 when x1 = I / \/Šand x2 = - 2 / \/E, (Observe that these extreme
values occur twice*in opposite directions*since -q1 and -q, are also unit eigenv
ct r ťor ň1 and i2r re pectively.}
§rnRiling 0unilrttic ]Ruations
The seneraI form of a quadratic equation in two variable .r and y is
aX2
where at least one oťa, b, and c is nonzero. The graphs ofsuch quadratic equations are
ca|7ed conic sections (ot conics), since they can be obtained bytaking cross sections
of a (double) cone (i.e., slicing it with a plane). The most important of the conic sections
are the ellipses (with circles as a special case), hlperbolas, and parabolas. These
are called thenondegenerate conics, Figure 5.14 shows how they arise.
It is also possible for a cross section of a cone to result in a single point, a straight
line, or a pair oflines. These are called degenetate conics, (See Exercises 59-64.)
Ttre graph of a nondegenerate conic is said to be in standard position relative to
the coordinate axes if its equation can be expressed in one of the forms in Figure 5.15.
Chapter Orthogonality
Cir*le
ílllí0§.l{
11l.L -.L
-'t!t
l. a,b >0
a- D,
Y* ilXL, ď}p0
]l uto S.lS
Y* &X2, a{0
Th* nonclegenerate coni
Ellipse or Círc I-ť; +
#
* L; a,b > ů
Ellipse Parabola Hyperbola
ct:b
y
a4b
Hyperbcla
* - *:1, tt,b> 0
b* a'
Parabola
X: oyL, a7a _?. -::"
a!2, a a 0
Ir_{ondegenerate conies in standard positinn
ection 5.5 Applications
If possible, write each of the following quadratic equations in the form of a conic in
standard position and identiff the resulting graph.
(a) 4xz + 9y' * 36 (b) 4x2*9y'*1*0 (c) 4x2 * 9y x fl
.,* ililiitj; (a) The equation 4x2 + 9y2 : 36 can be written in the form
t *ť:,94
so its graph is an ellipse intersecting the x_axis at (+3, 0) and the 7_axis at (0, -F2).
(b) The.equation 4x2 - 9 + 1 : 0 can be written in the form
ť-ť:,
i
so its graph is a hyperbola, opening up and down, intersecting the 1-axis at (0, +}),
(c) The equation +l - Sy: 0 can be written in the form
u=!*',9
so its graph is a parabola opening upward. ii
"q.,
If a quadratic equation contains too many terms to be written in one of the forms
in Figure 5.15, then its graph is not in standard position, When there are additional
termi but no xy tetm, the graph of the conic has been translated out of standard
position.
Identify and graph the conic whCI e quation is
x2*2y' 6x*&y+9š0
tII fitĚtllt We begínby srsuping the ,r end y term s*parately to get
{x2 6x} + Qy'+ sy) = *9
or
(x2-ex)+z(y'*4y):-9
Next, we complete the squares on the two expressions in parentheses to obtain
(x2 - 6x+ 9) + 2(y' + ay + a) = -9 t 9 * 8
or
(x-3)2+2(y+2)2:8
Wenowmaket}resubstitutions x' = x- 3andy' : y + z,turningtheaboveequation
into
{tI
(x,) 2
+ z(y,)z :8 or liJ *
(f'r)'
- 1
84
ki
Chapter 5 Orthogonality
Mjffir,t
This is the equation of an eliipse in standard positio.n in the x' y' coordinate sy tem,
intersecting the x'-axis at (+2\/1,0) and the7'-axis at (0, -ts2). The originlnthe x'y'
coordinate system is at x : 3, | : -2, so the ellipse has been tranďated out of standard
position 3 units to the right and 2 units down. Its graph is shown in Figure 5.16.
íIsillo,t0
A translated ellipse
Ifa quadratic equation contains a cross-product term, then it represents a conic
that has been rotated.
Identi and graphihe conic whose equation is
5x2+4xy*2y2:6
SOlUtlOn The left-hand side of the eguation is a quadratic form, so we can write it in
matrix form as xrÁx = 6, where
ls 21
Á:| l
L2 2]
In Example 5.23, we found that the eigenvalues ofÁ are 6 and 1, and a rnatrix Q that
orthogonally diagonalizes Á is
lz/xtl I/Ý51
u: l,r/* -z,tVr)
Observe that det Q = -1. In this example, we will interchange the columns of this
matrix to make the determinant equal to * 1. Then Q witl be the matrix of arotation,
by Exercise 28 in Section 5.1. It is always possible to rearrange the columns of an
orthogonal matrix Q to make its determinant equal to + 1. (Why?) We set
a* i I/Ý5 2/Ýi1l__l
L-2/ s t/Ýs],
instead, o that
[t 0-1
L0 6)
The change of variable x = Qx' converts the given equation into the form (x') TDx' : 6
by means of a rotation. If x' = 11,
.l,
** this equation is just
Ly,J
tx')'+ a(y')z_ 6 or S - {y')': 1
which represents an ellipse in the x' y'coordinate system.
To graph this ellipse, we need to know which vectors play the roles of e|
and ej _ [1l
'"
the new coordinate system.
t LlJ
of the x' andy'axes.) But, from x = Qx', w
Sectinn5.5 Applications
: *4x' * LTY'
Itl:
Lo]
positions
y'
(These two vectors locate the
have
t- I/\/i 2/Ýilrq izt\6l
2lÝí1
-lt/rls)
4li Il\/š 2l\/51[*,lllt
11
-1ll
\/=,j L- 2lÝštlÝi)Ly']:
,
X
ílgu]G 5.1l
A rotated ellipse
y
3
b
Ť
\\
*ž .\...,",/,"J'
\
\
*)-
\*,'\
\
w\*
2 Qeí :
|-{,|# ii*] L;]
:
|-|,|H]
and
QeL Ť
These are just the columns {r and qz of Q, which are the eigenvectors of Á! The fact
that these ďe orthonormal vectors agrees perfectly with the fact that the change of
variablc is just a rotation. The graph is shcwn in Figure 5.17.
You can now see whythe Principal Axes Theorem is so naned. If a real symmetric
matrix Á arises as the coefficient matrix of a quadratic equation, the eigenveďors
ofÁ give the directions ofthe principal axes ofthe corresponding graph.
It is possible for the graph ofa conic to be both rotated and tranďated out ofstandard
posiťon, as illustrated in Example 5,29.
Identify and graph the conic whose equation is
^2845x? * 4x1l * 2y' - ňx - VE
y + { x 0
But now rve also have
Bx _ BQx,-
L-*
SOlUtlOn The strategy is to eliminate the cross-product term first. In matrix form,
the equation is xl{x + Bx + 4 = 0, where
=|'2] and u:|_j1 _+1o=
L, 2J L V5 V5J
The cross-product term comes from the quadratic form xrÁx, which we diagonalize
as in Example 5.28 by setting x : Qx', where
r I/ š
a :
|r!,t\/i
Then, as in Exarnple 5-2B,
xrhx - (x')T.Dx' * (x')2 + a{y'}z
Chapt*r 5 Orthagorraliťy
i{}{Jť* . is
Thus, in terms of r' artd y' , the given equation becomes
(x')2 + 6{y')2 * 4x' - Lzy' * 4 : 0
To bring the conic represerťed by this equation into standard position, we need
to translate tlte x'y' axes. We do so by completing the squares, as in Example 5.27.
Wehave
This give* us the translation ťquation
X'' *#'*2 flnd y''*y'* 1
In the x"y" c*ordi ate y tem, the equetion íssimply
(x")2 + 6(y"}2 * 6
which is the equation of an ellipse (as in Exampte 5.28). We can sketch this ellipse by
first rotating and then translating, The resulting graph is shown in Figure 5.18. A
-4-
*1
*2
}))1:*
'\+#-+-1il- b" C,
ťIJ'2-ť-t
) | ,-) l ) t
a" D- C-
Z
-r2 \,Z
-I/
ilZ'b2
*Z ,,2 *2
-'1, r j-_ _. i
-1) | t) ? - l
(l- CI, C,
-r2 , }'Z
#-p
llipsoid:
x
Hyperboloicl 0f t\^/o sheets:
Elliptic parnbCIlCIid: z *
ff rs .lS
Quadric suďaces
Section 5.5 Applieatinn*
Hyperboloid oť *ne sheet:
Eltiptíc csne: z2 :
..2 ,.2
Hyperbolic parabolctid: ; * \ - #(l, D*
/
j
Chapter 5 Orthogonality
ponding orthogonal
to put it into standard position. Some quadrics in standard position are slrown in
Figure 5.19; others are obtained by permuting the variables.
Iďentify the quadric surface whose equation is
5x2 + LIy' + 2z2 + L6xy * ?,axr * 4yz : 36
36, where
li } 3l
Q:[qr q2 q,] :l? -? -lllr z zl
L3 a -íJ
Note that in order for Q to be the matrix of a rotation, we require det q = 1, which
is true in this case. (Otherwise, det Q : - 1, and swapping two columns changes the
sign of the determinant.) Therefore,
rTÁx corre
í[(
l
l
l
íh
,l
tI
lJ
1a
vith
2l
l
-1 ]
-?r _
{h:
íí1i
-1
l
l
I
l
Wi
i
*]
,,,.iilii,,:,,ll. The equation can be written in matrix fcrr
t-5 B 10
A*|B 11 *2
Lto *2 2
We find the eigenvalues of Á to be 18,9, and *9,
eigenvectors
Ll]L;],andL
re pectively. We normalize them to obtain
r+l r }l
Qr:lil,Q::| 1l,and
Lil L íl
and form the orthogonal matrix
L-1]
[ts 0 0l
QTaQ-J)-|o g 0l
L 0 0 -9l
veťiable x Qx', e se xrÁx : (x')I}x'
}2 9(z,)2: 36 or 9š * 9Ž _
24
cognize this equation a the equation of a
axes are in the directions of the eigenve,
is shown in Figute 5.2a.
,o
,t
i
rboloid cf one
{lt, {z, anrl q3,
36
,J'
ffi5
(z'J'
4
}rlrpe.
;tor ,
}er
{
afid, with the change of
18(x')e + 9{y'
From Figure 5.19, we re
sheet. The x,', /' , artď z'
respectively. The graph ,
Section 5"5 Applications
íiuurofi.?íl
A hyperboloid of one sheet in
nonstandard positinn
We can also identify and graph quadrics that have been translated out of standard
position using the 'tomplete-the- quares metlroď' of Examples 5.27 and 5.29, You
will be asked to do so in the exercises.
0unfiralic ts]ms
In Exercises ]*6, evaluate the quadraticfarmf(x) : x%,x
íu,the giveft A &nd x.
1.Á:
2, A:
3, A:
4,A=
5. Á ::i
6,Ax
In Exercises 7*12, find the symrnetric matrix A associated
with the given quadrntic form.
7, *? + 2*Š+ 6xrx2
8, .XiJí2
9" 3x2 * 3*y * y'
lS. x| x! + Lxrx, 6xrx,
11. 5xf - x} + 2*3 + 2xrx2 4xrx, + 4x2x,
!2,2x2 * 3y' + z2 Axz
Diagonalize the quadraticforms in Exercíses 13*1S by
Jtnding aíx orthoganal ynatrix Q such that the chtlnge aJ
variable x - Qy transforms the given form into ol"tc with no
cross-product terms, Give Q and the n*w quadratic form.
13" 2x| + 5r1 4x62
!4, x2 + Bx}, * y'
15.7xi + x] + x_| + lxrx, + gx,x, L6x2x,
t6. x| + x; + 3*i 4xrx2
17"x2 + z' 2*y + Zyr
l8" Zxy -L ]x: 2),z
Iz
L3
Is
Lt
i3
L*;
r1
lo
L-,
r1
lo
L -l
Ia
|2
Lo
3l
nJ,*
: ["l
Lrj
[;;]
x -: ['l
L6]
:
-il,X:
L;]1 3]
:i],X:Ii]
1 3j L lJ
1l
*1.J,*
rl
;I
2 0l [tl
0 1i,*: |z l
1 1l L:J
l a,
Chapter 5 Orthogonality
Classify *fl h af tLte quadratic forms in Exercisgs 19*26 as
positive dertnite, positive semidefi,nita, negative dertnfte,
n*gativ e semidefinite, ar indertnfte,
19. xf + 2x! 2 , x{ * *t 2x62
21. * 2x2 2y' + žxy ž2,x2 * y' * ilxy
23,2x| + Lxl + 2x * 2,xlx2 * 2xrx1 + 2x2x3
24, x| + x3 + x3 * uxpl 25. x! + x} * xŠ* 4xlx2
26, * x2 * y' tr2 Lxy * ?xz * 2yz
t
lfi. Let á
|t ;]
*e a ymmetric 2 x,2 matrix. prove
that Á is positive definite iť and only if a > 0 and
ďet Á > ,0.
|Hint: axz + \bxy * ďy' *
-(--*,)+(r-#),','
29. Let . be an invertible matrix. Show that Á .= ,STB is
positive definite.
S0. Let Á be a positive definite ymm tric matrix. how
that there exists an invertible matrix B such that Á
BT. . |Hint: tJse the pectral Theorem to write Á *
QPQT. Then show that,D can be factored as Crť ťor
ome invertible matrix C,]
3l. Let Á and B be positive definite symmetric n X n
matrices and let c be a positive scfllar. how that the
following matriees aťe po itiye definíte.
(n) cA (b} A2 (c) Á+B
(d) Á* 1
1First show that Á is nece arily invertible.)
32. Let Á be a positive definite qrmmetric matrix, Show
that there is a positive definite ymmetric matríx B
such that Á . 82. ( uch a matrix 8 is called a quere
root of Á.)
In Exerctses 33*36, rtnď the maximam anď minimum values
af the quadratic form f{x) in the givgn frxercise, subject
to the constrainr ||x|| L, and determine t\te values af xfor
which tLtese occ!Ár.
In Exercťses 45*5a, u trans\atiun af axes to put t\te eanit
in stanďard pasition, Iďentify ťh,e graph, give its equatiCIn in
the translateď coardinate system, and sketch the curve,
4,x2*y' 4x*4y+4 ů
46,4x2 + 2y' 8r* I}y* 6:0
&7,9xŽ 4y' 4y * 37 &8, x2 + 10x * 3y * * 13
49,2y2 * 4x * 8y :0
0,2y2 3x2 l8x - žfry* 11 = ú
In Exercťses 51-54, use a, rutation af axes to put the cartic in
standard position. Identify the graph, give its equation in the
rotated coordinate system, anď sketth the curve,
l. x? * xy * y'* 6 2_4xž+ LOxy * 4y, ffi 9
53. 4x2 + 6xy * 4y* ffi 5 4,3x2 yxy * 3y' * 8
In Exercíes 55-5& identify the conic with the given equa,
tion and give its equation in standard form.
. 3x? 4xy * 3y' 28ráx + LžÝIy+ 84 0
56. 6x2 * Lxy * 9y' 20x * 10y * 5 0
57,2xy*2 Žx* 1 ů
58. x2 # Lxy * y' + aÝÍx * 4 ffi 0
Sometimgs the graph af a quadratic equation is a straight
line, a pair af straight lines, CIr a single point. We refer to
such a graph s a degeneratp cgnic, J ťs also passible that
the eqiati-on is nut sati$ed far frny values af thu variables,
in which case there is no graph at all and we refer to the
conic & an imaginfrry canit, In xeretscs 9*64, iďentify
the rcnic with tlte given ěquetign as either degenerate or
imaginary and, where passible, skeťcbl the graph,
59,X? * y'* 0
61.3x2 *y' ff0
SO.x2 + }yr + 2 m 0
63. xŽ 2xy * y' + žÝŽx * 2\#y * CI
64.2x2 * lxy * 2y' + 2ťž* rr#y + 6 0
65. Let Á be a ymmetrie 2 X, 2 matrix nnd let k be a
scalar. Prove that the graph of the quadratic equation
xl4,x : fr is
(a) a hyperbola iť k * ůand det á < 0
(b) an ellípse, circle, or imagin*ry ccnic if k * 0 and
detÁ > 0
(c} a pair of straisht lines r an ímaginary conic íť
k*OanddetÁ,=0
(d) a pair of straisht lineg ťfr singlc pnint if lc * 0
anddetÁ * 0
(e) a straight line iť k : CI and det Á x S
|Hint: lJse the Principal Axes Thenrem,]
33. xercise 20
35. xercise 23
34. Exercise 22
36. Exercise 24
37. Finish the proof of Theorem 5.23(a).
38- Prove Theorem 5.23(c).
ffirmwx sffi fifirn*u m*m *n*
In Exercises 39-44, identify the graph ,íthe given gquation.
39,x2+ 5y': 25 4a.x2 _ y' _ 4= 0
4l.x2 * y*1 a* 0 42.2x' + y' B ff 0
43,3x2:yŽ 1 44,X*-},,2
In Exercises 66*73, identifu the quadric with the given
equatian and give its equation in standard form.
66.4x2+4y'*4z2 + 4xy*4xz*4yz: B
67,x2 * y2 + z' 4yr * 1
68 *x2 * y' z? * 4xy * 4xz * 4yz : 12
69.2xy*z-0
70. L6x2 + 100y2 + 9z2 24xz * 60x BOz : 0
7!.x' + y' 2z2 * 4xy Zxz * Zyz * x + y + z - 0
72, Iaxl + 25y' + Lazz 40xz * Za\/Žx + 50y +
\aÝžz: 15
73,IIx2 + LIyz + L4r2 + Lxy + \xz * \yz - IZx *
LZy+LZz:6
Chapter Review
74, LetÁ be a real 2 X 2 matrix with complex eigenvalues
ň n* bi suchthatb * nand la| l, Provethat
very trajectory of the dynamical system xk+t * Áxt
lies 0n an ellipse. |Hint: Theorem 4,43 shows that iťv
is an eigenvector corre panďing to A ffi a bi, then
the matrix "P
* [Rev Imv] is invertible and
f6 -b1A: pl: " lp*i. et t (pp?)*l- Show that the-Lb aJ r
quadratic xr8x tr k define &n ellipse for all k > 0,
and prove that if x lies nn this *11ips*, so does Áx.]
ff
K§Il n frinitllln tfiil BgnnnBts
fundamental subspaces
oť a matrix, 380
Gram- chmidt Process, 3S9
orthogonal basis, 370
orthcgonal complement
of a subspace, 378
orthogonal matrix, 374
orthogonal projection, 382
orthogonal set of vectors, 369
Orthogonal Decomposition
Theorem, 384
orthogonďly diagonalizable
matrix, 400
orthonormal basis, 372
orthoncrmal set of v*ctors, 372
pr pertícs of crthogonal
matrirc*, 74*376
QR factoriratínn, 393
Rank Theorem, 386
spectral decomposition, 405
pectral Theorem, 4ů3
lHcuiaw
0uegtions
1. Mark each of the following statements true or false:
{a} Hvery orthonormal set of vectors is linearly
independent.
(b) ,very nonzero subspace of Rn has an orthogonal
basis.
{c) If Á is a quare matrix with orthonormal ť?v }
then Á is an orthagonď matrix.
(d) very orthogonal rnatrix is invertible.
(e) If Á is a matrix with det Á : 1, then Á is an
orthogonal matrix.
(f) IťÁ ísan m Y ru matrix such that (rnw{Á))' : ffin,
then Á must be the rero matrix.
(g) TťW is a sub pace of Ro and v is a vector in Rn such
that projw{v) : 0, then y must be the zera yectoť.
(h) IťÁ is a ymm tric, orthogonal matrix, then A2 * L
(i) very orthogonally diagonalizable matrix is inverLible.
(i} iven any n real numbers A1, r . r , i* there exists
a ymffietric n X n matrix with Ar, .,,, hnas its
eigenvaltl .
2. Find all values of a and b such tlrat
{[i] [ ;] |l] },-anorthognnalsetafvct*r
3. Find the cocrdínate vector [v]n cf v *
respect to the orthogonď basis
B:
Il]*''
i}[i]}--,{Li][
e Chapter 5 Orthoganaliry
withť p ctto
ll}ll ůll 1ll
15. (a) Apply the Gram-Schmidt Process to
tiltil[l]xi :
I
, l,"r:
l , l,x::
I
, l
LrJ L;j Lrl
to finď an orthogonal basis for 1ť : span{xlr x2: xr}.
(b) Use the result of
|Tt
(a) to find a QR factori zatio
tll?lofÁ:|; ; ;l
[t 0 tl
16. Find an orthogonal basis for Ra that contains the
vector-[i]-"-[;]
L;] L-,:
17. Find an orthogonal basis for the subspace
TÁr- ít'"'l ta:=r} of*aw--
tLí]
-,*xz*xl*:-
,|
r 2 1 -tl
L*r 1 zJ
, (a} Orthogonal}y diagonalize Á.
(b) Give tň* *p.ctral decomposition of Á.
19. Find a symmetric matrix with eigenvalues At : iz :
ňr : *2 and eigenspace
" panffl]
Ll]) rE,:
pan(|
l])
20, If {r,, y2, . . . , yn} is an orthonormal basis for R" and
4. The coordinate vector of a vector v with respect to an
orthonormalbasis B - {r,,v2} of R2 is [v]r : [
-,'l
Lt lz )'
If v, :
|'_',:l,
OrrO all possible vectors v.
r 6/7
,_ 2l7 3 /7 _1
5. Show that | - r l\/5 0 2l\/Í |
i, *
L+/rÝí -n/rÝs 217Ý5) &
orthogonal matrix.
n
1,
6. If l'/,'
ol
,* an orthogonal matrix, find all possible
Lb cl
values of a, b, and c.
7,IfQis an orthogonal nX nmatttx and {rr,...,v1} is
an ortho*Órmal set ínRn, proyc that {Qr,} l . . , Qvrr}
is an orthonormal set.
8. If Q is an n X n matrix such that the angles
t_tQx, Qy) and L(x,y) are equa} for atl vectors x and
y in Rn, prove that Q is an orthogonal matrix.
In Questians g*12,
finď a basis for WL. .
g, 1ť is the line ín R2 with general equation
2x-5l-0
1{}. W is the line in R3 with parametric equations
X-:- t
y:2t
ry :
-íl*- L
11" Vť :
12,W:
pan
pan
bases
A
\ 13. Find
l"4. Find
A: c,v,v|+ c2vrv{ +",-+ c,rv,rv|
pro\re that Á is a s}rmmetric matrix with eigenvalue
c1: c2> , . . ) c,, and corresponding eigenvectors
YirV2l ..,rV,,'
the o
{Ll]Ll]}
{Li]Ll]}
for each of the four fundamental subspaces of
r 1 -1 2 1 3l
l-t 2 *2 1 -2l:l l
I z 1 4 8 9l
L 3 *5 6 -1 7J
rthogonal decúmpo itíonof
v = [ i]
| -r l
L zj
Algebra is gen*rous; she aften gives
mzrc than is asked af hw,
*Iean lc Rond dÁ"lembert
{L717 *I7s3)
: In Carl B. Boyer
A History af Mathematics
Wiley, 1968, p. 4S1
6.0 lnt]Oliluctloil: ]ibonacci in flectot) space
The Fibonacci sequence was introduced in Section 4,6. It i the sequence
0, 1, 11 2,3,5,8, 13, . . .
of nonnegative integers with the property that after the first two terms, each term
is the sum ofthe twotermsprecedingit. Thus 0 * 1 = 1, 1 + 1 : 2, L + 2 : 3,
2 * 3 :5, and so on.
If we denote the terms of the Fibonacci sequence by ío,ír, ír,. . . , then the entire
sequence is completely determined by speci$ling that
ío: O,fi = t and í,: ín-t * ír*z fot n > 2
By analogy with vector notation, let's write a seQu rrce.ípl x1l x2, x3l . . . &s
X : [ío, Xy X21 X31 , , .)
The Fibonacci sequence then becomes
f = W,ír,ír,ír,. . .) : [0, 1, 1, 2, . , .)
We nowgeneralize this notion.
For example, 1L,ÝŽ, t + {\ I + Z{Z, Z + 3..'/Ž,., .) is aFibonacci-tlpe sequence.
Pť &*s ii 1 Write down the first five terms of thre e more Fibonacci-tlpe sequences.
By analogy with vectors again, let's define the sum of.two sequences x : [.íg, J 1,
x2, , . .) and y : |yo, yr, yr,. , , ) to be the sequence
x * y : íxo * yo, h *:lyxz * yz,...)
If c is a scalar, we can likewise deíine the scalar multiple of a sequence by
CX - |CXg, CXt, CX2l " . .)
t
Chapter 6 \&ctor Spaces
: Yli,1l&i$ 3 (a) Using your examples from Problem 1 or other examples, compute
the sums of various pairs of Fibonacci-tlpe sequences. Do the resulting
sequences appear to be Fibonacci-type?
(b) Compute various scalar multiples of your Fibonacci-tlpe sequences
from Problem 1. Do the resulting sequences appear to be
Fibonacci-type?
i}iiirli};íit ii (a) Prove that ifx andy are Fibonacci-type sequences, then so is x * y.
(b) Prove that ifx is a Fibonacci-type sequence and c is a scalar, then
cx is also a Fibonacci-ťype sequence.
Let's denote the set of all Fibonacci-type sequences by Fib. Problem 3 shows that,
like R', Flb is closed under addition and scalar multiplication, The next exercises
show that Flb has much more in common with R'.
i:i,lt*i|ii;it ;.}: Review the algebraic properties of vectors in Theorem 1,1. Does Fib
satisfr all of these properties? What Fibonacci-tlpe sequence plays the role of 0? For a
Fibonacci-type sequence x, what is -x? Is -x also a Fibonacci-type sequence?
sequencef : [0, l, I,2,,..)canbethoughtofastheanalogueofe2becauseitsíirst
two terms are 0 and 1. What equence e in Fib plays the role of e,?
What about 3l 4, . . . ? Do these vectors have analogues in Fib?
itr$il}É*tň Letx : |xg,xyx2,,..) be aFibonaccitlpe sequence. Showtlratxis a
linear combination of e and f.
i}ls{iiuťiti ? Show that e and f are linearly independent, (That is, show that if
ce * dí: 0, then c: d: 0,)
ťí**i*i** Given your answer to Problems 6 and7, what would be a sensible
value to assign to the "dimensiorí' o Fib? ÍNhy?
frťi}i,iii,}í'! ii Are there any geometric equences *I Fib? That is, if
f!, r, 12, 13, , , ,)
is a Fibonacci-type sequence, what are the possible values of r?
iiy #ls# Ť* Find a "basis" for Flb consisting ofgeometric Fibonacci-type sequences.
Íij*il; ,1{,í! !.1 U ing your answer to Problem 10, give an alternative derivation of
Binet's formula |formula (5) in Section 4.6]:
í,,:*(
.,ť:"oi,,,.,i,.. for the terrns oťthe Fibonacci se
the basis from Problem 10.]
1 + \6\, t /t - Vg\,,
, ) .rc[-, J
quence f lfo, Ír, Íz,. . . )" |Hint: xpress f in t*rms of
T'he Lucas sequence is named after
Edouartl Lucas (see page 336).
The Lucas sequence is the Fibonacci-tlpe sequence
l = Uo, 11,12,\,..,) = í2,!,3,4,,,,)
trOnlen 12 Use the basis from Problem 10 to find an analogue of Binet's formula
for the nth term In oíthe Lucas sequence,
n0bl0mlff ProvethattheFibonacciandLucassequencesarerelatedbytheidentity
ín-r*ín+t=lnforn>L
l..} |Hint:TheFibonacci-type equence f- = [1, !,,2,3,...) andf* = [1,0,1, 1,.,.)
form abasis íor Fib. (t,Vhy|]
In this lrrtroduction, we have seen that the collection Fib of a[l Fibonacci-type
equence behaves in many re pect like R2, even though the "vectors" are acfirally
infinite equence . This useful analogy leads to the general notion oť a vector space
thx is the subject ofthis chapter.
#
Section 6.1 Yector paces and ubspaces
ailil
In Chapters 1 and 3, we aw that the algebra of vectors and the algebra of matrices
are similar in many re pects. ln particular, we can add both vector and matrices,
and we can multiply both by scalars. The properties that result from t}rese two operations
(Theorem 1.1 and Theorem 3,2) are identical in both settings. ln this section,
we u e these properties to define generalized "veďors" that arise in a wide variety
of examples. By proving general theorems about tlrese "vectorsj' we will therefore
simultaneously be proving results about all of these examples. This is the real power
of alg|ebra: its ability to take properties from a concrete setting, like Rn, and abstract
them into a general setting.
The German mathematician
,,]
' is generally credited with
frrst introducing the idea oť a
vector space (although he did
not call it that) in 1844. IJnfortunately,
his work was yery difficult
to reacl and did not receive
the attention it deseryed. one
p rson w}ro did stucly it was the
Italian mathematician,,,,,,,
, :,
",]i 1 ,In'lriS
188B book Calcalo Geametrico,
peano clarified Grassmann}
earlier work and laid tlown the
axiorns for a vector space as
we know them today. Feano's
book is alsa remarkatrle for
introducing operetioíl On sets.
His notations U, n, and (ťor
"unionj' "it]tersectionJ' ancl "is
an element tlf") are the one we
still use, although they were not
immediately aťcepte,l by other
mathematicians. Peano's axiomatic
definition of a vector space
also harl very little iní]uence ťor
many years. Ácceptanc came
in1918,after,, ,, ,::]
, , , ], :, , ] fťpťaterlit in
his book Space, Time, Matter, an
introducticn to insteint general
tlreory of relativity.
ffi# $ Se Let Vbe a et on which two operations, called addition artd scalar
miJUptiiiii,Ón, have been defined. If u and v are in V, the sum of l and v is denoted
by u * v, and if c is a scalar, the scalar multiple of u by c is denoted by cu. If the
following axioms hold for all u, v, and w in Vand for all scalars c and d, t}ren V is
called a vector space and its elements are called vectors.
2.u*v:v*u , :, .
3.(u+v)*w:u*(v*w)
4. There existsanelementOin 1/, calledazero ector, suchthatu * 0: u,
5. Foreachuin %thereisanelement -uin Vsuchthatu * (-u):0.
6. CUiSinV. ] l.,,., !,!.:l]iii]:'i;ť,|i,!i;ll,iil,,,:ilrl,i]",:
7. c{U* v) : cu * cv i::lii,il;l;iiilli,
S. (c * d)tt: crl* da i]::illill ;,,,,
9. c(da): (cd)a
10. lu: u
ntnill3
. By "scalars" we will usually mean the real numbers. Accordingly, we should
refer to V as a real vector space (or a vector Eace rver the real numbers), It is also possible
for scalars to be complex numbers or to belong toZo,wherc p is prime. In these
cases, V is called a complex vector space ot a yector space over Z, respectively. Most of
our examples will be real veďor pace , so we will usuďly omit the adjective "real]' ť
something is referred to as a "vector spacei a sume that we are working over the real
number sy tem.
In fact, tlre scalars can be chosen from any number ystem in which, roughly
speaking, lve can add, subtract, multiply, and divide according to the usual laws of
arithmetic. In abstract algebra, such anumber y tem is called afield..
o The definiťon of a vector space does not speci what the set v consists
oť. Neitlrer does it speciff what the operations called "addition" and "scďar multiplicaťon"
look like. Often, they will be familiar, but they need not be. See Example 6.6
and Exercises 5-7,
We will now look at several examples of vector spaces. ln each case, we need to
speciff the sď V and the operations of addition and scalar multiplication and to verify
axioms 1 through 10. We need to pay particular attention to axioms 1 and 6 (closure),
Chapter 6 Yector Spaces
For any n > 1, Rn is a vect r pace with the
multiplication. Axioms 1 and 6 follow from
the remaining axioms ťollow from Theorem
usual operations of addition and scalar
the definitions of these operations, and
1.1. ,:.
,":',
-,:-iaxiom
4 (the existence of a zero vector in V), and axiom 5 (eachveďor in Vmust have
a negative in V).
The set of ďl 2 X 3 matrices is a vector space with tlre usual operations of matrix
addition and matrix scalar multiplication, Here the "vectorď' are actually matrices.
V e know that the sum of two 2 X 3 matrices is also a 2 X 3 matrix and that multiplyinga}
X 3 matrixby a scalar gives another 2 X 3 matrix; hence, we have closure,
Tňe remaining axioms follow from Theorem 3.2. In particular, tlte zero vector 0 is the
2 X 3 zeromátrix, and the negative of a2 X 3 matrix Á is just the 2 X 3 matrix -Á.
There is nothing special about 2 X 3 matrices. For any positive integers m artd n,
the set of all m X n matrices forms a vector space with the usual operations of matrix
addition and matrix scalar multiplication, This vector space is denoted M^,
t+ŤLet.
2 denote the set of all polynomials of degree 2 or less with real coeffrcients.
Define addition and scalar multiplication in the usual way- (See Appendix D.) If
P@)
: ao * arx * arxz and q(x) = bg + brx + brx2
are in 02, then
p@) + q(x) = (ao * bg) * (a, * br)x * (ar-| b2)x2
has degree at most 2 and so is in 92. If c is a scalar, then
cp@):cag*carxlca2x2
is also in 92. This verifies axioms 1 and 6.
The zero vector 0 is the zero polynomial-that is, the polynomial a]l of whose
coefficients ate zero,The negaťve Óf a polynomialp(í) : ao| alx * a 2x2 is the polynomial
- p@) = - ag - alx - azx2.It is now easy to verify the remaining axioms. We
will check axiom 2 and leave the others for Exercise 12. Withp(x) and dx) as above,
we have
p(x) + q{x) 1Ť (a, + e$ * azx}} + (bo + bp * bzx'}
where the third
commutative.
* tao + b*) + (a, + b,)x * {a, + br)x'
(bg + fl + (b, + ar)x* (b, + frz)xz
-= (bo + b$ + brx') + (ao + alx + azxL)
q(x} + p{x)
equalit}r fo}lows from the fact that addition of real numbers is
ectian 6.1 Yect*r paces and Subspaces
In general, for any 6xed n ž0, the set 9n of all polynomials of degree less than or
equal to r is a vector space, as is the set 9 of all polynomials.
Let S denote the set of all real-valued funcťons defined on the real line. Ifland g are
two such functions and c is a scalar, then/+ gartd cf are defined by
(/+ gXx) : í(x)+ g(x) and (ď)(x) : cí(x)
In other words, the value oíf + g at x is obtained by adding together the values ofl
and g 3t r [Figure 6.1(a)] , Similarly, the value of cf at x is just the va]ue oflat x multiplied
by the scalar c [Figure 6.1(b)]. The zero vector in S is the con tant function
Á that is identically zero; that is,/g(x) = 0 for all r. The negative of a function/is the
function -/ defined by (-í)(x)= -í(x) [Figure 6.1(c)].
Axioms 1 and 6 are obviouslytrue. Verification of the remaining axioms is left as
Exercise 13. Thus, S is a veďor space,
(x,
"ť(x)
(;, g(i))
(x, * 3/(x))
lsurn §.I
The gťaph *f {a} f, g, andl + g (b) l 2f, anď * 3í,and (c} /and
*/
Chapter 6 Yector Spnces
." |,,I
Then, for example,
In Exarnple 6.4, we could also have considered only those functions defined on
some closed interval [a, b] of the real line. This approach also produces a vector pace,
denoted by @|a, b].
The set Z oí i*egerc with the usual operations is nof a vector space, To demonstrate
this, it is enough to find that one of the ten axioms fails and to give a specific
instance in whic.h it fails (a counterexample). In this ca e, we find that we do not have
closure under scalar multiplication. For example, the mulťple of the integer 2 by the
scahr } is (iX2) : !, which is not an integer. Thus, it is not true that cx is inZ for
gvcry x in tr" and_ every scalar c (i.e., axiom 6 fails).
Let v = R2 with the usual definition of addiťon but the following definition of scalar
multiplicationl
-L;] *
,[;]
Ll]-Li]
so axiom 10 fails. [In fact, the other nine axioms are all true (check this), but we do
not need to look into them, because V has already failed to be a vector space. This
example shows the value of looking ahead, rather than working through the list of
axioms in the orďer in which they have been given.]
Let C2 denote the set of aíl ordered pairs of complex numbers. Define addition and
scalar multiplication as in R2, except here the scalars are complex numbers. For
example,
i
1 +r;
l + [-l + r,1: |*z+ u,l
L2 3iJ L 4 J L6 3i]
[(t iXt+,)l [ 2
:l I:I
L(t ilQ 3,)j L*t ,]
and
tJsing prffperties of the camplex number } it is straightťorward tc check that
axioms hold. Thereforť, C2 is a cornple}r vťťtorpace,
In general, C* is a csmplex yector pacť for all ft t L
all ten
+
*+
IfP is prime, the set Zi fuildr' the usual definitions of addition and multiplication by
sealnrs frorn Zr) is a vector pace over Z-o far aLI n h 1 .
ection 6,1 Yector paces and ubspaces
Before we consider firther examples, we state a tlreorem that contains some u e_
ful properties of vector pace . It is important to note that, by proving this theorem
for vector spaces in gen eral, we are actuďly proving it for every specaíc vector space,
a"
b"
L.
d,
p il{i We prove properties (b) and (d) and leave the proofs of the remaining properties
as exercises,
(b) We have
cO:c(0+0):cOfc0
by vector space axioms 4 and 7, Adding the negative of c0 to both sides produces
c0 + (*c0) x (c0 + c0) + (*c0)
which implies
0 : c0 -| (c0 * (-c0)) i,: l:,:",:::":', l,:l,i
-]
:cO-l0
: c0 i.i ;!i.ll):,] i
(d) Supposecu: O.Toshowthateitherc _ 0orll:0, let'sassumethatc + 0. (If
c : 0, ihere is nothing to prove.) Then, since c * 0, its reciprocal l/c is defined, anci
u ::*
x
:
1u
ít \
[;,l"
l(r*r)
l*
1
*0
0
by axiom 10
by axiorn 9
by properťy (b)
We will write u * v for u t- (-v), thereby defining srbtraction of vectors. Wb will
also exploit the associativity property of addition to unambiguously wite u * v * w
for the sum of three vector anó more generally,
ťtltt+íeilz+",{
for a linear cambinatian of vectcr .
ubs0aces
We have seen that, in R', it is possible for one vector pace to sit inside another one,
giving rise to the notion of a subspace. For example, a plane through the origin is a
subspace of R3. We now extend this concept to general vector pace .
4t8
Chapter 6 Vector Spaces
itself a vector space with the same scalars, addition, and scalar multiplication as v.
As in R', checking to see whether a subset lď' of a vector qpace v is a subspace of
Vinvolves testing only two of the ten vector space axioms. We prove this obseivation
as a theorem.
Theorsm S.2 Let V be a vector space and let I4r be a nonempty subset of V. Then 1ť is a subspace
of Vif and only if the following conditions hold:
a. If u and v are in I4l, then u * v is in Iď'.
b. If u is in W and c is a scalar, then cu is in I,ť'.
PíOO| Assume that W is a subspace of V. Then I4l satisfies vector space axioms 1 to
10. ln particular, axiom 1 is condition (a) and axiom 6 is condition (b).
Corrversely, a ume that Iť is a subset of a vector space V, satis$ring conditions
(a) and (b). Byhypothesis, axiom 1 and 6 hold. Axioms 2,g,7,8,g, and t0 hold
in Wbecause they are true for a//vectors in Vand thus are true in particular for those
vectors in }ť. (We say that W inheňts these properties from V.) This leaves axioms 4
and 5 to be checked.
Since l,ť'is nonempty, it contains at least one vector u, Then condition (b) and
Theorem 6.1(a) imply that 0u : 0 is also in }V. This is axiom 4.
If u is in V, then, by taking c = -7 in condition (b), we have that -u : (- l)u is
alsoin l44usingTheorem6,1(c). $uf
n0mrlt Since Theorem 6.2 generalizes the notion of a subspace from the context
of R' to general vector space , all of the subspaces of R' that we encountered in
Chapter 3 are subspaces of Rn in the current context. In particular, lines and planes
through the origin are subspaces of R3.
}Ve have already shown that the set Sn oť all polynomials with degree at most
Yector Pace. Hence, #o is a subspať Of the vector spa ce W oť al|polynomials.
Let Tť be the set of }rínmetric ffi X í?mfftríce " Show thx Iť is a subspe{;e of Mnn.
$0l[tlon Clearly, lť is nonempty, so we need only check conditions (a) and (b)
Theorem 6.2. Lď Á and B be in I4l and let c be a scalar. Then Ár : Á and Br =
from which it follows that
(a+,a)?=AT+Br:AlB
Therefore, A + B is s;rmmetric and, hence, is in Iť'. Similarly,
(cÁ)r= cAr:cA
so cá is sYmmetric and, thus, is in W. \ťe have slrown that Wis closed under addition
and sca}ar rnultiplication. Therefore, it is a subspace oť Mnn,by Thesrem 6.2.
ruisa
^I
Section 6.1 Vector paces and Subspaces
Let be the set of all continuous real-valued functions defined on R and let 9 be
the set of all differentiable real-va]ued functions defined on R. Show that % and 9 are
subspaces of 9, the vector space of ďl real-valued functions defined on R.
0tutl0n From calculus, we know th at if f and g are continuous functions and c is a
scalar, then/ f g and c/are also continuous. Hence, % is closed under addiťon and
scalar multiplication and so is a subspace of S. Iíf and,garedifferentiable, then so are
/ + gand cf Indeed,
(í* g)' : í' -| g' and (cf)' = c(f')
SoSis
fw.
a}so closed under addition and caler multiplicatíon, mekins it a *ubsPace
It is a theorem of calculus that every differentiable function is continuous. Consequently,
9 is contained in (denoted by 9 C
(6),
making 9 a subspace of
( . It is also
the case that every polynomial function is differentiable, so 9 C 9, and thus 9 is a
, g Cg C<3 C9F
This hierarchy is depicted in Figure 6.2.
l!0ur. 6.2
The hierarchy of subspaces of S
There are other subspa ces of ? that can be placed into this hierarchy. Some of
these are explored in the exercises.
In the preceding discussion, we could have restricted our attention to functions
defined on a closed interval [a, b]. Then the corresponding subspaces af gF|a, b]
would be % [a, b), Qi|a, b], and 9 ía, bl.
Let . be the set of all functions that satisfy the diťferential equation
f" * í:0
Thatis, -S isthe solution set of Equation (1),Showthat - is asubspace of ff"
Chnpter S Vector paces
&lrll0r is nonempty, since the zero function clearly satisfies Equation (1). tet
/andgbe in S and let c be a scalar. Then
{í+g)' + (í+ g) = U" * g'') + V+ s)
=(f''+fl+@'+s)
:0*0
-0
which shows:"'-sisin s' Similartr'
:r!::- ď
-"v'' +í)
::o:U
so c/is also in S.
Therefore, S is closed under addition and scalar multiplication and is a subspace
oíW.
f
Ť
The differential Equation (1) is an,example of ahomogeneous llnear differcntial
equation, The solution sets of such equations are always subspaces of S. Note that in
Example 6.12 we did not actually solve Eguation (1) (i.e., we did not find any specific
solutions, other than the zero function). We will discuss technigues for finding solutions
to this type ofequation in Section 6.7.
As you gain experience working with vector spaces and subspaces, you will notice
that certain examples tend to resemble one another. For example, consider the vector
spaces Ra, 93, and M22, T!ryical elements of these vector space are, respectively,
, p(x}: ťt+ bx * cx\ + ,lxr, and A_
|: n
Any calculations involving the vector space operations of addition and scalar multiplication
are essentially the same in all three settings. To highlight the similarities, in
the next example we will perform the necessary tep in the three vector spaces side
by side.
{*} $how that the set W qť all vect r of the form
L1]
a1
bl
_bl
frJ
is a subspace of Ra.
(b) Show that the set l{'of all polynomials of the form a * bx - bxz + ax3 is a
subspace of 93.
(c) Show that the sď Iť' of all matrices of the fot l a b1
- L_, o]*"subspace
of.M22.
eetion 6.1 Yector pac*s nnd rrbsp*ces
"i|
i| i; ii':: ; il i,,
(a) 1ť is nCInempty because it contains
the zero vector 0. {Take a : b :
0.) Let u and v be in fir*- ay,
ra1 rc1
l, | ,l ^^, Ýr-| ol
u:|_bl *"o v-|_al
L a) L ,]
** Á + "B is nlso in
the rigbtt farm),
ímilarly, if fr is
r
kA: l
Lt
;i:lu*Y:|-, dl
L tl+ c ]
t a + c l
I b + d l
| -t, + d} |
L a + c j
0 tl * v is also in W (because it has
the right form}.
Similarlp if k is a scalar, then
i ko1
ku:
I II|
L ko)
Then
so ku is in trť.
Thus, 1ť is a nfinempty subset of
ffi4 th*t is closed under addition and
scalar multiplication. Thcrefore, W is
a subspace of R4, by Theore m 6,2,
(b) W is nonempty because it ffntains
the zero polynomial, (Take *
b - 0,) Letp {x) and q{,x} be in W*.*ry,
p(x) ":* fr + bx*bxz + ax'
and
q(ň: c + dx- dxZ + cx3
'rrr*r,
pk) + q{,x}:{a +c}
+ tu + á}x
tb + d)x2
+ b +c)x3
so p(x) + q{x) is also ínIM {b*cause it
has the rightform},
ímilarly, iť k is a scalar, then
kP{x} ffi ka*kbx*kb# + kax'
t fl+c b+d1
L_-tr+ d) a+r_]
(c) lť is í}ffnempty bec&u it contains
th* uer* m*trix #, {Take a
b "- ů.)Let Á and ^B
be in tr4Á* ayl
r ab\4, - L-, ,)
and B .-=
i '. d1
L-ď cJ
Then
A+ *
W {bec us íthas
a scalar, then
ka kb1
kb ku)
so kp (x) is in W,
Thus, W is a nsfi mpty subset nf
#3 that is closed under adáition and
scalar rnultíplic*tion" Therefore, Iť ís
a subspace of Wlby Theorem §,ž.
so kÁ is in 1ď,
Thus, W ísa ncnempty subset of
Mz, t}rat is elosgd under adclition and
scalar multiplicatían" Th*refore, 14/ is
a subspace of Mm, by Theoťem 6,2. .i:,
fixmrtt$t* S- &
Example 6. 13 shows that it is often poseible to relate oramples that, on the surface,
appeaí to have nothing in common. Consequently, we can apply our knowledge of
Rn to polynomials, matrices, and other examples. We will encounter this idea several
times in this chapter and will make it precise in Seďion 6.5.
If V is a vector space, then V is clearly a subspace of itself. The set {0}, consisting of
onlythe zero vector, is also a subspace of V, cďledthe zero subspace, To showthis, we
sirnply note that the two closure conditions of Theorem 6.2 are satisfied:
0+S*0 and c0*0 foranyscalarc
The subspaces {0} and 1/are called the triuial sub p$ce * V.
Chapter 6 Vector Spaces
An examination of the proof of Theorem 6.2 reveals the follorrying useful fact:
If l'lZ is a subspace of a vector space V, then W contains .the zero vector 0 of V. : ,, ',,, ,. '
This fact is consistent with, and analogous to, the fact that l,ines and planes are subspaces
of R3 if and only if they contain the origin. The requirement that every subspace
must contain 0 is sometimes useful in showing that a set is not a subspace.
Let W be th'e set of flll 2 X
Is 1ť a subspace af M22?
olutt0n Each matrix in I4lhas the property that its (1, 2) entry is one more than its
(1, 1) entry, Since the zero matrix
o
does not have this properť}r, it is not in 1ť is not a subspece of M22,
Let W'be the set of all2 X 2 matrices with determinant equal to 0. Is I4l a subspace
of M22? (Since det O : 0, the zero matrix is in 14/, so the method of Example 6.15 is
of no use to us.)
:1,1!]]jil!ii;i Let
1ťis not closed
lannin0 ets
The notion of a spanning set ofvectors carries over easily from Rn to general vector
sPaces.
2 matrices of the form
Ia a + 1l
Lo b ]
[o 0lllll
L0 0l
V7. Hence,
[t 0l [o 0l
^ -' I and B_ l lÍ1'
Lo 0]
clll(J L)
Lo 1 j
Then det A * det B - 0, so Á and B are in 1ť. But
A+ B_ [t 0l
L0 1 ]
so det(Á + B) : L + 0, and therefore Á + B is not in 1ť. Thus,
addition and so is not a subspace of Mzr,
und"er
+
-_+
ection 6.1 Yector paces and Srrbspaces
Show that the polynomials 1, .lť, and x2 span 2,
{ By its v ťy definition, a polynomial p (x} x a
nation CIf 1, ff, and x2. Therefore, {í'z: span(1, x, fr2)"
* bx + cxz is a linear connbiExample
6,17 can clearlybe generalized to show tbat*n = span(l, x, i, , ,
x').,However, no finite set of polynomials can possibly span g, the vector space
all polpomials. (See Exercise 44 in Section 6.2.) But, if we allow a spanning
to be infinite, then clearly the set of aílnonnegative powers of x will do. That
.)
of
set
iS,
, haw that ňď23 : ,r), whereZZ,
Io
L0
Io
Lo
ppfuI3} b21l
Er, :
En,
0l
0]
0l
0]
SPSn( rr,
D
Llt
Er, : Err:
Ě!1J
D
L23
[o 0 1l
LoO0J
[000l
Lo 0 1]
0l
0]
0l
0]
[t 0
Lo 0
[o 0
Lr 0
with a 1 in row ť,column j and uero elsewhere.)
observe that
1
0
s
}
(That is, . 4.is the matrix
SOlutIí}n }\Ie need only
I or, &lz aru1
l l:
Lar,, frzz azl )
al"t tl + iln n + {"tnUt3 + &ztF,zl + QzzEzz + G::Ez:
t
*+
Extending this example, we see that, in general, M *, is spanned by th e mn matri
cesEl;, where i = 1, .,., m andj = L,,,,, tl,
In S2, determine whether r{x} * 1 * 4x * 6x2 is in span( p(x), q(x)), where
P(x)*l*Jf *x2 and q(x}*2*x*3xz
1i{,Jil*ii il We are looking for scalars c and d such that cp(x) + dq(x) : r(x). This
means that
c(l - x + x2) + d(2 + x - 3x') : l - 4x * 6x2
Regrouping according powers of x, we have
(c + 2d) * (-c + d)x + (c * 3d)x2 : L - 4x * 6x2
Equating the coefficients of like powers of r gives
c+2d: 1
-c + d- -4
c-3d: 6
f,hapter 6 Vector paces
whic}r is easily solved to give c - 3 anď d
,,.,,,i,
:::,,, , r(x) is in span(ptx), q{x)). (Check this.)
In S, determine whether sin 2x is in span(sií} Jí,cos x).
80Ill$0n Wesetcsinx * dcosx = sin2xandtrytodetermine candd sothatthis
equation is {ue. Since these are functions, the equation must be true for all values of
x. Settins tr * 0, we have
csin 0 + dcos 0 * sinp or c(0) + d{L} * ů
from which w ee that fl r 0. ettins Jí:Ť lL,we g t
c sintrrll) + d cos(rr/Z} ffi sin(rr) sr fi(l) + d(CI) ffi 0
giving c : 0. But this irnplies that sin 2x = 0(sin r) * O(cos x) = 0 for all x, which
is absurd, since sin 2r is not the zero function. we conclude that sin 2x is not in
pan(sin x, co .r).
iomrrt It rs true that sin 2x can be written in terms of sin x and cos x. For
example, we have tlre double angle formula sin 2x = 2 sin r cos x. However, this is not
alinear cambination.
In M22, describe the span of Á : [t 1l [t 0l
L; ;],u
:
Ln ij,*a
f; :_
+e1
d-l
,z,andÉ*"x+ y*
Il
{lltii;3it Every linear combination of A, B,and C is of the form
cA ]- dB* ..: .[i ;]
- ,l; ?]
- .[l i]
lc * d c * ef
:L.*
e d ]
This matrix is symmetric, so span(Á, B, C) is corrtained within the subspace of symmetric
2 X 2 matrices. In fact, we have equaliry; that is, every spmďric2 X 2 matrix is
in span(Á, B, C), To show this, *" t., |' |] *
"
symmetric 2 X ? matrlx. Setting' Lyz)
lx v1 lc+d c
l, 'r)L.
+ e
and solving fur c and d,we find that c = x - z) fl,-;
r :): (x - rll ;]
+,L; :]
+ ( x + y+ -,[l
:
(Check this.) It follows that span(Á, B, Q is the subspace of symmďric 2 X 2 matrices,
g. Thergťore,
1l
0]
ection 6.1 Yectcr p*ces and ubspaces
As was tlre case in Rn, the span of a set of vectors is always a subspace oťthe vector
space that contains them. The next theorem makes this result precise. It generalizes
Theorem 3,19.
TheO]sm 6.E Let v1 , y2, r . . , va be vector in a yectoť pace Y.
a. span(vr, y2,.. ., yk) is a sub pace Of Y.
b. pan(v1, y2, . . . } vi) is the smallest subspace of Vt}rat contain y,, y2, ,. . 1y1.
P1O0í (a) The proof of properť},(a) is identical to the proof of Theorem 3.19, with
Rn replaced by V,
(b) To establish property (b), wc need to sho,w that any subspace of Vthat contains
y
1;
y2l,. ., v1 also contains span(v1, vz, . . ., yk). Accordingly, let 1ť be a subspace of V
that contains ylly2l ,, , , vk. Then, since lť'is closed under addition and scalar mulťplication,
itcontainseverylinearcombination c 1* c2y2 + ",+ clv1 ofv1,y2l ... r
v1. Therefore, span(v1, v9, . . . , vl) is contained in I4l.
In Exercťses 1*11, ďetermine whether the given set, toggther
with the speci ed aperations oíeddition frnd scalar multiplicatian,
is a vector space, Iíit is not, list aII of the axioms
that fail to holď.
1. The set of ď1 vectors in R2 of the form ['l, with the
Lxl'
u uml ve tor addition frnd scalar multiplication
2. The set of all vector, [*l in R2 with Jc }* a,y= 0 (i.e.,
the first quadrant), *r?'rne u ual vector *aaition and
scalar multtplication
ual scalar multiplication but addition
7. The set of all pnsitive r fil nuťnb r , with addition fi
defined by fr O y *.ť/ and scaler multíplication fi
definedbyrffx*ffd
8. The set of all ratinnel nuínber , with the usual addítion
and multiplicatinn
9. The set nťall upper triangulnr 2 X žmatrices, with the
u ual matrix flddition and celer rnultiplication
10. The §t of all 2 X žmatrices oť the form r- *-l,
Le dJ'
where ad * 0, with the usual matrix addition a-nd
scalar multiplication
l1. Th et of all skew-symmetríe ru X n rnatrices, with the
u ual matrix addition anď scalar multiplication
(see page L62),
12. Finish veriŤrrg that #2ísa vector pace
(see Example 6.3),
l3. Finish verifyrrg that S is e y ctor pace
(see Example 6.4).
6. R2, with the ug
defined by
[,,l
Lh)
+
|-;,1: [;; :-;,i i]
3, The set of ďl vector- [-l in R2 with xy ?0 (i,e., the
|,y J
uníanoťthe first and third quadrants}, with the usual
vector additíon and scalar multiplication
The set of all vectors [-l in R2 with x *,y, with the
LyJ
usuď vector addition and scnlar multiplication
ffi2, with the usual addition but scalar multiplication
defined by
4,
5.
.[] :
[;]
Chapter 6 Vector paces
ffi In Exerctses 11*1?, determine whether the given set, together
with títespecirt,ed operations af addition and scalar multiplication,
is a ůmp\ex vector space" trí iťis ytot, list all of )
the axizms that fail to kald,
14. The set of all vectors in C2 of the form
L;],
with the
usual vector addition and scalar multiplication
15" The set M*n(C) of I m X ncomplex matrices, with
the usual matrix addition and scalar multiplication
16- The et,C2, with the usual vectcr addition'but scalar
multiplication defined by ,^
|
"1
['",
l'Lrr] Lerr)
tr- 7. Rn, with the usuatr vector addition and scalar multiplication
In Exercises 18-21, determine whether the given set, together
with the specffied operatinns af addition and scalar multtptlcation,
is a ugctor space ol}er the ináicateá Zp, If it is nat, Iist
all af títeaxiz?ns that fail to hold,
18. The set of all vectors rnL with an ven number of
ls, over Zzwith the usual yector addition and scalar
multiplication
19. The set of all vectors inT with an odd number of
ls, ove r Zzwith the usual vector addition and scalar
multiplication
20. The set Mn,r(Zp) of all m X n matrices with entries
from Z* ure, Zpwith the usual matrix addition and
scalar multiplication
2|.7*6,over T4withthe usual addition and multiplication
(Think this one through carefully!)
22, Prave Theorem 6.1(a).
23. Prove Theorem 6.1(c).
In Hxercises 24-45, use Títorem 6.2 to determine whet\ter
W is a subspace af V
24.v :R3, l,y _
{LÍ]}
25. v *R3, 14/ -
26, V : R3, T4l :
27,V: ffi3, YY:
28, V * M22, |t\f *
\,r t Mno, |T y {Á in Mnn: det A - 1}
$ :r Mno, }lr is th* set of diagonal n X n matrices
Y -., Mn,r, Iď is the set of idempotent n X ru matrices
given (frxed) matrix
v * *r, !d/ * {&x + cx}}
f *:Pr,!d/:{a+ bx*cx}:&+ b + ť:0}
1,r =Sn,Jrlrx {a+ bx* cJtr2:abc:0}
.ry W, W is the set of all polynomials of degree 3
ť p s, Iť : {/ in W :f{*x} * y(x)}
\í= s, W * {/ ir, %,í(*x) * ť{x)}
pro s,W* {/ins:/(s) - 1}
p r W,W = {/ in W:f {a} * 0}
ž9, \r : M2z,
30.
3l.
32,
4a)
JJ.
34.
itF
a
3s.
4n
l,
38.
39"
40,
4l.
{L: ';
{[;]}
{[,+ |-,]}
{L1]}
{L; :,1}
[t 21
5l.C: l l
L3 4)
i- ,r
lJ
52. C* l
lrl"a
L-t
M &ž. V * W, W is the set of all integrable functinns
]fl 43. y : S, W : {/ irr' :f'(x) > 0 for all x}
]ffi 44, V : W, W .." (6('},
thc set of alt functions with
continuous second derivatives
]#_45"
y: W,l,y * {/ in S: _ tiry f k}-1Ť o}
46. Let be a vector pa e with subspaces U and Ýť. Prove
that u n fi/ is a subspace of v.
47 . Let V be a vector space with subspaces U and Ý11. Give
an Kample with Y : R2 to show it at U U Iť need not
be a subspace of V.
48. Let Ybe a vector pace with subspaces {/and W.
Deťine the sum aíUanď Iť to be
, {.I + },[ :* {u + w : u is in t/, w is in 1ť}
(a} If Y * ffi3, U is the x-axis, and 14/ is the y-axis,
whatisU* W?
(b} If {/ and Iť are subspaces of a vector space 1/,
prsve that U + Iď is a subspace of V.
49. IťU and ť are vector spac l ďefinc the Cartssian
producf of U and Y to be
U X lf -- {(u, v) : u is in Uand vis in V}
Prove thet U X Y is a vector pace,
50. Let Iťbe a subspace of a vector pace V. Prove that
A * {(w, w) : w is in 1ť} is a subspace of Y X Vr
t-r _1l L*l lJ
fi : l " ^|.
Determine whether C is in span A, J.
Ll 0] -ť_,-,\-
In Exercťses 5j and 54, Iet ?{x} - 1 *
andr(x) : *2 + 3x * x2. netermine
pfrn(p(x), t(x), r{x))"
53.{s(x) * 3 * 5x * x2 54. s(r) *
In Exercíses 55*58, Iet f{x}
fr*ťermine whether h(x} is
55. h(x) : 1
57. h(x) : sin 2x
It
59. Is Mrzspannedbr
L*
ection 6.2 Linear Independence, Basis, and Dimension
60. Is Mrr panned by
61. Is 92 spanned by
ó2. Is S2 spanned by
*1 *x +2x2?
63. Prave that very vector pace has a unique uero
vector.
6{. Prove that for every vect ť y in a vector space Y,
there is a unique v' in F such that v + v' = 0-
2X, q(x) * x ,* X2,
whether s(x) is trl
1*x+x2
Ll3],Il;],[li],[?-l],
1 * Jc,x * xT,L + x2?
1*x *2x2,2*x*2x2,
: sinTx ands(x) * ťCIs?x.
i,n spantf{xJ,s{x)).
56. h(x) - cos 2.r
58. h(x) * sin x
i]L?l],L1,1],[?-;],
lfilíitins pr0iact The Rise of Vector Spaces
As noted in the sidebar on page 429, in the late 19th century, the mathematicians
Hermann Grassmann and Giuseppe Peano were instrumental in introducing the
idea of a vector space and the vector space axioms that we use today. Grassmanďs
work had its origins in barycerrtric coordinates, a technique invented in 1827 by
August Ferdinand Móbius (of Móbius strip fame). However, widespread acceptance
ofthe vector pace concept did not come until the early 20th century.
Write a ťeport on the history of vector spaces. Discuss the origins of the notion of
a vector space and the contributions of Grassmann and Peano. Why was the mathematical
community slow to adopt these ideas, and how did acceptance come about?
1, Carl B. Boyer and Uta C, Merzbach, A History of Mathematics (Third Edition)
(Hoboken, NJ: Wiley, 201 1),
2. |ean-Luc, Dorier (1995), A General Outline of the Genesis of Vector Space
Theory, Historia Mathematica 22 (L995) , pp . 227 *26l .
3. Victor |,Katz, A History of Mathematics: An Introduetion (Third Edition) (Reading,
MA: Addison Wesley Longman, 2008).
.í*sfirlltilsssn$gllst, ft l , t!|d ll msn inll
irr this seďion, we extend the notions of linear independence, basis, and dimension
to general vector paces, generalizing the results of Sections 2.3 and 3.5, In most cases,
the proofs of the theorems carťy over; we simply replace R'by t}e vector space V.
llnGat lndG[Bnd8ncG
* l$it tm A set of vectors {y1, v?, .
pendenf if there are Scalars C1, Cx, . , . , ck,
ctyt + czyz + ",+ cď*.- 0
:
A set of vectors that is not linearlydependent is said to be linearly independent.
Chapter 6 Vector paces
As in R', |vyy2, . . . , v7,} is linearly independent in a vector space V if and only if
crvr* c.rv, + ",+ Clryp:0 implies c' :0, C2: a,,,,,Ck:0
Wb also have the following useful alternative formulation of linear dependence.
ThcOtGn 6.{
Prmm Thc.proof is identícal to that of Theor m 2,5"
As a special case of Theorem 6.4, note that a set of, two vectots is linearly dependent
if and only if one is a scalar multiple of t}re other.
Tn?}r,the set {1 * x + x2,I * Jť +
z(t*x+x2}
1 + 3x * Jr2} í linearly ďependentl since
.r+3x2) 1+3x*x2
3X2,
(t
In M22, let
A:L; i], B:Ll -;]
, C:Ll ?]
Then Á + ft * C, so the set lA, B, C} is linearly rlependent.
In S, the set {sin2x, ťQ§?x, cos žx} is linearly d*pendent, since
cos 2x * cosŽx * sin2x
Show that the set IL, x, , . , . , x'| is linearly independent in 0r.
*;si,"l F}'t i Supposethatca, c7,..,,c,arescalarssuchthat
c6,1 * crx * crx2 + ", + cr{' _ 0
Thenthe polpomialp(r) : c0 * clx * c2i +,.. + c,x" is zero for allvalues ofx, But
a polpomial of degree at mo t r cannot have more than n zeros (see Appendix D).
So P(x) must be the zero polyrromial, meaning that ca : cl : c2: ", = cn : 0.
Therefore, {I, x, ť, , . . , x'; is linearly independent,
JK
nlut nn Yťb begin, as in the first solution, by a uming that
p(x} ffi Cg+ ť +c2x2*...f CnX' ffů
ince this is true for all ff} w can substitut§ Jí: $ to obtaifi f,o * 0. This leaves
ClX+CzX2 + "u + CrrX":0
Section 6.2 Linear Independence, Basis, and l)imension
Taking deriva.tives, we obtain
c1 + 2c2x * 3cax2 + " , + ncl,Xn*l * 0
andsettingx = 0,weseethatct - 0.Difierentiating 2c2x* 3cax2 + ",* ncrx'-L = 0
and setting í= 0, we find that 2c2 = 0, o cz = 0, Continuing in this fashion, we
findthat klcp= 0fark= 0,..,, n. Therefore, c0= ct= Cz= ",- c,, = 0, and {1,r,
x2 r,,,, xo! íslínearly independent,
In #2, ďetermine whether the set { 1 + x, x * x2, 1- + x'I is }inearly independent,
t I * Let c1 , ž,and ca be calar uch that
ci(l + Jc) + c2(x + x2) + c.(1 + x2}* 0
tr, + c:} + {r, + c2)x * (r, + c)xZ ,,=
0
Then
Thísimplie* that
C1 +
C1 +
the sotrution to which is c1 *: c2 x c3
linearly independent,
Q*0
C2 :0
C2 + Cj_0
--: 0, It follows that {1 * ':c, x * x2,1 + x2} is
Ét}l3 { Compare Example 6.26withExample 2,23(b). The system of equations
that arises is exactly the same, This is because of the correspondence between 92and
R3 that relates
Ll],1+xzsLi]
matríx oť the linear y tem that we haye
+ x'I is linearly independent is equiva-
{Li]L;]Li]}
is linearly independent, This can be done simply by establishing that the matrix
[t 0 1l
l, 1 0l
Lo 1 1]
has rank 3, bv the Fundamental Theorem of Invertible Matrices,
Chapter 6 Vector Spaces
In 9, determine whether the set {sin x, cos x} is linearly independent.
S0lutl0n The functions.f(r) : sin x arrd g(x) = cos x are linearly dependent tf and
only if one of tlrem is a scďar multiple of the other, But it is clear from their graphs
tlrat this is not the case, ince, for example, any nonzero multiple otf(x) = sin x has
t}re same zeros, none of which are zeros ofg(x) : cos í.
This approach may not always be appropriate to use, so we offer the following
direct, more computational method. Suppose c artd d are scalars such that
csin x + dcos x:0
Settingrí + 0, we obtain il * 0, and setting )c: Ťl\,we obtain c:
set {sin ,c, cos x} is linearly independent.
0. Thereťore, the
Although the definitions of linear
in terms of finite et of vcctor , we
follows:
dependen e and independence are phra d
can extend the con ept to infinite ets a
Note that for finite sets of vectors, this is just the original definition. Following is an
example of an infinite set of linearly independent vector .
olutl0n Suppose there is a finite subset 7 of S that is linearly dependent. Let xm be
the highest power of x in T and let x'be the lowest power of x in 7. Then there are
scalars Cn1 Cn,ll , . . 1 C*, not all zero, such tlrat
CnX" + C,ri_lL'''*I + + C*X*: 0
But, by an argument similar to that used in Example 6.25, this implies that cn =
cnl! : " , : cftl: 0, which is a contradiction. Hence, S cannot contain finitely many
linearly dependent vectors, so it is linearly independent.
t*5*
Saser
The important concept of a basis now can
space .
be extended easily to arbitrary vťctor
íf;+rl'Í$.$ f$ri g A subs et Bof a vector pace V is a basisfor V if
1. 6 spans 1,/ and
2, B íslinearlv independent.
Section 6.2 tinear Independence, Basis, and Dimension
Ife;isthe ťthcolumnafthe nX nidentítymatťix,then {e1, *2,. r r ) er} isabasisfo
catrled the standard basis for R,'.
{ 1, x, x2, ,. . , n}
is a basis ť*r W * called the sícnd.ard basis for Sn.
The set t: ÍErr,..,,E!r,Ezr,,,,,E2n,E^1,...,E^n) is a basis for M*r, where the
matrices 4
are as defined in Example 6. 18. á is called the standard basis for M*n.
We have already seen that á spans Mlnn,ít is easy to show that t is linearly independent.
{VbriSr this!) Hencs, t is a basis ť r M*n,
Showthat B: {| * x,x * x2,L + 12}isabasisfor92.
show that 6 spans 92, let a * bx + cxz be an arbitrary polynomial in 92, We must
show that there are scalars cr, c2, and cr such that
c,(t + r),l c2(x + x2) + c.(1 + 12) : a * bx * cx!
or, equivalently,
(c, * ca) + (c, * c2)x * (c2 * ca)x2 : a l bx * cx2
Equating coefficients of like powers of x, we obtain the linear system
4I
clŤ C3_a
cl +
f [R',
+
--+
C2
C2
[r o tl
whichhasasolution,sincethecoefficientmatrk l t' O I rrurrank3 and,hence,
Lo , ,]
is invertible. (We do not need to know what the solution is; we only need to know that
it exists.) Therefore, 6 is a basis for'I2,
* j
[t 0 ll
Observe that the *ut.L* | t r O
I
i, the key to Example 6,32, We can
Lo 1 t]
immediately obtain it using the correspondence between 92and R3, as indicated in
the Remark following Example 6,26,
Chapt*r 6 Yector pn*es
Show that B = {I, x, x2, . ,.} is a basis for 7.
i} *'áisťt In Example 6.28, we saw that 6 is linearly independent, It also spans O,
since clearly every polynomial is a linear combination of (finitely many) powers of x,
Find bas*s for the three vector space in Example 6"13:
(b) W? -* {a + bx * bx? + a,x3l. (c) W3 x
SO! tton Once again, we will work the three examples side by side to highlight the
similarities among them, In a strong sense, they are all the same exarnpie, but it will
take us until Section 6.5 to make this idea perfectly precise.
(b) ince (c) ince
{[; :] }
(a) Since
Il],[i]+b[I]
we hav e Wt: pan(u, v), where
u:[l] and yffi[ i]t,*
L:]
f,lr** ,
L ij
ince {u, v} is elearly linearly indepenďent,
it is also a basis far W7.
(7+bx-bx}+ax3 r- a b1
* a(I + x3) + b{x - x2) L*u o]
[t 0l:al l
L0 1]
+bl 01l
L*1 0]
we have Wz * pan (u{x), y(x)),
where
u{x) * 1 * x3
ince lu(x}, v(x)} is ctearly lint
early independent, it is also & basis
for W2.
we have Wl * pan( U, V), whcre
u - L; l] and J,r - L_l i]
Since { t{ 4 is clearly línearly independent,
it is also a basis for W3-
GOOídlnOtG
Section 3,5 introduced the idea of the coordinates of a vector with respect to a basis
for subspaces of R', We now extend this concept to arbitrary vector spaces.
infinite, since linear combinations are, bydefinition, finite.
ection 6.2 Line*r Independence, Basis, and Dimension
The converse ofTheorem 6.5 is also true. That is, ÉB is a set ofvectors in a vector
space Vwith the property that every vector in V can be written uniquely as a linear
combination of the vectors inB,then6is abasis for V(see Exercise 30). In this sense,
the unique representation pťoperty characterizes a basis.
Since representation ofavectorwith respectto abasis is unique, the next definition
makes sense.
observe that if the basis B of Vhas n vectors, then [v]r is a (column) Yect r in R'
Find the coordinate vector [P(x) J r oí p@) = 2 - 3x + 5í2 with respect to the standard
basis B = {1, r, x2} of P2.
[p(x) ]n :
This is the correspondence between 92 and R3 that we remarked on after
Example 6.26, and it,carr easily be generalized to show that the coordinate vector of a
polynomial
P(x) * fig + a , +
with re pect to the standard basi s B :
+ fr*Xn in S,
, . . r , Xn} is just the vector
ínffin+l
,i,: : The order in which the basis vectors appear in 6 affects the order of
the entries in a coordinate vector. For example, in Example 6,35, assume that the
The polynomial p(r) is alreaďy u linear combination of 1, x, and í2, so
r21
l -, l
L 5J A
"**==''
a
{,,
,,2X2 +
L, X, X2
|1:]
|p(x) ]r :
4{
---
40 í
Chapter 6 Vector Spaces
standard basis vectors are ordered & ff' :
F (x) * ! * 3x * 5x2 with respect to 8' is
Find the coordinate vector tÁ ]6 oťÁ
ffi -- {fin,, ;, ,,, ,r} ať í 22"
{ &ffi ince
It-- 1l
-[0
:)HuuIL
2,
x,1}. Then the cocrdinate vector oť
5l
_3
l
2)
í^-
L,t
t-
I
L
lz -1l- Ln i J
*i*t respect to the standard bnsis
0l [o 1l , ,[o 0l ^[o
0J Lo 0-l
* nL,
0J
+
'Lo
* ,r + 4En + 3En
t-?ltÁ] B:
L ;]
0l
1J
we have
This is the corr,espondence between M22 and Ra that we noted before the introduction
to Example 6,13. It too can easily be generalized to give a correspondence
between M^nartdR".
^
*+
Find the coordinate vector [p(r) )3 of p(x) : I l 2x - x2 with re pect to the basis
C = IL * x,x * x2,1 + x21 ofg r.
i ť} íjiišei lvty'e
need to find c1, c2,lfid, ca such that
c,(1 * x) * cr(x + x2) + cr(I * x2) : l * 2x - x2
or, equivalently,
(c, * cr) * (c, * c)x * (cri cr)xz : | * 2x - x2
As in Example 6.32, this means we need to solve the system
cr * cs: l
"*|,, _:
2
c, l- c,: -l
whose solution is found to be c, : 2, c2: 0, c3 : - 1, Therefore,
t- 21
tp(x)]c:I ol
L-,]
Section 6.2 Linear Independence, Basis, and Dimension
fSince this result says that p (x) = 2(1 + x) - (1 + l;, it is easy to check that it i
cor.ect.] J*-Ť
The nefi theorem shows t]rat the process of forming coordinate vectors is compatible
with the vector space operations of addition and scalar multiplicaťon.
§,s
,,:,
: ..;, ,::. We begin by writing u and v in terms of the basis vector
- a)r:, a
u:-ctvt+Czyz+"N+C,ryn and v-d t+dzyz+ "+dnyn
Then, using vectCIr spa e properties, we have
and
s0
u * v: (., + d,)v1 + (r, + d2)v2 + t t t + (,:u + d,r)v,,
]u : (cc,)v, + (cc2)v2 +,,, + (cc,,)v,,
[u + v]n * [u]n + [v]n
and cIu]n
An easy corollary to Theorem 6.6 states that coordinate vectors preserve linear
combinations:
You are asked to prove tlris corollary in Exercise 31.
The most useful aspect of coordinate vectors is that they allow us to transfer
information from a general vector space to Rn, where we have the tools of Chapters 1
to 3 at our disposal. We will explore this idea in some detail in Sections 6.3 and 6.6.
For now we have the following useful theorem.
ld,
lo,
Ii
Ia,
:L::
.r:l:Ll]
+
cu8:L:l:],L1]
(1)
Chapter 6 Yector paces
ffi ThrOrBm S.I Let B * {o,, y2, ..,,y,r} be a basis for a lrector space V and,let u1,
be vectors in V. Then {u,, . . . ) ut} is linearly independent in V if and
{[u'}B,,r.,[ut]*}iu1inearlyindependentinRn.
8 {}* Assume that iu1, . . . , ul} is linearly independent in Vand let
crlul]r+",+cl[u1]6:0
in R'. But then we have
[c,u, *...* clruolu:g
using Equation (1), so the coordinates of the vector clu1 + ... + clu7. w.ith respect to
B are all zero, That is,
Cttti +'' * *l CklťIk * ůyr + Syz +,,, *| avn
The linear índependcnce oť{ur, . , , , ut} now ťgrces c1 ff c,,2
{ [u, ]r, . . ., [u*]r} is linearlyindependent.
The converse implication, which uses similar ideas, is left as Exercise 32.
Observe that, in the special case where ui = Ý, we have
vi = 0.v, + ", + l.v, f ... * O.vo
so [v1]6 = ;aod {[v,]r,..., [v"]n} : {e,,, . ., e,} isthe standardbasis in Rí.
0inenslon
The definition of dirnension is the same for a vector space as for a subspace of Rn-the
number ofveďors in a basis for the space. Since a vector space can have more than one
basis, we need to show that this definiťon makes sense; tJrat is, we need to establish
that different bases for tíre sarne vector space contain the same number of veďors.
Part (a) ofthe next theorem generalizes Theorem 2.8.
TfieOtem 6.8 Let b' : iv,, v2, , . . ,y,i be a basis tbr a vector space V-.
a. Ary set of more than fi vťctors in tr/ must be linearly dependent.
b. Á*y set of ťewer than ru yector in 1/ cannot span 1/.
PlaOí (a) Let{ut,..., u,n}beasetofvectorsin % vnthm } n.Theu{[o,]r...,
lu*la]r is a set of more than n vectors in R' and, hence, is linearly dependent,
by Theorem 2.8. This means that {ur, ..., a*l is linearly dependent as well, by
Theorem 6.7.
(b) tet {ur, . . . , ur} be a set ofvectors in V,withm< n. Then S = { [u, 7s,. , , , [u. ]u}
is a set of ťewer than n veďoí in Rí. Now span(u1, .,., a*) = V iťand only if
span(S) : R' (see Exercise 33). But span( ) is just the column space of tlre n X m
matrix
[ [u, ] n lu,,,] sJ
m < r,. Hence, S cannot span ffin, o {ur, . . * } U,r,|
Ax
so dim(span{ )} : dim(col(Á)) Ť{
does not span Y.
Now we extend Theorefi} 3,23, ]
ecti*n 6.2 Lingar Independence, asig, and Dimension
Tngnr$m 6.9 The Basis Theorem
If a vector space V has a basis with ,x vectors, then every basis for V has exactly n
vectors.
The proof of Theorem 3.23 also works here, vifirally word for word, However, it
is easier to make use of Theorem 6.8.
Pl0ql Lď Bbea basis for Vwith ,4 vector and let 6' be another basis for V with m
veďoť . By Theorem 6.8, m s n; otherwise, 6' would be linearly dependent.
No.w use Theorem 6.8 with the roles of B and B' interchanged. Since B' is a
basis of Ywith ž? vector , Theorem 6.8 implies that any set of more than m vectors
in Vis linearly dependent, Hence, n šll7, since B is a basis and is, therefore, linearly
independ,ent.
Since n rn affiďm < n,we mu t have n ffi m,a required.
The following definition now makes sen e, since the number of vectors in a
(finite) basis does not depend on the choice ofbasis.
DtÍinitiOn A vector space V is ca|ledfinite-dimensional if it has a basis conistiná
or-finitely many veciors. The dimeision oí V, denoted by dim % is the number
of vectors'in a basis for V. The dimension of the zero vector space {0} is defined'
to be zero. A vector space that has no finite basis is called infiniie-dimnsional.
Since tlre standard basis for R' has ,4 veďor , diínRn = rr,Inthe case of R3, a onedimensionď
subspace is just the span of a single nonzero veďor and thus is a line
through the origin. A two-dimensional subspace is spanned by its basis of two
linearly independent (i.e., nonparallel) vectors and therefore is a plane through the
origin. Any three linearly independent v ctor mu t pan R3, by the Fundamental
Theorem. The subspaces of R3 are now completely classified according to dimension,
as shown in Table 6.1.
elimY
pl
Flanc through the orisin
Line thraugh the orisin
{0}
The standard basis far #,, contains rt + 1 vectors (see ,xample 6.3ů), o dim
?, + 1.
Y
ó
J
ž
1
0
6b:Ltn
tl
l
I
Chapter 6 \fuctor Spaces
The standard
dim M*n = rnrr.
Both 9 and 9 are infinite,dimensional, since they each contain the infinite linearly
independent t {1 , K, #,. . .} (see xercis e 44}.
Find the dimension of the vector space Iď' of symmetric 2 X 2 matrices (see
Example 6.10).
tl tiOn A symmetric 2 X 2 matrix is of the form
: ol' 0l
+ a[o 1l
+ ,[o 0l
L0 0l Ll 0l L0 1]
la b1 [o 0l
L, ,J
:
Lo 0]
from which it immediately follows that ia :"7,
b , c :* 0. Hence, S is
dent and is, therefore, a basis for 14/. We conclude that dim |4l x }|a
b1
La ,]
so W is spanned by the set
{L;
:],[l i],L; l] }t-- Jl
If S is linearly independent, then it will be a basis for Iť. Setting
ol' 0l
+ a[o 1l
+ ,[o 0l : [o 0l
L0 0] Ll 0] L0 1] L0 0J
we obtain
The dimension of a vector space is its "magic number]' Knowing the dimension
of a vector space Vprovides us with much information about V and can greatly simpli$
the work needed in certain types of calculaťons, as the next few theorems and
examples illustrate.
linearly indepen(see
Example
Section 6,2 Linear Independence, Basis, and l}inncnsion
P OOí The proofs ofproperties (a) and (b) follow from parts (a) and (b) ofTheorern
6.8, respective$
(c) Let be a linearly independerrt set of exactly nvectorsin V. If S does not pan Y,
then there is some vector rr in Vthat is not a linear combination of the vectors in .
Inserting v into produces a set ,S' with n * 1 vectors that is still linearly independent
(see Exercise 54). But this is impossible, by Theorem 6.8(a). We conclude that S must
span V and therefore be a basis for I/.
(d) Let S be a spanning set for V consisting of exactly /, vectoťs, If S is linearly
dependent, then some vector y in , is a linear combination of the others. Throwing v
away leaves _a sei S' with n * 1 vectors that still spans V (see Exercise 55). But this is
impossible, by Theorem 6.8(b). We conclude that S must be linearly independent and
therefore be a basis for v.
(e) Let S be a linearly independent set of vectors in Y. If S spans V, it is a basis for
V and so consists of exactly n vectors, by the Basis Theorem. If S does not span V,
then, as in the proof of property (c), there is some vector v in \/ that is not a linear
combinaťon of the vectors in S. Inserting v into S produces a set S' that is still linearly
independent. If S' still does not span l/, we can repeat the process and expand it into
a larger, linearly independent set. Eventually, this process must stop, since no linearly
independent set in Vcan contain more than n vectors, byTheorem 6.8(a). When the
proces stops, we have a linearly independent set S* that contains S and also spans V.
Therefore, S* is a basis for V that efiends S.
{ť) You aťe fr. ked to pr*ve this property in xercis 56"
You should view Theorem 6.10 as, in part, a labor-saving ďevice. In many
instances, it can dramatically decrease the amount of work needed to checkthat a set
ofvectors is linearly independent, a spanning set, or a basis.
In cach ca e, determine whether , is a basis fcr V.
2x2, *L + 3x * r?}
(b) $ -l Mrr, S ,*
(c) 1 * S2, .r.
{t * tr, rť + xL,I +
$olutlon (a) Since dim(9J - 3 and S contains four vectors, S is linearly dependent,
by Theorem 6.10(a). Hence, S is not a basis for P2,
(b) Since dtm(M : 4 and S contains tfuee vectors, S cannot Wffi Mzz,by Theorem
6,10(b}. Hence, S is not a basis íorM22.
(c) Since dim(Or7 : 3 and S contains three vectors, S will be a basis for g2 if it is linearlyindependent
or if it spans 92, by Theorem 6.10(c) or (d), It is easier to showthat
S is linearly independent; we did this in Example 6,26. Therefore, S is a basis íorP2,
(This is the same problem as in Example 6.32-but see how much easier it becomes
iusing Theorem 6,10!)
<-Ťffiffi
*ffiss-dl xtend {1 * x, 1 - ,c} to a basis ťor 92.
#@ i,j;*lá !ít.llFirst note that {1 * x, I - x} is linearly independent, (\tIty?) Since dim
(Vr1 : 3, vye need a third vector-one that is not linearly dependent on the first two,
-l]}
{t *x,2*x*x',
{[l?],[:-i]
(a) Y: S2, : 3x*
T-
It
' Lo
J('}
Chapter 6 Vector Spaces
We could proceed, as in the proof of Theorem 6.10(e), to find suďr avector usingtrial
and error. However, it is easier in practice to proceed in a different way.
We enlarge the given set of vectors bythrowing in the entire startdardbasis for 92,
This gives
s_{t+x,I-Jc,I,x,xzl'
Now S is linearly dependen| by Theorem 0.10(a), so we need to throw away ome
, vectors-in this ca e, two, Whicfi ones? We use Theorem 6.10(f), taťting with the
first vect9r that was added, 1. Since 1 = |(1 + x) + á(t - x), the set {1 * x, 1 - a 1}
, islinearlyflependent, sowethrowaway 1. Similarly, ,r=r(L + r) - |(t - x), so
{1 + a !' - i, xI iílinearly dependent also. Finally, we check that {1 + x, I - x, x2!
}+ is linearly independent. (Can you see a quick way to tell this?) Therefore, {1 + x,
1 * .rr, .ť} is abasis for S2 that extend {1 * x, 1 * Jú}.
In Example 6.42, the vector pace lťof symmetric2 X 2 matrices is a subspace of
t]re vector space M22 of ail, 2 X 2 matrices. As we showed, dim 1ť = 3 S 4 = dim Mrr,
This is an example of a general result, as the final theorem of this section shows.
TheOíGm 6.11 Let W be a subspace of a finite-dimensional vector space 1/. Then:
a. 1ť is finite-dimensional and dim W s dim V.
b. dim W : dim V iíand only if W : V,
In xercťses 1*4, test ťhe sets af matrices far linear indepenďence
in M22, or those that are linearly dependgnt, expťe
frn af the mutrice. #$ a linear rcmbination af th* CIthtgr ,
l {L; -l],[l
-l],[l
:] }
2,{
[;
-;],[l -i],[-i
;] }
3
{ [:l l], [i l], [_l i], L: l l] }
P OOí (a) Letdim V = n,lf,Iť = {0}, then dim(lť) = 0
=
n = dim 1z.If Wis
nonzero, then any basis B for V (containing r vectors) certainly spans lť', since lť'is
contained in V. But 6 can be reduced to a basis B' for W (containing at most fl vectors),
byTheorem 6.10(f). Hence, I4lis finite-dtmensional and dim(l{r) s n = dtmV,
(b) If W= ,thencertainlydjm W= dim V. Ontheotherhand,if dim W= dim
V = n, then any basis B for W consists of exactly ,, vectoť . But these are then n linearly
independent vectors in Vand, hence, a basis for 1z, by Theorem 6.10(c). ThereIn
Exercťses 5*9; test tha sets af palyn*mials for linear independence.
Far those that arc linearly ďependgnt, express one
af the polynumials as linear eambinatiun oI the oth&rs.
5.{x, 1+x}in91
6. {1 * x, 1 + xz,L* rú+ *'} in#2
? , lx,2x * x2,3x + žx2}in #2
Lll],[li],[li]}
lav
l;;
8, |2x,x * x2,L * x3,2 - x2 + Jí3}inP3
9. { ! * žx,3x * x2 *J 3,
1 * x2 * 2x3,3 + 2,x *3x'} in S3
In.Exercisas ru*M, test the sets offunctions for \inear in,
dependence in ?F, For those that are linearly dependent,
gxprgs onť of thefunctions ils ňline,ar combinatian of the
others,
10, {1, sin r, cos x} 11. {1, sin2 x, coszx!
12, |e*, e-*I 13. {1, tn(2r),l*(r')}
14, {sin x, sin 2x, sin 3r}
15. If/and g are in (6 {r),
the vector space of all functions
wíth continuou derivatives, then the determinant
w(x) |/(x) g(x)
t:
|/t*l g'(*) l
is calle d the Wronskian of/ and g [named after tlre
Polish - French mathematician
],,,,"i ,: ],, :, ,', , , who worked on the theory of
determinants and the philosophy of mathematics],
how that/and are linearly indepenclent if their
!\konskian is not identically zera (that is, if there is
ome x such that }ť(x) * 0).
i* 16. In generel, the Wronskian of ír1. . .,frin'(,,Í"* 1)
is the
i"--- ' cJeterminant
ection 6,2 Linear Independence, Bnnis, nnd Dimension
of M22,
2S. Find the eoordinate vector of p (x) * 1 * 2x * 3x2
with respect to the basis , x {t * x, 1 * x, x2l| oťP2.
29. Find the coordinate vector af p(x) : 2 - x * 3x2 with
respect to the basis ffi - {t, t * x, - 1 * x'} of *2.
30" Let §be a set of vectors in a vector pace Ywith
the property that ev ry vector in Y eaí} be written
uniquely as a line ar combination of the vectors ínB.
prove that B is a basis fur v.
31. Let Bbe abasis ťor avector spac F, let ll1, . . , l llk
be vectors in 14 and let c1, r l ., cg be scalars. Show that
[cru, +,,, *l- Cpq]n * 61 [ur]B +,,, * ct[ut]n.
32. Finish the proof of Theor rn 6,7 by showing that if
{ [u, ] , . " . , [ut]*} i* linearly indepnndent in Rn then
{r,, r r l ) u1} is linearly independent in 1/.
33. Let {u,, r r l , a*} be a set oťvectors in an
n-dirncnsional vector pace Vand let fr be abasis ťor ť.
Let, * { [rr,] ). ",7lu*]r} b* the set of coordinate
vectors of {u1,,. r * rr*} with respect ta B, Prove that
span{u!} . .,, il*) : Y ífanď only if span(S) : Rn.
In Exerctses 34*39, rtnd the dimansion of tLte vector space V
and tve a basis for V
34,y : {p(x) inff2;p(0) 0}
35. ť: {p{x) inL2l p0) r 0}
M36. : {pk} lrr.ffrI xp'(x) * p(r)}
20, V : M2},
8 -{[; ?],[? i] Ll i],Ll ?]}
2l, V * M22,
E {[; i], Ll l], L_; i], L: i], [i ;]}
22,V : #r,B -* {x, 1 * x,x * tr?}
23,\r:? r,ffi-.- {t ff, 1*xt,X* "T'}
24,V : ? r,B : {t,t + ?x * 3*r}
25,V: Wr, § - {t,Z * x,3 xŽ,x + 2*')
26. Find the coordinate vector of Á * t i
2l
,"itr,
L3 4)
respect to the basis ffi -=
t&rr, Err, Err, Ě,,} of Mrr,
27 . Find the coordinate vector of Á : t 1 1l *Un respect
L3 4)
tothebasis B-{[; ;],[; ;],Il ;],Ll l]}
W{x}
18. 1/* Mrr,E*
fr(x)
/ť(x)
ír{x)
íik)
nndfi, l , l , fn are linearly independent, provideď \,V{x}
is not identicďly uero. Repeat xercises 10-14 using
the \,Vronskian test.
17, Let {u, v, w} be a linearly independent set oťvectors in
a vector space 1/.
{a} Is {u * v, v * w, il + w} lin*arly indep*ndent?
ither prove that it is or give a counterexample
to show that it is not.
(b) Is {u
* yl y * w, tl * w} linearly independent?
ither prove that it is or give fr c§unterexample
to show that it is ltot.
ine wítether the set B is a basisdeterm
V
{[i
In x*rcises 18*25,
ío, the vector space
l], Il ;], L-l
-l]
}
l9,V:M,,,B-{i; :],L:
-j],[i
i],Ll l] }
],]] Chapter 6 Yector paces
37, V : {a in Nírr:Á is upper triangular}
38" 1/ : {e m Mrr: Á is skew-symmetric}
39. v : {e in Mrr: AB- BA},where B =* t1 1l
40. Find a formula for the dimenr'-;oíthe
j:.r":1ou.*
of symmetric n X rr matrice ..
4l. Find a formula for the dimension of the vector pace
of skew-symmetric n X n matrice .
42, Let U and W be subspaces,of a finite-dimensiohal
vector space V'. Prove Grassmfrnft's ldentity:
dim(U+ W}: dimy+ dimlď* dim(ryn W)
|Hint: The subspace U + W is ďefined in Exercise 48
ofSection 6.1" Let B : {yt, . . ., v1} be abasis for
U n W. xtend B to a basis ť oť U and a basis D oť W.
Prove that C U fi is a basis for {/ * W"}
43, Let U and V be finite-dimensional vectcr pace .
{a} Find a formula for dim(U X V) in terms of dim t/
and dim . ( ee Exercis e 49 in Section 6.1.)
(b) If W is a subspace oť , show that dirn A *
dim trť, where A : {(w, w} l w is in 1ť}.
44. Prove that the vector space 9 is infinite-dimensional.
{Hint: uppose it has a finite basis. Show that there is
om polynornial that is not a linear combination of
this basis.]
45. xtend {1 * x, 1 * x + *'!to abasis for W2,
46. Extend
{[; i], [l l]} to abasis ťorM22,
4?,Extend{[
l],Ll ;],[l
*i]
}*abasis
fotM,,.
48. Extend
{ L; :], [? ;] }
to a basis for the vector
pace of symmetric 2 X 2 matrice ,
49. Find a basis for pan(l, 1 * x,2x} in$1.
50- Findabasisforspan{l * Lx,žx* xŽ,L * x2,I + *')
ínff2,
51. Find abasis for span(l - x, x, - x2, L * x2, L * 2x *
x21 inPr,
{- Let r {or, r . r , vr} be a linearly independent set in
a vector pace ť,how that iťv is a vector in Y that is
notin span{ }, then ' : {vr,... lynlv} is still linearly
independent.
5. Let ,- {or, . . . , vr} be a pannins set for a vector
pace Y. Shor,v that iťvris in pan{vr, . r , yr,*,), then
, ' : {vr}. r ., yn*r} ir still a pannirs setťbr F.
S. Prove The rern 6.10(f).
7. Let {or, r r ! ! vn} be a basis for a vector space V
and let ci, . r * c,, be nonuero scalar . Proye that
{clv1, . . . , cnvr} is also a basis for V.
58. Let {"r, . . t } vn} be a basis for a vector space V. Prove
that
{Vr,V, * V2,V1 * V, * Y3,...1Y1 + "'* Yn}
is also a basis for 1/.
Let ag, &t, . . . , frnbe n * I distinct real numbers. Define
polynomiaís po{x}, ?,,(x),. . . , p-{x) by
Fi(r) aŤ
(x * #o)",(.ť * di*r)(x * ili*t}-..(g * iln}
These are calíeď ttte Lagrange po$lnomia s associated
with &ry á1> l . . , an, f|oseph-Louis Lagrange (1736*1S13)
we born in Ita$ but spent mast af his life in Germany and
rance. He mnďe important contributi*ns fp such fields a,s
number theory, algebra, astronom!, mechanics, and the
calcuÍus af vari*ti*ns. In 1773, Lagrange w&s the rst to give
the valuma interpretation of a ďeterminarut (see Chapteť 4).}
5 . {a} Compute the Lagrange polynomials associated
with&a*tr&t*žrfr1*3,
(b) haw, íngeneral, that
60" (a} Prove that the set ffi ,
fuo(x), p,(x), . . . } p"(ň}
of Lagrange polynomials is linearly independent
in Sn. {Hint: et cnpo(x) + . . . $,FnU) 0 and
u s ,xercise 59{b),]
(b) Deduce that S is a basis for #n.
6l. If q(x) is an arbitrary polynomial ínWn it follows from
Exercise 60(b) that
q{x):copg(x)*",* cnpn{x} (U
for some scalar Cgl, . . 1 Cn,
(a) Show that c; : q(aiffor i : 0, . . ., ťl,and deduce
that q{x} : q(ao}patx} + " . * .{- qta*}p*(x) is the
unique repťesentation of q(x) wiirr ,*ip*.t to the
basis ff,
Fl(ui}a:
{l 'rI',!''
52. Find a basis for **(Li
t- 1 *1l\
L*, I))inM22,
?],[l;],L-l-i],
53. Find a basis for span{sín?x, ťos?x, cos 2x} in S.
(b) Show that for any n * 1 points (ag, ct}, {at,cl)r . . . l
(a*cn) with distinct first components, the function
q{x} defined by quation (1) is the unique
polynomial of degree at most rt that pa e
through all of the points. This formula is known
as the Lagrange interpolation fornnula, (Compare
this}or*ula witň Problem 19 in Exploration:
Geometric Applications of D*terminants in
Chapte r 4.)
{c) tJse the Lagranse interpolation formpla to rur*d
the polynomiď oť degree at most 2 that pa e
through the points
ection 6.2 Linear Independence, Basis, and l}imension
(i) (1, 6), (2, - 1), and (3, *2}
(ii) (- 1, 10), (0, 5), and (3, 2)
62. U e ths Lagrange interpolation fnrmula to show that
if a polyno*tul in Pn has ru * 1 reros, then it must be
the zero polynomial.
63. Find a formula for the number oť invertíble matrice s
in Mnr[p). |Hinh This is the ame as determining the
number Óf different bases for Z#. (VrW?) Count the
number of Ývay to construct a basis far 7$, one vector
at a time.]
[xpl ratl0
Magic Squares
The engraving shown on page 461 is Albrecht Diirer's Melancholia I {l5I4). Among
the many mathematical artifact in this engraving is the chart of numbers that hangs
on the wall in the upper right-hand corneť. (It is enlarged in the detail shown.) Such
an array of numbers is known as a magic square,We can think of it as a 4 X 4 matrix
32
ls 11
67
15 L4
Observe that the numbers in each roq in each column, and in bot}r diagonals have
the same sum: 34, Observe further that the errtries are the integers !,2, . , ,, 16. (Note
that Diirer cleverly placed the 15 and 14 adjacent to each other in the last row, giving
the date of the engraving.) These observations lead to the following definition.
tr. \ťM is a classical n X
13l
8l
L2
l
1J
16
5
9
4
n ma1ic quare, show that
wt(M) * n{nz!I)
2
|Hint:Use Exercise 51 in Section 2.4.]
2. Find a classical 3 X 3 magic $raíe. Find a different one. Are your two examples
related in uny way?
{lI
3 magic squares. L,et
cll
íl
i)
(,/)
ffi
ír
C3
(J
-a
.*)
:c)
(-)
,;--
aa
m
U)
ÉO
cr
C)
CJ
c
cJ
(J
:()
LJ
C.,6
u)
p
nfi
3. Clearly, the 3 X 3 matrix with all entries equal to I is a magic square with
weight t. Using your an Wer to Problem 2, find
13 _1
3 magic square with weight 1,
all of whose entňes are different. Describe a method for construďing a 3 X 3 magic
quare with distina entries and weight w for any real number r.
Let Magndenote the set of all n X n mag,ic guare , andLet Mag| denote the set of all
n X n magic quares ofweight 0,
4. (a) Frove thatMaguis a subspace of My.
(b) Prove thatMaďisasubspace of Mag.
5. Use Problems 3 and 4 to show that ťM is a 3 X 3 magic quírre with weight
w, then we can write Mas
M= Mo* kI
where Mp is a 3 X 3 magic square of weight 0, / is the 3 X 3 matrix consisting entirely
of ones, and k is a scalar. What mu t kbe? [Hint: Show that M _ kI isin Ma fot an
appropriate value of k.]
Let's try t* íindff" wff"y of describing fllI 3 X
M*
be a magic guare with weight 0. The conditions on the rows, columns, and diagonals
give risi to a system of eight homogeneous linear equations in t}e variables n,
br,,,,i,
6, Write out this system of equations and solve it. |Note: Using a cAs Wilt
facilitate the calculations.]
Ia b
|a e
Ls h
t s -s t t l
fuí:l-r+f 0 s t|
L -t .+t -s J
8, Find the dimension of Magu, |Hint: Combine the results of Problems 5 and 7.]
9. Can you find a direct way of s}rowingthat the (2, 2) entryof a 3 X 3 magic
square with yeight rv must be wl3? |Hint: Add and subtract certain rows, columns,
and diagonals to leave a multiple of the central entry,.]
10. LetMbe a3 X 3 magic square ofweight0, obtainedfromaclassical 3 X 3
magic square as in Problem 5, IíIuI has the foim given in Problem 7, wťtte out an
equation for the sum of the quares of the entries of M, Show that this is the equaťon
of a circle in the variables s and í,and carefully plot it. Show that there are exactly
eight points (s, í)on this circle with both s and f integers. Using Problem 8, show that
these eight points give rise to eight classical 3 X 3 magic squares. How are these magic
squares related to one another?
42
#, {fr as*
Section 6"3 Change of Basi*
In many applications, a problem described using one coordinate system may be
solved more easily by switching to a new coordinate system, This switch is usually
accomplished by performing a change of variables, a process that you have probably
encountered in other mathematics course . In linear algebra, a basis provides
us with a coordinate system for a vector space, via the notion of coordinate vector ,
Choosing the rigtrt basis will often greatly simptify a particular problem. For example,
consider the molecular structure of zinc, shown in Figure 6.3(a). A scientist studyrng
zinc inight wish to mea ure the lengths of the bonds between the atoms, the angles
between these bonds, and so on. Such an analysis will be greatly facilitated by introducing
coordinates and making use of the tools of linear algebra. The standard basis
and the associated standard xyz caordinate axe aíe not always the best choice. As
Figure 6.3(b) shows, in this cáse {u, v, w} is probably a better choice of basis for R3
than the standaťd basis, since these vectors align nicely with the bonds between the
atoms of zinc.
(a) O)
íl0u10 6.3
Ghanoc-oí-Basis MatlIGG$
Figure 6.4 shows two different coordinate systems for R2, each arising from a different
basis. igure 6.4(a) shows the coordinate system related to the basis 6 : {u1, u2},
while Figure 6.4(b) arises from the basis C = {v1, v2}, where
The same vector x is shown relative to each coordinate system. It is clear from the
diagrams that the coordinate vectors of x with respect to B and, C ate
[x]n :
respectively, It turns out that there is a direct connection between the two coordinate
vector . One way to frnd the relationship is to use [x]6 to calculate
tl1 :
L-l],
ll2 :
L_i],
Vi :
Ll],
Y2:
Li]
L]]
and [x]c:L_Í]
Chapter 6 Vector paces
(a)
línrg.{
Then we can find [x]c by writing x as a linear combination of v1 and v2. However,
there is a better vyay t0 proceed-one that will provide us with a general mechanism
for such problems. Wb iJlustrate this approach in the next example.
tJsing the bas es Band C above, find [x]c,given that [x]r :
l l
oluilOn Since x = t * 3u2, writing u1 and u2 in terms of v1 and v2 will give us the
required coordinates of x with re pect to C.We find that
tl1 :
[-;]
: -,
Li]
+ 2[
l]
lI2:
[_i]:,L;] [l]
* 3y, + }vz
x3vr*V2end
CI
x-tl1 +3ul
:6vr-Y2
Thís gives
[x]c *
in agreement with Figure 6.4(b),
This method may not look any easier t]ran the one suggested prior to Example 6.45,
but it has one big advantage: We can now find [y]6 from [y]r for any veďor y in R2
ection 6.3 Change of Basis
with very little addiťonal work Leťs look at the calculations in Example 6,45 from a
differenťpoint of view. From x = ut * 3u2, we have
[x]c = lur * 3uz]c = [ur]c * 3[uz]c
by Theorem 6.6. Thus,
[x]c :
*
where p is the matrix whose columns are [u1 ] a and [u2 ] g. This procedure generalizes
very nicely.
Think of 6 as the
,blď basis and C as the "neuť basis. Then the columns of, Pc<_s
are just the coordinate vectors obtained by writing the old basis.vectors in terms of
the new ones. Theore m 6.L2 shows that Éxampleo.+s is a special case of a genera1
result.
[[u,]cIu,,-,Il]
L-li][l]
P [x]r
§'12
P]Otlí(a) Let x be in V and let
Thet is, x : Ctltl + ", * cnun. Then
[x]r *
[x]c _ [c,u, +
_ [[u,]c
: Pcn-,r [X ] n
L:,]
""'f Cn nlc
+,",{ cnlun]c
Ir, l
, , , [u*]-'L;,]
Chapter 6 Yector paces
(b) Suppose thatPis aurtn X n matrixwiththe propertythatP[x]n: [x]6for allx
in V, Taking x : tl;, th ith basis vector in B, we see that [x] a
: [ul]r : e;, so the Jth
column of p is
pi: Pei = P[ui]n = [ul]c
which is the ith column of Pc*r,by definition. It follows thatP : Pg*3.
(c) Since {ut,..., un}islinearlyindependentin 1{theset{[ul]c,..., [un]c}islinearly
independent in R', by Theorem 6.7. Hence, Pc*B : [[u, ]c . . . t"i. i, invertible,
bythe Fundamental Theorem.
For all x in % we have Pg*3|x]6: [x]c. Solving for [x]6, we find that
l lxjn : (P.*r)-l[x].
for all x in V. Therefore, (Pg*s)-' is a matrix that changes bases from C to 6. Thus, by
the uniqueness property (b), we must have (Pg*3)-' : PB*c. .,
nOmOtl8
o You maYfind it helpfrrl to think of change of basis as a transformation (indeed,
it is a linear transformation) from R'to itself that simply switches from one coordinate
system to another. The transformation corresponding to pr*uaccepts [x]r ffi
input and returns [x]6 as ouputi (pc*s)-' : pr*rdo"rlust the'Jfposite. Figure 6.5
gives a schematic representation of-tháprocess.
^
V
-rl\
[]c,/ \t]s
ť' Multiplicaticrn
\
by Fg*3
*-
Rl?
-**-*.**_} R/,
Multiplication
bY Fg*c * (Pg*g}*L
o The columns, of pc+6 t the coordinate vector of one basis with respect to
the other basis. To remember which basis is which, think of the notation c <- B as
saying "B in terms of C." It is al o helpfut to remember th at Pg*g[x] 6is a linear combination
of the columns of P4*6. But since the result of this combination is [x]a, the
columns of pc*s must themselves be coordinate vectors with respect to c.
Find the ďrange-oÉbasis matrices P6+5 and P6*a for the bases 6 : {t, x, x2} and C :
{1 + a x* x2,L + xzlofg2.Thenfindtheioordiout.vectorofp(x) = 1 * 2x- x2
with respect to C.
s0lml0n changing ío a standard basis is easy, so we find p6*c first. observe that the
coordinate vectors for c in terms of 6 are
[1 * x)s: [x + xz]B : [1 + x?Jn:
íigure S.S
Change of basis
[,l
L?]
Iol
L1]
Il],
LOJ
ection 6.3 Change of Basis
(Look back at the Remark following ,xampLe 6.26.) It follows that
[t 0 1l
PB*c: |1 1 0l
Lo 1 1]
To find Ppe,,we could express each vector in 6 as a linear combination of the vectors
in C (ao tnis), but it is rnuch easier to use the fact that Pc*B: (Pr*J-', by
Theorem 6.I2(c). We find that
Pc*B * (Pg*c)*'
It now follows that
[p(x) Jc
which agree with Example 6"37.
nonaít If we do not need Pg*6 e{plicitly, we can find [p (r) ]c from [p (x) ] r and
P6+g using Gaussian elimination, Row reduction produces
lps*cltp(x)lr!
-+ [1|(P6*c)-' [p(x)]r] = |t|Pc*s|p(x)]r] : tl| tpk)]c]
(See the neť section on using Gauss-|ordan elimination.)
It is worth repeaťng the observation in Example 6,46: Changing fo a standard
basis is easy. If ťis the standard basis for a vector space V and 6 is any other basis,
then the columns of Pe*sare the coordinate vectors of B with respect to á, and these
are usually "visible]'We make use of this observation again in the next example.
In M22,1et B be the basis |Ep E2t, Err, Ezal- and let C be the basis {Á, B, C, D}, where
r ! ! *rl
I z 2 2l
l_r ! !l
|-1 1 žl
L i *l iJ
: Fc* s;P(x) 1n
:Llii]Li]
L-,]
It
L0
It
B- l
L0
It(_ l
L1
l) = [t 1l
Ll 1l
change-of-basis matrix Pg*s and veriry/ that tX]ť * P**n [X]s ťor :
Find the
[t 21
Ll 4J'
Chapter 6 Yector Space*.
$0lrtl0n l To solve this problem directly, we must find the coordinate vectors of 6
with respect to C. This involves solving four linear combination problems of the form
X - aA + bB + cC * dD, where X is in 6 and we must find a,b, c, and d, Howevet
' here we are lucky, since we can find t}re required coefficients by inspection.
Clearly, Eu- A,Ezt: -B + C,EI2 = -Á i B,andE22 = -C * D.Thus,
[f;,, ]c : , IErr}ť, *
l], IE,,]c:I
i]
l',,]c:
0jLi]
r1
l0
Io
Lo
0]
CIl
*1 l
1J
0l
0l
1l
1J
If x : [t 2l.
*h.r,
L3 4I
Fť*B: [[
',]r,
{Err}r: lf.rr}c {Err}c}*
tX] #:
Dl ť+-I3
* and P n_c :
0
*1
1
0
1l
11
01
ůú
*1
1
0
ů
end
Li]
PcsxB:[i:il]Li]L:i]
This is the coordinate vector with respect to C of the matrix
lr ol tr ,]_[, ,l*n[, rl
-A-B-C+4D:-l |-|
L0 0] L0 0J Ll 0] -Ll l]
[t z1
-l
l-v
- L: 4)-^
as it should be.
S*ltJt!i}í, We can compute Pg*3 in a different way, as follows. As you will be
asked to prove in Exercise 2l,if t is another basis for M22,then Pc-s: Pg-5P5*3:
(pr*r)*1p.*r.Iíťisthestandardbasis,thenP;*6and Pg?gcalbefoundbyinspection.
we have
r1
lo
lo
Lo
[t 0 0 0]
lo 0 1 0l
Io 1 0 0l
Lo 0 0 1]
ection 6,3 Change of Basis
4. (Do yoll see rt}ry?) Therefore,
F(* * t?w
*'Fr*r
r1
l0
Io
Lo
r1
Io
|0
L0
r1
lo
l,L0
11
11
01
úů
*1
1
ff
ff
ff
*1
1
ů
*1
1
fi
fi
*'r1 0 0 0]
lo 0 1 0l
lo 1 0 0l
Lo 0 0 ,J
0lr1 0 0 CIl
0ll0 s 1 0l
l]L;i;i]
0l
0l
*1
l
1l
t]
1l
1l
tJ
s
*1
t
fi
which a ree with thc first soluti*n"
l3milX The second method has the advantage of not requiring the computation
of any linear combinations. It has the disadvantage of requiring that we find a
matrix inverse. However, using a CAS will facilitate finding a matrix inverse, so in
general the second method is preferable to the first. For certain problems, though,
the first method maybe just a ea yto use, In aíry event, lve are about to describe yet
a third approach, which you may find best of all.
TfiG a[ s-lordan fifinOil íorGom[lltlng t Ghange,oí,Basi Mauix
Finding the change-of_basis matrix to a standard basis is easy and can be done
by inspeďion. Finding the change_of-basis matrix from a standard basis is almost
as ea y, but requires the calculation of a matrix inverse, as in Example 6,46. If we do
it by hand, then (except for the 2 X 2 case) we will usually find the necessxy inverse
by Gauss-Jordan elimination. We now look at a modification of the Gauss-fordan
method that can be used to find the c.hange-of-basis matrix between two nonstandard
bases, as inBxanple 6.47.
Suppose E = {l1,...,un} and C = {or,...,vn} are bases for a vector space V
andPc*ris the change-of-basis matrix from Bto C. The ith column of P is
[u,] (:
$o u; : pt t +, , , * Fn ,r,If tis snybasis íbrY, then
[u,]ť: |prr, + * pnř,']e: p,i[v,]g * " " + Fui[vn]g
This can be rewritten in matrix form as
[ [v, ]s Iu,]r
Io,,l
[r,,]r]l : l:
Lp,, )
{fi$
Chapter 6 Vector Spaces
which we can solve by applying Gauss-|ordan elimination to the augmented matrix
[ [v, ]s [*-]u | [uu]g]
There are n such systems of equaťons to be solved, one for eaďr column of P4*6,
but the coefficient matix [ [v, ]s , , , [v, ] 6] is the same in each case. Hence, we can
solve all the systems simultaneouslyby row reducing the n X 2n augmented matrix
[v,]rlIur]g[ [v, ]g [u-]a] ": lciB]
Since {v1, . . . , v,} is linearly independent, so is { [v, ]9 . . . , [vn]e}, by Theorem 6.7,
Thereforp, the matrix C whose columns are [v,]a ..., [vn]5 has the n X n identity
matrix / for its reduced row echelon form, by the Fundamental Theorem. It follows
that Gauss-fordan elimination will necessarily produce
1c I 11 -+ 1l I11
where P : Pc*B,
We have proved the following theorem,
If t is a standard basis, this method is particularly easy to use, since in that case
B = Pg*3 and C = Pg+-a. We illustrate this method by reworking the problem in
Example 6.47.
Rework Example 6,47 using t}re fiauss-}ordan mcthod"
& * ffi T'aking á to be the standard basis for Mn, we see that
[1
lo
Io
L0
i-
l
-B: Pt*B*
Row reduction produces
and f, -Pt*ť*
111
111
011
001
10
0ů
01
00
000
1ů0
filů
001
11
l1
ů1
00
10
0 *1
01
00
0l
0l
l]
00
01
10
00
l]
1J
[1
lo
Io
Lo
0l
0l
0l
1l
*1
1
0
0
0 0l [1
1 0l l0
0 0l--*|o
0 lj Lo
*1
1
_]L
0
0
Ii
L;
tCl ]:
.,,i,:u,--.,," (Verify this row reduction.) It follows that
Da C<-B
0
*1
1
0
0l
0l
*1 I
1J
a we ťbund before.
Secticn 6.3 Change cf asis §
In Exercťses 1*4:
(a) Findtíte ffioráinate vectors [x]r and [x]c gíxwiťLl
respect to the bases B and C, respe#ivety,
(b) Finá the change-af-basis matrix ?g*7fram fr to C.
ťď] LIse yCIur frfisw r to part (b) to {oťnputg [*]r, anď
compare your an w r with the one fourud in part (a).
(ď} inď the thange-af-basis matrix Pn*círa{yl C t* B.
(e) Use your ú,nsw rs to parts {c} and (d} ta cornpute [x jr,
anď compfr.re your &n wer with the one found iru part (a).
l.x:
{7*
L,
2-§(:
J''
t"
3.x:
ť-
4.x:
C*
In Exercises 5-8, folíow the instructions for Exercises ]*4
using p(x) instead af x.
5, p(x) * ? x,B : {t, r},C : {", 1 + x} in9,
6.p(,x}*1+ 3x,B: {t + xol x},
C*{2*,+}mp,
7. ptx} : 1 * x', B- {1 + x * x2, x * x', x'},
C - {1, *, x2} in iP,
8.p(x):4 2x_'x',3- {x, 1*x',x* x'},
C - {t, t * x, x2} in?,
In Exercises 1}" and 12, follow the i.nstructions ío,
Exercises l *4 usin /(x) instead aíxr
t 1. /(x} 2 sin Jť * 3 cos x, #: {sin x * cos Jť} cos x},
C -- {sin x * co x, sin.ť * cosx} in span(sinx, cosx)
12, f {x) * sin x, B - {sin x + cos .s, cos r},
( -:.
{cos Jr * sin x, sin x * cos x} in span(sin x, cos x}
13. Rotate the xy-axes in the plane counterclockwise
thraugh an angle 0 *-
60n to nbtain neT,ť x' y'-axe .
[Jse the methods of this section to find (a} the
x'y'-caordinates of the point who e Jfl-coordinates
are {3, 2) and (b) the xy-coordinates *f the point
whose x'y' -coordinates are {4,
*4}-
14. Repeat xercise 13 with § ,: 135o.
In ,xercťsss 9 aná fi, follow the instructions for
Exercises 1*4 using A instsad at x.
9. Á : [+ '1,U: the standard basis,
L0 * 1l,
C - { [i -i], Li l], L; l], Li :] }
[t 1l
Lr 'J'
{ L; ?], L: i], L; ;], Ll ;] }
{ Li i], [i i], [l ?], L: l] }
[i],
ffi { Li], Ll] },
{ Ll], L- l] }
,* *,
L -l],
u :
{ Ll], Li] },
{ Ll], [i] }
in R2
Li],u:{Li]L:]Ll]}
{Li]L;]Ll]}**,
{Ll],L;]}
:oťis
Ll],B{L;]Ll]Li]}
{Ll]L;]Li]}*o,
10. Á ::
,y
11 *
L/
in M22
in M2
15. Let 6 and ťbe bases for R2. If ť :
the change-of-basis matrix ťram t
Pc*B:t i -:l
L*l 2]
frnď B,
16. Let6andCbebases farW2,If B * {x, 1 * x,
1 * x * x2} and t}re change-of-basis matrix
from§to*is
and
t- 1 0 0l
p(*B: l 0 21l
L-r 1 lJ
find ť.
Chapter 6 Yector paces
In calculus, you learn that a Tayltr polynaminl af ďrgree n
about a is u polynornial af th* farm
p{x) * ilg + ar(x * a) + a2(x * o}'+ ., * * an(x * frJ"
wlrere an * 0., In other wards, it is a polynomial that hns
been expanded in terms af powers af x *,il,insteaď af powers
of xr Taylor polynomials ilre v ry useful for approxirnatingfunctions
that are "well behav ď" nefrr x a ň,
The set B : {1, * * a, (x * G}2,. . r, (x - o)'} is a basis
forWn ío, ilny real number a. (Do yCIu ee fr quick"way to
sktaw this? Try using Theorem 6,7,) This fact allows us ta use
the techniqaes af this sectian tg rewritg a palynamial .s il
Taylor palynomial about a given a.
17. xpre sp(x) * 1 * 2x * 5x2fr fl Taylor polynomia1
about fr re 1,
\ďe encnuntered linear trfl"nsforrnations in
formations from ffi'to R*. In this section,
maticns between arbitrary vector space .
18. xpr s p(x) : 1 * 2x * 5x2 a a Taylor polynomial
about fl : -Z,
19. Exprf,s P(x) * ff3 as fr Thylor p lynomial about # ffi * l.
20. xpr F(r) * Jí3as a Taylor polynomial about a ffi
*,
2l. Let ffi, C,qnd Dbebases for a finite-dimensional vector
space V. Prove that
n*ďc+* * PB*n
22, Let be an rr-dímensian*l vgetor pŘ ť with basis
B * {or, . . l ,yn}. Let P be an invertible n Y, nnratrix
and set
U;:PtiVr+",*F* o
forl* 1,.. ,1ťl,Provethetfr x {nr, ll.,ur}isabasis
for Y and show that P z F *ť.
ection 3.6 in the eantcxt of matrix transwe
ext*nd this cffnťept to linear transforffiffitr
e ffiffiffi A linenr transformation fram a vector
W ;r*ňping r l V--+ Tťsuch that, for all u and v in
1. 7(u * v) * 7(u) + 7(v)
2- T(cu) * cT(u)
It is straightforward to show that this definition is equivalent to the requirement
that Tpreserve all linear combinations. That is,
T : V * W is a linear transformation if and only if
T(c,v, + czyz + .,. + clv1) _ crT(vr) + czl(vz) +
for all yl,, " ", yfr in Y and scalars Cl, .,, 1 C17,
$í#
Every matrix transformaťon is a linear transformation. That is, if Á is arl m X n
matrix, then the transformaťon T,a : Rn --r R' defined by
ra(x) * Áx for x ínRo
is a linear transformation. This is a restatement of Theor nt 3,30.
t
ry
ection 6.4 Lin*ar Trnnsťormations
Define T : Mrnl Mnnby T(A) : A'I. Show that Tis a linear transformatiůn"
i.,,,,..i,,,;:,.;.i.: We check that, for Á and fi in Mnnand scalars c,
r(A + B) : (e + B)7"* A1" + "BT: r{A} + r(3)
T(cÁ)ffi{cA}ť*cÁť*c (Á)
Therefore, T is a linear transformation.
t
Let D be the differential operator D , g -+ ?} defined by nV) = /', Show that D is a
linear transformation.
calculus, we know that
D{f + s) - ry + ň' * í'* g' * D{í)+ D(g)
D(cf)* {cf)' * cí': cD(í)
Hence, -D is a linear transformaťion.
In calculus, you learn that every continuou function 0n la, b]is integrable.
next example shows that integration is a linear transťormation.
Define S z%ía,b] -+ Rby a _ ÍIít*ldr,Showthat, is alineartransformation.
úIlltl0n Letlandgbe in.(o|a, b]. Then
{/+ g) tr gXx)
) +s(
ďx*
(g)
and {c/) cf)(x}
(x) ďx
tx} dx
)
It follows that S is linear.
t
The
r_,
ť_,
ť,
:
x) ďx
dx
x}} ď
f-,
dx
(/+ r
ť(x)
f(x) d
f)+t
"ť,o,
,|_u,
1b
,rJ
^í
, cS(,ť
f {*",
f)
,ť,
=ť_
=,|
:cS
(c) T: R *+ R defined by T(x) : r + 1
Chapter 6 Yector Spaces
Show that none of the following transformations is linear:
(a) T: Mzz-+RdefinedbyT(l; = 4"rn
(b) 7: R + R defined by T(x) : 2*
(c) T: R-+ Rdefinedby T(x) :
"
a'
$0lútl0nIn each case, we give a specific counterexample to show that one of the
properties of a linear transformation fails to hold.
?], 'nen
Á
' [1 0l
(*) Let Á :
L; oJ
*"a r í-ol
Lo
[t 0l
*B:
Lo ,],'o
It 0|
r{Á+ }_det(Á+fi)
But
T(Á)+r()*detÁ+detB:li :l +|0 0|*0+0 a0
l0 0| |0 1l
so T(Á + B) * r(Á) + r(B) and T is nat linear"
(b) Letx:landy* 2.Then
T{x + y} * T(3) * 23 * 8 * 6 : 2\ + 22 * 7(x} + Tty)
so T is not l"inear.
{c} Letx: landy:2.Then
Ttx +y): r(3):3 + 1* 4* 5* (l + 1} + {z + 1}* T(x) + r{y}
Therefore, T is not linear.
Romtll Example 6.53(c) shows that you need to be carefi,rl when you encounter
the word "linear]' As a function, T(x) : x t 1 is linear, since its graph is a straight
line. However, it is not alinear transformation from the vector space R to itself, since
l-+ it fails to satisS the definition. (Which linear functions from R to R will also be linear
transformations?)
There are two special linear transformations that deserve to be singled out.
(a) For any vector spaces V and lV, the transformation Tg: V -+ l,Y that maps every
vector in V to the zero vector in Iť is called the zero transfurmation, That is,
?'o(v) : 0 for all v in V
(b) For any vector space V, the transformation r : V -+ V that maps every vector in V
to itself is called the iilentity ttansformatiorr. That is,
/(v) = ., for all v in V
(r it i important to idenť the vector space V, we may write rn for clarity,) The
proofs that the zero and identity transformations are linear are left as easy exercises.
m
l
ection 6.4 Linear Transformations
ProRoítiss 0ílinstí TransíOrmntion
In Chapter 3, all linear transformations were matrix transformations, and their
properties were directly related to properties of the matrices involved. The folH,
lowing theorem is easy to prove for matrix transformations. (Do it!) The full
proof for linear transformations in general takes a bit more care, but it is still
straightforward,
m s.l4
a.
b-
c.
rrol - 0
r(*v) :
T{u * v)
- T(v)
_ T(u)
a|ťijiii' We prove properties (a) and (c) and leave the proof of property (b) for
Exercise 21.
#*Ť (a) Letvbe anyvector in 1/, Then T(0) : 7(0v) : 07(v) : 0, as required. (Can you
give a reason for each step?)
(c) 7(u-v) : 7(u+ (-1)v) : 71o; + (-l)T(v) : T(u) - T(v),._,,,_,.,'_:,is,i
nsmr t Property (a) can be usefirl in showing that certain transformations are
notlinear,As an illustration, consider Example 6.53(b). If T(x) = 2", then 7(0) = 20 =
1 * 0, so Tis not linear, by Theorem 6.14(a). Be warned, however, that there are lots
of transformations that do map the zero vector to the zero vector but that are still nof
linear. Example 6.53(a) is a case in point The zero vector is the 2 X 2 zero matrix O,
so T(O) : det O : 0, but we have seen that flÁ) = det Á is not linear.
The most important properry of a linear transformatio n T : V-+ I4l is that T is
completely determined by its effect on a basis for V, The next example shows what
this means.
Suppos e T is a linear transformation from R2 to S2 such that
ffiF--M, S0lutl0n Since *
pan {B}. Solving
"1
t*Xa
basis for K2 (wtry?), eYery Yect r in R2 is in
r['l :2 3x*xz and
Llj
lz1
Tl |_
L3l
Find ,L-i] *o ,L;]
{[l], [i]}
is
r.[,] _L r t2-1 :_ [-,l"'Lr] "'L3_]
*L2]
Chapter 6 Yector Spaces
+ -Li])
ll1
- 3Tl'Il?lL"J
c*x2}+3{I
y * Laxz
}re,
1l
I
1J
l
l+j
3x
2Ie
X
txX
:efo
_[t
/l
L1
-1l
_1J
Ťl
sr,
í
-?|
It
rl
Ll
he:
I
7rl
I
7tz
t1
-,Li]]
,,,Ii]
+{u
ix * {4,
yer the
)
l
').l
}v
7-
d
v')
6b",
eco
X*.
6,
:c
we find that c1
* x')
b
imílarly, we discorrer that
la1
L;]
: (la
{
b{u
l
-r}
+
7e r,
q
I
l+
l
+{t
3x*
-9a
l, we
*:
)
l1
Lt
I
{-
m)L
It
rl
L1
í ,.|
\á
+{
fo-:
2I
T
{:
{la,|;]:,(
* {Sa
* (Za
:--1
tsa
aXt * x2}
a 3b}x2
solution r[
L
zb}
2b}
3b}
and 11 +
t
*tl
z)
ffiry"&- (},{ote that by settins a - * 1
LLx * Lax?j
The proof of the senereI theorem is quite straightforward.
P1O0í The range of T is the set of all vector tn W tbat are of t}re form T(v), where
v is in V. Let T(v) be in the range of T. Since 6 spans % there are scalars cy , . . ; cn
such that
v=clvl *...*cnv,
Applyrng Tand using the fact that it is a linear transformation, we see that
T(v) = T(c,v, * ... * c,vr) = c,T(v1) + ... + cnT(v,)
In other words, T(v) is in span(T(6)), as required, i
Theorem 6. 15 applies, in particular, when 6 is a basis for V. You might gu,ess that,
in this case, T(B) would then be a basis for the range of T. Unfortunately, this is not
always the case. We will address this issue in Section 6.5.
c0ma0 lti0n oílinenr Transíormation
In Section 3,6,we defined the composition of matrix transformations. The definition
extends to general linear transformations in an obvious way.
o
Section 6,4 LinearTransformations
observe t}rat S o T is a mapping from Uto lť(see Figure 6.6). Notice also that for
the definitíon to make sense, the range of 7 must be contained in the domain of .
iguto S.S
Cornposition of linear transforínations
Find ( - nL_;] and (, " ,,L;]
Smlní gn l,\re oínput
( -n[_;] * -(rL
_;]) : {3 + {3 2}x}:.(3 * x) * x{3 + x)
Let l ffi2 --* Sr and : S1 *+ S2 be the linear transťormations defined by
la1
'L;-|
Ha+b+b}x and ,(p(x)):x?{x}
(s.rL;] : r(rl;il: s{a
:ax*{a+b}*'
:3x*x2
and
+ (a + b}x) * x{a + {a + b)x}
Chapter 3 showed that the composition af two matrix transťormatíons
nn*ther matrix transformatioí}. Iil general, ws have the following theorcm.
;lt]Gm
SoT
Chapter 6 Yect*r paces
Frmg Let u and v be in [/ and let c be a scalar. Then
(, -rXu + v) : S(T(u + v))
: s(r(u) + r(v))
-.(r(u))+.(r{v))
since T is linear
since is linear
and (S"TXcu): S(r(cu))
'1T (cr{u}) since T is linear
* c (r(u)) sincc is linear
TŤ c( o rXu)
Therefore, S o Tisalineartransformation. _ ...jffi
The algebraic properties of linear transformations mirror those of matrix transformations,
which, in turn, are related to the algebraic properties of matrices. For
example, composition of linear transformations is associative. That is, if R, S, and T
are linear transformations, then
fto( oT): (Ro,S)o7
provided these compositions make sense. The proof of this property is identical to
that given in Section 3.6.
The next example gives another useful (but not surprising) property of linear
transformations.
Let,S : U-+ V andT l V -+ Wbe linear transformations and let J: V-+ Vbe the identity
transformation. Then for every v in V, we have
(r"JXv)* r(J{v))* T(v}
Since T o I and have the ame value at
T o J : 7. Similarly, J o,S =-,S.
very v in their clomain, it follslť that
nGmrrl The rnethod of Example 6.57 is worth noting. Suppose we want to show
that two linear transformations 7, and 7z Goth from Vto I4z) are equal. It sufÉcesto
showthat Tr(v) = 72(v) foreveryvin V.
Further properties of linear transformations are explored in the exercises.
lnrerses 0íllnGar Trrnslotm0tlons
:;J;* ,
Section 6.4 Lin*ar Transformatians
ROntlls
o The domain vand codomain lď'of Tdo not have to be the same, as theydo in
the case of invertible matrix transformations. However, we will see in the next section
that Y and 1ť must be very closely related.
o The requirement that T' be linear could have beerr omitted from this definition.
For, as we will see in Theorem 6.24, if T' is any mapping from I4l to V such that
T' o T = Iy and T o T' : lw, then T' is forced to be linear as well.
o If T' is an inverse for T, then the definition implies that T is an inverse for T'.
Hence, T' is inverjible too.
Y*rify thet the mappings T: R2 *+ Si and T' : #r --* ffi2 deíined by
rl:1: a+ G+ b)x and T"(c * dx), t ,' l
Lb ) Ld c_]
&re lnver e .
fiI tltln Ure compute
(r,-r)L;] : r,(rL;])
[;]
and
(r. T'){c + dx} * r(r'(c * dx))
Hence, T' " T : /n*; and T o T| :
other.
_ C + lc + G c))x: C + dx
7 and T' are inl.erses of each
t
.+
ť,
As was the case for invertible matrices, inverses of linear transformaťons are
unique if they exist. The following theorem is the analogue of Theorem 3.6.
PíOOíThe proof ie the same as that of Theorem 3.6, with products of matrices replaced
by compositions of linear transformations. (You are asked to complete this
proof in Exercise 31.) .,litwt
Thanlcs to Theorem 6.|7, tf T is invertible, we can refer to ílteinverse of 7. It will
be denoted by 7-t (pronounced "7 inversď'). In the next two sections, we will ad,
dress the issue of determining when a given linear transformation is invertible and
finding its inverse when it exists.
r c l": Tl l
Ld c)
Igr
t Therefo
s.1l
Chapter 6 Vector pnces
In Exercisas 1*12, determine whether T is n linear
transformation.
1. TiMzz4 Mmdefinedby
_lo b1 |a+b 0 l
'lr d) L0 c+d)
2, T i Mzz a Mz?defined by l
rl* "l Ťt 1 w*z1
|,y zJ Lx*y 1 J
3. r I Mnn 1 Mnndefined by r(Á) * Ať,where -B is a
fixed nx nmafttx
4, T i Mrn 1 Mnndefine,d by T{A) : AB * BÁ, where .B
isafixed nx nmatrix
. r i M,n *+ ffi defined by r{Á) * tr{A}
6, T i M*, -+ ffi defined by r(Á) = flnfrz, , ,
dnn
7. T I Mn,-+ R defined by r{Á) * rank(Á)
8. r :#z+#zdefined by T{a + bx * cx\ : (a * 1) +
(b+L)x*(c*l)xz
g, T:§z+Wtdefinedby (c + bx * cx,z) : *
b(x+1)+b(x+1)'
10, T : W *S defined by r(fl * 7(x2)
l1. r: ffi *+ S defineď by r{/) f(x))'
12. r: S.+ R defined by r(fl * /(c), where e is a fixed
scalar
13. Shnw that the transťorrnations - and in xample
6. 6 are bath linear,
14. Let T: R2 -.+ R3 be a linear transformatíon for which
[-i]nnd Li]
16, Let T : W 2 a # zbe a linear transformation for which
Find r(6 * x - 4*') and T(a + bx * cx').
17 " L*t T : W 2 a W zbe a linear transformation for which
r(1 *x)xl+x2
r(1 + x2)
T(x+ x') x*x2
:1*x+x?
Find T(4 * x * 3x2} and T{a + bx t ex?),
lS. Let T : M22 -+ R be a }inear transfnrmation ťor whích
:1, T a
k1
3,T
:1*X, : Z x* x2
Findr[i ll *u ,l: 'r1L4 2J
19. Let T : M22--+ R be a linear transformation. Show that
there are calars a, b, c, and d such that
,l:, :1: aw * bx * cy + dr
Ly zJ
ro, uu [' "l ,r, Mr.,
Ly zJ
20. how that there is no linear transformation : ffi3 4 ffz
such that
[t 1l
Lo 0j
[ll]
r[, 0l
L0 0J
r[, 1l
Ll 0J
,Li]
{1]
1 . Let l R2 4 W zbe a línear transformation for which
r[1l :1 žxand r[ 3l :x* 2x2
LlJ -
L*lJ
t 0l
rl 6l : *2+2x2
ll
L -8J
2l. Prove Theorem 6.14(b).
?]?.Let.{or,. í,,vn} be abasis for avector pace Fand
let T : V--+ 1/ be a linear transformation. Prove that if
T(vr) : yl, T(vz) : y2. . . , 7(vn) : yn, then Tis the
identity transformation on V.
ffi23. Let T :Ln+Lnbe a linear transformation such that
T(xfr) -- kxk-'for k - 0, 1, r.l, m. Showthat Tmust
be the differential operatot D.
{l]
r[1l ffi
L0]
Find r[;] *u ,[;]
Find ,L-í] *u ,[;]
3{. Let yl, . ,.. 1yp bn vectors in a vector spmce Y and let
T; 1/ ** Iť be a linear transformation.
(n) If {r(y,), , , . , T{vn)} is linearly independent in W,
show that {o,, r r r t vn} is linearly independent ínV,
{b} hnw that the canver of part (a) is fal e.
Thet is, it ísnot nece arily true that if
{or, r r , } vn} in línearly independent in Y, then
{r(v,)! i N r , ť(vn)} t* 1inearly independent ínW.
Il1ustrate this with &n *xaínpl*
r:
K2 ** RŽ,
2 . efine linear transťarmations , l K2 a Mezand
t
T: R2 ** R2 by
,L;] :
L,;
b
a'_,] and ,L;] :
|"
joo1
Compute (S " T)
Li]
and (,s
" -
[]
, can you
compute (T * )
L;]?
Iť o, compute it,
26. Define linear transťormations , , 9, + #zand
T:W2+W,by
,S(a+ bx): fr+ {a+ b}x+Zbxz
and Tk + bx* wž}* b + acx
Compute( -rX3 *2x *x'}and
{ - T}ta + bx * cx2). C&n you compute
(T " , Xn + bxf Iť so, e*mpute it,
Section 6.5 The Kernel and Range of a Linear Transformation
30" , l Sr -+ 91 defined by (a + bx) :
(.*4n + b) + Laxand Tl S1 *-t S1 defrnedby
Tfu + bx) * bll + ta + Zb)x
31. reive Theorem 6,17,
32. Let T : V --r V be a linear transformation such that
To T * Ir
(n) hnw that {v, ť(v}} i linenrly dependent if and
cnly iť T(v) ffi ty.
(b) Give an e)ffimple oť such a línear transťormation
wíthfmm.
T : V *+ V be n linear transformation such that
f r r.
how that {v, T{v)} is linearly dependent iť and
only if T(v) y or T(v) * 0.
Give &n xample af such a Iinear transťormatíon
with 1/ m m?.
The set uf all linear transf*rmations from , vector space V
ta a vectar pft.t W is denoted by tr(V, W). ť and T arg
in ,(y, w ), wg ca,n defi.ne the sum, + T af S arcd T by
(s+rXv}=,(v)+T(v}
ía,aLl v in V If c is a scalan wg defr,ne the scalar multiple
cTufTbyctabe
(crXv) ffi cT(v)
ío,allv in V Thgn + T and cT are bot\t transformations
from Y ta W
34. Frovc that , + T and cT are linear transformations.
3 . Frnve that ffitv, W} is a vectnr pace with this addition
and scalar nrultiplication,
36- Let ft, , and T be linear transformations such that the
follawins operations make eí}e. Prove that;
(a}-R-( +r):fto, +RoT
(b} c(R " ) * (cn) o : p
" (c ) for any scalar c
In Exercises 29 and 30, verifu that S and T are inverses.
29. s:R2--+R2defined*rr] :
Lj| :|and7:R2
+R2
definedrrrL;] :
|_:;l n,1
33, Let
fn
(a)
(b)
jfr 2? ,I)efine linear transformations : W nl W ,, and
T:W,r4W"by
, (p(x)) * p(x * 1) nnd r(p(x)) = p'(x)
Find ( " fl(p(x)) and (T o Xp(x) ). |Hinr; Remember
the Chain Rule.]
ifr" 28. Define linear transformations , : ?? na W n and
T|Wr+"Ťoby
(p (x)) * p(x * 1) anď rk(r)) * xp'(x)
Find( "flb(x))anď( * Xp(x)).
Tile l(ernol aníBange 0l a llnear TítnsíOmatlon
The null space and column space
with a matrix. In this section, w
line ar tran s forntation.
are two of the fundamental
extend these notions to the
subspace a ociated
kernel and range oť a
Chapter 6 Yector paces
Let Á be an m X n matrix and l et T : T 1be the corresponding matrix transformation
from R'to R'defined by ?(v) = Áv. Then, as \.r'e saw in ChJpter 3, tlre range of ?is
the column space ofÁ.
The kernel of T is
{vinffi*:Áv: 0}
: null(Á)
In words, the kernel oť a matrix transforrnation is just the null space of the
ponding matrix.
Find the kernel and range of the differential operator D:99 -+ 92 defined by
D(p(x)): p'(x).
S0lutl0lt SinceD(a * bx* cx2 + dx31: b +zcx+ 3dx2,wehave
ker(D) = {a * bx * cxz + dx3:D(a-| bx * cx2 + dx3) :0}
_ Í- l l--- l 2 , 1 1 ,
= |a l bx * cx2 * dx3:b * 2cx + zdx2 : 0}
But á f Zcx * 3dx2 : 0 ifand only í b : 2c : 3d = O,which implies that b : c =
d: 0. Therefore,
ker(D) : {a -| bx * cx2 * dx3:b: c: d: O}
: {alainR}
In ot}rer words, the kernel of D is the set of constant polynomials.
. Th:
,*g.. ofD is all of 92, since every polynomial in g2 is the image under D (i.e.,
tlre derivative) of somepolynomial in 93, To be specifrc, ía * bx + ň'irin g2, tiien
corre*
+
*-t-
/
\\
(:)-,+ (;)",)
\
ection 6.5 T}re Kernel and Range of a Linear 'Transforrnation
Let : S r --+ ffi be the linear transformation defined by
rt
S(p(x)) * | p(x)dx
Jg
Find the kernel and renge of {
ffitfrít i In detail, we have
Therefore ,
1|
S(a + bx} -::-
Joro
+ bx) dx
t b ,l,
L2]o
\ /,/ 2
ker( ) * {a + bx:S(n + bx):
:
{,+
bx:a+*:r}
( bl
*
t"
+ bx:a - -;}
0}
ffsure s.I
b
If y: *: + bx,.f2
r1I
then l yd* - 0
J0
: { -!+ b*}
L2 )
Geometrically, ker( ) consists of all those linear polynomials whose graphs have the
property that the area between the line and the x-axis is equally distributed above and
below the axis on the irrterval [0, 1] (see Figure 6.7),
The range of is R, since every real number can be obtained as the image under
S of some pó nomial in s1. For example, if a is an arbitraryreal number, then
rI
Iadx:lax]á=a_O:aJ6
0á*,(a).
Let 7 : Mzz + M2, be the linear transformation defined by taking transposes:
T(A) :.eT. r'irrd the kernel and range of 7.
í,iiii*i,i*lt We see that
ker(T) : {AinMrr: T(Á) : o}
: {Á in M22: A,T : O}
nut if Ár : O, then Á - (Á')' : OT : S. It ťollows that ker(T)
Since, for any matrix Á in Mzz,we have Á -a- (Á')' : r(Á?)
we deduce that rense{T} * Nízz.
- {o}.
(and Ár is in Mer},
In all of these examples, the kernel and range of a linear transformation are subspaces
of the domain and codomain, respectively, of the transformation. Since we are
gineralizing the null space and column space of a matrix, this is perhaps not surprising,
Nevertheless, we should not take anything for granted, so we need to prove that
it is not a coincidence.
Chapt*r 6 Vector paces
TngOrsm .lS Let T : V + trť be a linear transťormation. Then:
&, T}re kernel of 7 ísa subspace of V.
b, The range clf T is a subspa ee nť W,
* * ** e, Let T : V --> Wbe a linear transformation. The rank ofTis the
dímensión of the range of T and is denoted by rank(T). The nutlity of T is
the dimension of the kernel of T and is denoted by nullity(T).
P1O0t (a) ince 7(0) = 0, the zero vector of Vis in ker(T), so ker(T) is nonempťy.
lct u and v be in ker(T) and let c be a scalar. Then 7(u) : T(v) = 0, o
T(u*v) T{u)+r(v) 0+ *0
end T(en) cT(u) x c0 * 0
Therefore, u * v and cu are in ker(T), and ker(T) is a subspace of V,
(b) Since 0 = T(0), the zero vector of ilr is in range(T), so range(?) is nonempty.
tet 7(u) and T(v) be in the range of T and let c be a scalar, Then (u) * T(v) =
T(u + v) istheimage of theveďoru * v. inceu andvarein V, sois u * v, and
hence T(u) + T(v) is in range (T), Similarly, c(u) = T(cu). Since u is in % so i fi,
and hence cT(u) is in range(T). Therefore, range(T) is a nonempty subset of l{'that is
closed under addition and scalar multiplicaťon, and thus it is a subspace of w.
iiil
Figure 6.8 gives a schematic representation of the kernel and range of a linear
transformation.
p*si ť* *,#
The kernel and range of T : V --> W
In Chapter 3, we defined the rank of a matrix to be the dimension of its column
space and the nullity of a matrix to be the dimension of its null space, We now extend
these definitions to linear transformations.
ffÁ is a matrix and T = 7n is the matrix transformation defined by T(v) : Áv, then
the range and kernel of T are tlre column space and the null space of Á, respective$
by Example 6.59. Hence, from Section 3,5, we have
rank(r} ]:,: ťank(Á} and nullit1,,(T) -: nullity(Á)
ection 6.5 The Kernel *nd Range of a Linear Transffirmatign
Find the rank and the nullity of the linear transformation D : 93 -r S2 defined by
D(p{x)) : p'(x),
"!*lti is } In Example 6.60, we computed range (D) : 92, so
rank(D) : dim 3z: 3
The kernel of D is the set of all constant polyromials: ker(D) : |a : a in R} : |a, | : a
in R}, Hence, {1} is a basis for ker(D), so
nu}lity{ }ff dim{ker( }}x1
Find the rank and the nullity of the linear transformatisn : i , + R defined by
rI
(p(x)) *
Jrnk)
dx
.r,i:ii]iii,iirii From Example 6.61, rang (S) * R andrank(S) : dim m - 1. AlSo,
ker(S): {-?*bx:binm}
L2 )
:
{I,(*i * x),binR}
pan(*i + x)
so {*} + x} is abasis forker(S). Therefcre, nullity(S) : dim(ker(S)) : 1.
Find the rank and the nullity of the linear transformation T : M22-+ Mz, defined b,v
r@): .l,r.
*lxlír:* In Example 6,62,we foundthat range(T) : M22andker(T) : iO}, Hence,
rnnk(T} dim Mrn q { end nulliťy(T) dim{fi} ffi ů
In Chapter 3, we saw that the rank and nullity af an m X n matrtxÁ are related
by the formula rank(Á) + nullity(á) = r. This is the Rank Theorem (Theorem 3.26).
since the matrix transformation T = Tlhas Rn as its domain, we could rewrite the
relaťonship as
rank(Á) + nullitfiÁ) = dim Rn
This version of the Rank Theorem extends very nicely to general linear transformations,
as you can see from the last three examples:
rank(D} * nullity(p) _ 3 + 1 --T
4: dimp* Exarnpl e 6.64
Examp}e 6.65
Example 6.66
Chapter 6 Vector Spaces
Let T : V --+ 14/ be a linear transformation from
V into a vector space 1ť. Then
rank(T) + nullity(T) * dimV
In the next seďion, you will see how to adapt the proof of Theorem3.26 to prov
this version of the result. For now, we give an alternative proof that does not use
matrices.
l
PlOoí Let dim V : n andlet {v1,. . ., vt} be abasis forker(T)[so that nullity(T) :
dim(ker(T)) : k]. Since {vr, . . . , v1} is a linearly independent set, it can be extended
to abasisfor V,byTheorem 6.28. LetB = {or,...1yllytr4lr...,vn} be such abasis.
If we can show that the set C : {T(vla1), . . , , T(vn)} is abasis for range(T), then we
will have rarrk(T) : dim(range(T)) = rr - k and thus
rank(T) + nullity(T) = k,| (n- k) = n - dimV
as required.
C.riuirrty C i, contained in the range of T. To show that C spans the range of T; let
T(v) be a vectoí, in the range of T. Then v is in V, and since B is a basis for % we can
find scalars c1l . . . , cn such that
Y : ClV1 + "' + ClVtr * C1.,1V1*, * "' * cnv,
Sincev1, ,.,,ykareinthekernelof T,wehave T(vr): ",- T(vr) = 0, so
T(v) = T(c,v, * ", * cpv1,* c1l1vtra1 + ", * c,vn)
: crT(vr) +...+ q,T(vl,) * c1*,Ť(v1*,) +...* crT(v,)
: c1..1T(vla1) + .,. + cnT(vn)
This shows that the range of T is spanned by C.
To show that C is linearly independent, suppose that there are scalars cllp , , , t cn
such that
ťí.+tT(vx*,) + + cnT(v) ffi 0
Then 7(c1.,iv1*, * ", * crvr) : 0, which means'that í&+tvt+t + ", + crvn is in tlre
kernel of T and is, hence, expressible as a linear combination of the basis vectors
vl, . .,, v1 ofker(T)-say
Ck+tVft+t + ", + CrYn : ClY1* ", * ClV1
Butnow cl\*,"*cpvp-Cr+tyk+t cryr=0
andthelinearindependenceof6forces cl : ", : crl:0. Inparticuláí, ck+t : trr:
cn = O,which means C is linearly independent.
"
We have shown that C is a basis foithe range of T, so, by our comments above, the
proof iscomplete.
We
have verifi ed the Rank Theorem for Examples 6,64, 6,65, and 6.66, In practice,
this theorem allows us to find the rank and nullity of a linear transformation with
only half the work. The following examples illustrate the process,
ffi@ection
6.5 'Ihe Kernel and Range of a Linear Transťormaticn
Find the rank and nullity of the linear transformation T : 92 -+ 93 defined by
r (P@)) : xP{x). (Check that 7 really is linear,)
oluU0n In detail, we have
T(a + bx + cxz) : Ax * bxz + cx3
It follows that
ker( ) : {a + bx * cxT : T(a + bx * cx?) * 0}
- {a+ bx*cx2:ax+bxz+ cx3:0}
' : {a+ bx*cxz:a: b_ c- 0}
_ {o}
o we have nullity(r) : dim{ker(T)) : 0. The Rank Theorem implies that
rank(r) : dim92 nulliry(T) - 3 0 : 3
Rcmltr In Example 6.67, it would be just as easy to find the rank of T first, since
íx,Í, i| is eaůseen to be a basis for the range of T. Usually, though, one of the two
(the rarrk or the nullity of a linear transformation) will be easier to compute; the Rank
Theorem can then be used to find the other. With practice, you will become better at
knowing whiďr way to proceed.
Let I4z be the vector space of all rylrnmetric 2 X 2 matrices. Define a linear transformation
T :W --+TzbY
la
'1
: ro- U) + (b- c)x-| G- a)x:
'L, c)
(Check that 7 is linear.) Find the rank and nullity of T.
.;í,jítt ** The nullity of 7 is easier to compute directly than the rank, so we proceed
as followsl
(|o
ker(T) :
t L,
:
{L;
ír,:
lla
íl,:
t La
íl,
1L,
(c * a}x, * -}
(c - a):
'}
'n***" { [l 1] }
*
"O^,.forthekernelof
7, so nullity(T) : dim(ker(T)) : t.
The RankTheorem and Example 6.42tellus that rank(T) : dim V - nullity(T) :
3-1*2. t
<+
Chapter 6 Vector Spaces
0nt-t0-0nG rnd 0nto llnor] T]ansíO]mltlons
We now investigate criteria for a linear transformation to be invertible. The keys to
the discussion are the very important properties one-to-oíre and onto.
ffimmmr xg
The definítíonof one-to-one may be writtcn mCIťe formally as fnllows;
The above statement is equivnl*nt to the following:
Figure 6.9 illustrates thesc two statement .
v yW
ft) r is not one-to-one(a) T is one-to-on
íl0uto 6.s
i Another way to write the ďefinition of onto ísgs folIows:
In other words, given w in I4l, cloes there eťst some v in V such that w = 7(v)? If,
for an arbitraryIv, we can solve this equation for v, then Tis onto (see Figure 6.10).
Section 6.5 The Kernel and Range oť a Linear Transformation
v
{a} r is onto
]lil10 E.10
ft) r is noť onto
Hence, T is onto.
It turns out that there is a very simple criterion for determining whether a linear
transformation is one-to-one.
Which of the following linear transformations are one-to-one? onto?
f zx1
(a) T: R2--+R3definedOrrl]l :
l, -, I
L/J
L 0 ]
(b) D : 9 3 --+ T2defined by D(p(}c)) : p' (r)
(c) T: M22--> M22defined by T(A) : nr
-[",l : ,[r,'l,rn.r,t'ltťl,i]n (a) Let r[r'.,
LyzJ
|,*, l tZxz l
L,,;,,]
:L",;"]
so 2x1 : Zx2and x1 - lt : xz *
?z, Solving these equations, we see that x1 : x. and
[x'l
l"'], ro 7is one-to-one.lt: lz. Hence,
|,i,]: ,rr,
T is not onto, since its range is not all of R3, To be specific, there is no y ctor
["l ,n *, such that r['.l : lll,,*r,.,,,Ly) Ly) L,.]
(b) in Example 6.60, we showed that range(D) - 02, so D is onto. D is not oneto-one,
since distinct polynomials in 9, can have the same derivative, For example,
x3 + x3 + 1,butD(x3;: zl: D(x3 + t).
(c) Let A and Bbe in M22, with 7(Á) : T(B).Then ÁT = 87, so A: (!T)T =
@' )' : B. Hence, Tis one-to-one. In Example 6.62, we showed that range(?) : Mzz.
t
<-+
GOí8m 6.20 A hnear transformation T : V +Wis one-1
Chapter 6 Yector paces
Pt00l Assume that T is one-to-one. If v is in the kernel of 7, then T(v) = g, 3o'
we also know that T(0) = 0, so 7(v) : 7(0), Since T is one-to-one, this implies that
y = 0, so the only vector in the kernel of T is the zero vector.
Conversely, assume that ker(?) = {0}. To show that Tis one-to-one, let u and vbe
in Vwith T(u) : T(v). Then T(u - v) = T(u) - 7(v) : 0, whichimpliesthat u * v
isinthekernelof T. Butker(T) = {0}, owemusthaveu - v:0 or, equivalently,
ll :-:, v. This pťoye that T is ůne-to- íle .
is in the kernel of í,then
* + {a + b}x
+ b : 0. Hencc, S ,-:^
0, and thereťore
* dimffi2 nullity(T} * ? 0 _ 2
Therefore, the range of T is & two-dimensional subspace of Pa
ťange{f ) : R2. It fo11ows that T is ontů.
,,L;]
lal
0 :
'Lr]
and T is one*to*slle, by Theor m S.2S.
It f ll urs that 6l - 0 and a
quently, ker( ) { L;] },
By the Rank Theorem,
rank(T)
1-ol
:
Loj
Conse_
enď hence
For linear transformations between two n -dimensional vector spaces, the properties
of one*to-one and onto are closely related, Observe first that for a linear transformation
T : V -+ 1ť, ker(T) = {0} if and only if nullity(T) : 0, and Tis onto if and
l.,}, only ťrank( T) = dim W. (\ťhy?) The proof of the next theorem essentially uses the
method of Example 6.70.
how that the linear tťansformation T : R2 -+ S1 defined by
lal
'L;]
& + {a + Wx
is one-to-one and onto"
PlOoíAssume that Tis one-to-one. Then nullity(T) : 0 by Theorem 6.20 and the
remarkpreceding Theorem 6.21. The RankTheorem implies that
rank(T) : dimV - nullity(T) : n - O : n
Therefore, Tis onto.
Conversely, assume that T is onto. Then rank(T) : dim W = n. By the Rank
Theorem,
nullity(T) = dim V - rank(T) : n - n : 0
Hence, ker(r} - {S}, and T is one-t *o11e.
l
I
Section 6.5 The Kernel and Range of a Linear Transformation
In Section 6,4,wepointed out that if T: V+ W,is a linear transformation, then
the image of a basis foi V under T need not be a basis for the range of T. We can now
give a cJndition that nsrríes that a basis for Vwill be mapped by Tto a basis for 1,V"
ffiffi
Theorom 6"22
í;,i:itLet cy, ,. . , c1 be sca]ars such that
c17(v1) + ",+ clT(v1) : 0
Then T(clv1 + ", + ctyt) :0, whichimpliesthat clv, * ", * clv1isinthekernelof
T. But, since 7 is one-to-one, ker(T) : {0}, by Theorem 6,20, Hence,
clvl +",-lclvo:6
But, since {vr, . . . , vt} is linearly independent, all of the scalars c; must be 0. Therefore,
{T(vr), , . ,', i(.rn)i i. tirr"urty independent,
gr$$§ s. Letdim V= dim W - n.Thenaone_to_onelineartransformation T:V__> Ý
maps a basis for V to a basis for 1ť.
iť:i**l LetB : {o,,...,v,} be abasisfor V. ByTheorem 6.22,T(B) : {T(vr),, ",
7(v")} is a tineariy independent set in W, so we need only show that 7(6) spans
w. r.rt, by Theorern ali, r(s) spans the range of T, Moreover, T is onto, by Theo_
rem 6.21, so range(T) : 1,t/, Therefore, T(B) spans W, which completes the proof,
Let T: R2 --+ 9, be the linear transforrnation from Example 6.70, defined by
la1
'L;.]
:at(a*b)x
Then, by Corollary 6.23, the standard basis t = {e,, er} for R2 is mapped to a basis
T(E) : {r(e,), T(e)} oíTl.We findthat
T(er):r|-:l -1*x and T(eJ*
L0]
It follo\,vs that {1 * x, x} is abasis for 91"
L1]
we can now determine which lineaí transformations T:
+
*+
7 *+ 1ď are invertible,
Theorom 6.24 A linear transformation T : \; + lť is invertible if and
and onto. , ], ,
only if it is one-to-one
, 1-- . . - ". -_.*. -
Chapter 6 l/ector paces
r*i;i Assume that T is invertible. Then there exists a linear transformation T-1 :
w + vsuch that
T-1 "T=Iv and T"T-I:lw
To show that T is one-to-one, let v be in t}re kernel of 7, Then T(v) = 0. Therefore,
T*l{r(v}) ffi r*1(0) -+ {7'*t " TXv} : 0
=* í(v) * 0
which establishes that ker(fl : {0}. Therefore, Tis one-to-one, by Theorem 6,20.
To show that T is onto, let w be in 1ď and let v : r-1 (w). Then
T(v) = ?(T-l(w))
= (f o T-lXw)
= 1(w)
=w
which shows that w is the image ofv under 7. Since v is in % this shows that Tis onto.
Conversely, assume that ?is one-to-one and onto. This means that nullifiT) = 6
and rank(T) = dim t\r. we need to show that there exists a linear transformation
T' z W + Vsuch that T' o T = /yand T " T' = Iw,
Let
w be in ilr. Since 7 is onto, there exists some veďor v in V such that T(v) = 1,.
There is only one such vector v, ince, if v' is another vector in V such that 7(v') : *,
then 7(v) = T(r');the fact that Tis one-to-one then implies that v = v'. It therefore
makes sense to define a mapping T' : W + Vby setting ?(w) = y.
It follows that
and
{T' , rXv) x ť'{ť{v)}'(w) Ť, y
(r * T'Xw} § r{T'(w)) T{v) Ť 1ť
It then follows that T' o T = Jy and ?o T' = Iw. Now we must show that T' ísalinear
transformation.
To this end, lď w1 and w2 be in W and let c1 and c2 be scalars. As above, let
Ť(v,) = wl and T{v): w2. Thenv1 = T'(wr) andv2 : T'(w) and
T'{clwl + c2w2) * T'{c1 I(v,) + crT(vz))
T'{T{c t + c z))
* I{crv, + c2y2}
:: CtYt + CzYz
ctT'(*,) + czT'(*r)
Consequently, T' is linear, so, by Theorern 6.17, T' : T*I,
Section 6.5 'ťlre Kernel and Range of a Linear 'Transformation
lsomorlhlsns 0íUG8t0] aaces
We now are in a position to describe, in concrete terms, t tat it means for two vector
pace to be "essentially the same."
Show that 9n-rand Rn are isomorphic,
sn m Ťhe process of formin the coordinate veďor of a polynomial provides us
with one possible isomorphism (as we observed a]ready in Seďion 6,2, although we did
not use the term isomorphism there). Specificatly, define T : 9 n_t + R' by T(p(x) ) =
[Pk) ]a where = {I, n . . . , ť-l} is the standard basis for 9n-,. That is,
T'(aa + aIX, + " , + fr,r_tX""'l):
Theorem 6.6 shows that Tis a'|inear transformation.If p(x) = ao * ap l",*
&n-1x''l is in the kernel of T, then
* T{ag+ a +,1,+ fl,r*rXn*L)
Hence, a0: Q!: ",: an*l:0, sop(x):0.Therefore,ker(T) : {0},and Tisoneto-one.
Since dim 1ln_l : dim R' : fl,T is also onto, by Theorem 6.21. Thus, T is
an isomorphism, and 0, r = R'. Á
,{-., 1-,
how that M*nand re*n are i omorphic.
0luti0n Once again, the coordinate m appingfrom M*, to R*n (as in Exampl e 6,36)
is an i*omorphism. The details oťthe proof are left á an exerci .
In fact, the easiest way to tell if two vector paces are isomorplťc is simply to
check their dimensions, as the next theorem shows.
a0
lfr.y
l
:I
o,',,-r)
il: ]
:L:]
Chapter 6 Yector Spaces
sm 6.2
PIOOí Let n = dim Y. If V is isomorphic to Iť', then there is an isomorphism
T l V -+ 1ť. Since T is one-to-one, nullity(T) : 0, The Rank Theorem then implles
that
rank(T) : dim V - nullity(7) : fl _ 0 : n
Therefore, th'e range of T is an ,í-dimensional subspace of 1,ť'. But, since T is onto,
W : ralge(T), so dim W : n, as we wished to show.
Conversely, assumethat Vand Whavethe same dimension,n,Let9: {yt, . . . , vn}
be a basis for Vand tet C = iwr, . . ., w,} be a basis for W.We will define a linear
transformaťo n T : V --> lť' and then show that T is one-to-one and onto, An arbitrary
vector v in T/ can be written uniquely as a linear combination of the vectors in the
basis 6-say,
l^fu define by
v:Ctyi+,lt+Cnyn
T(v):ct\ťt+",+ cnwn
It is straightťorward to check that 7 is linear. (Do so.) To see tírat T is one-to-one,
suppose v is in the kernel of 7. Then
r\fi*",*cnwn=T(v)=0
and the linear independence of C forces cL = . . ,
= cn:0, But then
Y:ClV1 +",+cnYn=a
so ker(T) : {0}, mearringthat 7is one-to-one. Since dim V: dim }ť, Tis also onto,
by Theorem 6"2L Therefore, T is an isomorphism, anď Y x W.
Show that Rn and W n are not isomorphic.
Theorem 6.25"
Let lť'be the vector pace of all syrnmetric2 X 2 matrices. Showthat Wis isomorphic
to R3.
$0lfil0n In Example 6,42, we showed that dim W : 3. Hence, dim W : dim R3,
so W = R3, by Theorem 6.25. (There is an obvious candidate for an isomorphism
T:W --+ R3. What is it?) t
.+
c,bfl
t
ection 6.5 The Kernel and Range of a Linear Transformation
n0mr t Our examples have all been real vector spaces, but the theorems we
have proved are true for vector pace over the complex numbers C, or Zo, wherep is
prime. For example, the vector space M22(Zr) of all 2 X 2 matrices with entries from
Z2hasdimension 4 as a vector space over 22, aldhence M22(Zr) x l|,
1. Let T: Ní"22a Mzzbe the }inear transformation
defined by t
la b1 la 0l
'L, d): Lo d)
{a} Which, iť any, sf the following matrices are in
ker(r)?
r 1 il ,/..\ [o 4l I": 0l
(i) l _, :| (ii)
L; ;] (iii)
L; _;]L-t 51 Ll U_
{b) Which, if anyo of the matrices in part (a) *re in
rerrs*(r}?
(c) Deseribe ker{r) and ranse(T).
2. Let T : M22 *} R be the trinear transf rmation defined
by r(Á} * tr{Á).
(a} Which, iťany, of the following matric* are in
ker(r)?
1121(i)[_,3j (ii}L;
í] (iii)[l _i]
(b} Which, if any, of the following scalars are in
range(r)?
(i) 0 (ii} 2 (iii) Ýitz
(c} Describe ker(r) and range(T}.
3. Let T : W 2 -+ K2 be the linear transťormation
defined by
rx b1
T(a+ bx* cx|)- li l
La Ť c)
(a) Which, if any, of the foltowing potynCImials are in
ker(r)?
(i)l*x (ii) x*xŽ (iii)l*x*x2
(b} Which, iťany, of the followingvectors are in
range(r)?
(i}
[;] (ii)
Li] (iii)
L?]
{c} Describe ker(T) and range(T).
Let T :Wz*Wzbe the linear transfCIrmation defined by
r(p(x)) ; xp'i*).
{a} Which, iťany, of the ťo}lowing polynamials are in
ker(T)?
(i) 1 (ii) x, (iii} x2
(b) Which, if any, of the polynomials in part (a) are in
range (r)?
(c} Describe ker(r} and range(T).
In ,xercises 5*8, rtnd basesfotr the kern*í and range af the
linear translormations T in the indicated exercť. 6 , In each
case, state thg nullity and rank ůíT and verify the Rank
Tkeorem.
5. xercise 1
7. Exercis* 3
In Exercises 9*14, firud either the nullity or the rank of T
and then wse the Rarlk Theorem to rtnd the other.
g. T i Mzz*+ F&2 defined uv rlo ',|_ i' ',|**] -Lc d) Lc d)
10. T :wz*> ffiZ defined by r{p(r)) * [olill
[p(t)_J
6, Hxercise 2
8. xercise 4
ffi Á , where
* Á- * BÁ, where
1l. T: Nízza,Mzzdefinedby r(Á)
l- 1 *1l
B-.. l l
L-l lj
l2. r : Mzz a Mzzdefined by r(Á)
[t 1l
B --=
Ln lj
ff_ 13. 7 : .$z-+ ffi rlefrned hy r{p(x)) * p'(o)
|4. T i M:: - MErdefined by r(Á) *- A * ÁT
In Exercises 15-20, determine whether the linear transfor,
rnatiott T is (a) one*to-ons and (b) onto.
15" T: R,-* R2 definedbu T["l - l'- - ll-*] ^LyJ Lx+ 2y)
]& 4.
!? r 7" z W z
*> R3 defined by
r 2a*b l
Tta*bx+r*')*| a*b*Sr|
L t--t. ]
18. r : Wz -+ [RŽ defined by r(p(x)) *
19. r: R3 --+ NÁ22 dcíined by r[;] *
' L.]
Chnptcr 6 Yect*r pnces
16. T: R2 -+ #2 ďefined by
la1
'L;j
Ť:-(a žb} +(la+b)x+b+b}xz
3l" howthat%[0, 1] # q[0,2].
32. howthatÝr|a,b] # S[r, dJforaLlaa bandc1d,
33. Let : 1/ +W and T: [/-+ Vbe linear transformatíong,
(a} Prove that if and T are both one-to-onel so is
SoT.
(b) Prove that if and 7 are both onto, so is o T.
34; Let ; Y*+ W aná T l {J * Ybe linear
transformations.
{a} Prove that if , o T is one-to-one, so is T.
(b) Prove that iť " T is onto, so is .
35. Let T : V + Tť be a linear transformation between two
finite-dimensional vector pace$.
(a) Prove that if dim V < dim 1ť,then T cannot be
onto,
(b) Prove that if dim V > dim Iť, then T cannot be
one-to-oí}e,
36_ Let il6, frl, r r * anbe, n + 1 distinct real numbers.
Defrne T:Wn-*+ Rn{-t bI
rtp{x)
[::1]
Prove thgt T is an isomorphism,
37. If Yis a finite-dimensional vector pfrce anď T : V --* is
a linear transformation such that rank(T) : rank{r'),
prove that range(r)n ker(T) * {0}. |Hint: T2 denotes
T o T,IJse th* Rank Theorem to help show that the
kernels ať T and T2 are the same.]
3fi. Let {"/ and W be subspaces of a ťinite -dimensional
vector pece \/. ileťine T l U X '[ť --t Y by fiu, w} :
ll * Iť,.
(a) Prove that T is a linear transformation.
(b} how that range(T) * t"/ + W,
{c} Showthatker(T) x {J n W. |Hint: See Exercise 5S
in Section 6.1,]
(d) Frove Grassmann' Identity:
dim([/+ 1ť): dimt/+ dimw * dim(Un W}
|Hint:Apply the Rank The rem, using results
(a) and (b) and Exercise 43(b) in Section 6.ž,)
[r(ol1
Lptrl l
|a b
La + b
bcl
b + ,]
lo1
?0.'r: R3 + W definecl Uy r| a l :
l o + b + c b
"1,*i*r* fi.r is the vector space ofI
L b 2c c)'
all symmetric 2 Y,2 matrices
In Exercises 2].*26, determine wLtether V and W are
isomorphic. If they t},re, give (ln ťxplicit isomorphism
T:V4 W.
2!, V : .Do (diagonal 3 X 3 matrices), W : P:
72,V: _.(symmetric3 X 3matrices), lt{ * tlr(upper
triangular 3 X 3 matrices)
23. V : S, (symmetric 3 X 3 matrices), \r _ Sj (skewymmetric
3 Y,3 matrices)
24.V:(I2,W: {p{x)in#o:p(0) * 0}
a+bi 25, V : C, 1rť : PZ
26,V: {em Mrr: tr(Á) 0},W* pz
E 27. Show that T:{ ,,*ffndefined by r(p(x) ) : p{x} +
p'(x) is an isamorphism.
28. how thx T,,Wna #o defin*d by r(p(x) } * p(x * 2)
is an isomorphísm.
29. how that T: ffn a ffn defined by r(p(x)) - *o(*)
is an isomorphísm.
30. (n) howthat*6[0, 1] e'fi[2,3].lFlint:Defrne T:
ffi [0, 1] --+% [2, 3] by leffiing r(/) be the functian
whose value at x is {r(/)Xx) * /(x
* 2) for x in
[2, 3].]
(b) how that * [0, 1] # '6 [a, a * lJ for all a.
Section 6,6 The Matrix of a Linear Trarrsformation
Thn l}tnt ilt 0ít l,inffií Tran ítllttffti n
Theorem 6.15 showed that a linear transformation T : V ,+ W is completely deter_
mined by its effect on a spanning set for V. In parťcular, if we know how T acts on a
basis for V, then we can compute T(v) for any vector v in V. Example 6.55 illustrated
tlre process. We implicitly used this important property of linear transformations
in Theorem 3.31 to help us compute the standard matrix of a linear transformaťon
T; R'+ Rm. h this section, we will show that every linear transformation between
finite-dimensional vector spaces can be represented as a matrix transformation.
Suppose that V is an n-dimensional vector space, 1ť'is an m-dimensional vector
uPu.., *d T:V -+ ilris a linear transformation,LetB and C be bases for V andW,
rispectively, Then the coordinate vector mapping R(v) : [v]r defines an isomor_
phism R : i/+ Rn. At t}re same time, we have an isomorphism S : W+ R' given by
31*; = [w]g, which allows us to associate the image T(v) with the vector [T(v)]c in
R'. Figure 6.11 illustrates the relationships.
!l!!l0 0,1l
Since R is an isomorphism, it is invertible, o we may form the composite mapping
oToR*l:Rn_>R,
whic"h maps [v]6 to tT(v)]c. Since this mapping goes from R' to R', we know
from Chapter 3 that it is a matrix transformation. What, then, is the standard
matrix of S oT oR- 1?
We would like to find the m X n rnatrix Á such that Á [v]6 =
( o 7. n-1)([v]r). Or, sinée(S
"
|. :-''''.rj:'
= [7(v)]o we require
Á[v]r = [7(v)Jc
It turns out to be surprisingly easy to find. The basic idea is that of Theorem 3 , 3 1 . The
columns of Á are ttre image of the standard basis vector for Rn under S o T o R-1.
But, if 6 = {vr, . .,, vn} is abasio for V, then
R(v;) [v, ]r
+ ith entry
l-
L:]
;
á
Chapter 6 Vector paces
.2s
0 R*'{*,) : yi. Therefore, the ith column of the matrix Á we seek is given by
(s " T o .R-')(e,; - s(r(n-'(*,)))
Ť. (r (o,)):::
[ (v;) ]c
which is the coordinate vector of T(v;) with respect to the basis C ať W.
$r umm&rize this discussion a a theorem.
satisfies
ťor *lrery vector v in V.
The matrix Á in Theorem 6.26 is called the matri,x of T with respect to the bases B
and C, The relationship is illtstrated below. (Recall that T,a denotes multiplicaťon
byÁ.)
r(v)
+
T^
tv]r+Álvjr= 1r(v)Jc
nOmllls
o The matrix of a linear transformation Twith respect to bases B and C is someťmes
denoted by [7]a*6. Note the direction of the arrow: right-to-left (not left-toright,
as for T: V--+ Vl/), Withthisnotation, the final equation inTheorem 6,26 becomes
Observe thx the Bs in the subscriPs appeaí side by side and appeaí to 'tancel" each
other, In words, this equation says, "The matrix for Ttimes t}re coordinate vector for
v gives the coordinate vector for T(v)i'
In the special case where V: 1ť and B : C,we write IT]6 (instead of IT]r*r).
Theorem 6,26 then states that
o The matrix of a linear transformation with respect to given bases is unique.
That is, for every vector v in V, tlrere is only ore matrix Á with the property specified
by Theorem 6.26-namely,
á[v]r = 1r(v)ic
(You are asked to prove this in Exercise 39,)
Y --*+
I
Ý
Section 6.6 The }1atrix of a Linear 'Transťormation
o The diagram that follows Theorem 6,26 is sometimes called a commutathle
diagrambecause we can start in the upper left-hand corner with the vector v and get
to [T(v)]c in the lower right-hand corner in two different, but equivalent, ways. If, as
before, we denote the coordinate mappings that map v to [v] 6 and w to ffia by R and
,S, respectively, then we can summarize this "commutativity''by
of - fuoft
The reason for the term cornmutativebecomes clearer when V : Iť and 3 = (, fat
thenR = Stoo,andwehave
ftof:lioft
suggesting that the coordinate mapping R commutes with the linear trarrsformation Í
(provided we use the matrix version of T-namely, Td = Tínr*where it is required).
o The matrix lT\c*s depends on the ord,er of the vector in the bases B and
ť. Rearranging the vectors within either basis will ďfect the matrix [T]c_g. [See
Example 6,77(b).l
R3 --+ R? be the linear transformation defined by
t x-2y l:l l
Lx + y - 3zl
and Let B : d*,, *2, e3} and C ,- { 2, 1} be bases for R3 and R2, respe
matríx of r with rť pect to B and ť and veriťy Theorem 6-26 for v :
§ f, í First, \,ve compute
T(er} T(ez)
Next, we need their oordinate vector with re pect to C" Since
we have
,L;]
ctively. Find the
r 1l
L-;]
:
L-i],
T(eE) :
L_:]
ffiLi],
L_i], IT(e:)]c:
[-í][T(e,} ]c :
Li] [r(ez) 7c :
Therefore, the matrix ůfT nlith re pect to B and ť is
A: tr]c<_Lj: [[r(e,)]c IT(er)]1, IT(e,)]c]
|-t 1 -3lll
Lt -2 0.]
fihapt*r 6 Vector Spaces
[-sl
LroJ
To veriff Theorťm 6,26 for v, we first
r
I
]'(v) * Tl
l
L
compute
tl r
3l: I
*,) l '.LJ
r 1l
,| 3l
L-z]
Then [v]n :
*nd Ir{v) Jc *
,,il|i1,,,-.,,l.., (Check these.}
tising all af these facts, we confirm that
[;]-: [jl]
: t,:lL -5J
Á[v]n:
Ll -: ;] [ i-l
L- 2 J
Ir(v) Jc
Let D : 99 + 92be the differential operator D(p(x)) = p'(x),Let B = {L, x, x2, *3}
andC = |L, x, x2l be bases for 93 and gr, respeciively.
(a) Find the matrix Á of D wtth respect to B and, C.
(b) Find the matrixÁ' of D with respect to B' and C, where B'
(c) [Jsing part {a), compute D(5 * x + 2x3) and fr(a + bx *
Theorern 6,26.
.i,*ii,i;'iti': Firstnotethat Dta* bx+ cxz * dxt): b + lcx* 3dx2
A: tD] cny: {x}lc [D (x'}]c
LOJ Lot Lo]
Consequently,
* {*', x2, x, L}cxT
+ d*') to veri$r
. (See Example 6.60.)
(a) Since the images of the basis B under Dare D(1) = 0, D(r) = !, D(xz) = 2x, and
D(x3) = 3xZ, their coordinate vectors with re pect ta C are
Il]
D{"
0l
:]
r
ID
0l
0l
0J
0
0
3.
Y
[,
0
0
ú
=
(b) Since the basís B' is just
A' -: tD] ť*B, *
:T:
tt;:(1)]c i t
[0 1 0 |
|0 0 2 |
ff in the reve
ttn {x'}3c I
[o 0 1
lo 2 0
L, 0 0
[ {xu} JcJ
order, we ee that
(*')], [fi(x) ]c {n(r) jc]
Sectian 6.6 The Níatrix of a l.inear 'Iransformation
(This shows that the order aíthe vectors in the bases B and, C affects the matrix of a
transformation with respect to these bases.)
(c) First we compute D(5 - x + 2x3) : -1 + 6x2 directly,getting the coordinate
vector
[-,l
tD(s* x*2x3)](:: [-1 +6xlc=| oI
L o]
on the other hand,
t
[5 Jí+ 2x3 ]r: :
[D(5-x+zx'}]c
Since the linear transformation in Example 6.77 is easyto use directly, there is really
no advantage to using the matrix of this transformation to do calculations. However,
in other oramples-especially large ones-the matrix approach may be simpler,
as it is very well-suited to computer implementaťon, Example 6.78 illustrates the
basic idea behind this indirect approach.
Let T : W z*+ Sz be the linear transformation defined by
r{p(x}} *p{Zx* 1)
(a) Find ttre matrix of T $rith ťe pect tn ď * ÍL, *, #}.
(b) Compute T(3 * ?x * x2; indiťectly, u inspart {a}.
;,,tiil;;.ri"i;'i {a) We See that
r(1)*1, T{x)*2x* l, T{x'}xtLx* l}':1 4x*4x2
so the coordinate yector are
Therefore,
t7] : ttr(1)]r,, IT(x)],, lT{x')]il *
[0
|0
L0
5l
*1 I
0l
2J
i;l]L;]:Li]:
ž6.}1re leave proof of the genťral ťasc & an exerci e.which asťec with Theorem 6.
tr(l)]s:
[i],
I T{x)Js :
L i] [T(x,)]g:
I i]
[t -1 1l
lo 2 *4l
Lo 0 4)
Chapter 6 Yector Spaces
(b) We apply Theorem 6,26as followsl The coordinate vector ofp(x) : 3 + 2x - x2
with respect to á is
[p (x) ]s
Therefore, by Theorem 6-26,
Ir(3 + 2x * lc2) ]s
:Li]
lr@@))]g
[r -1 ,l[- 3l t 0l
L; 3
-ijl_i] :Lj]
M Itťollowsthat r$ + 2x* x?} ffi 0,1+ B,Jí* 4-x2 * Bx * 4x2
computing7(3 *2x* x2) * 3 +2{2x- 1) * {2x_ t)2dire*tly.]
[VeriSr this by
The matrix of a linear transformation can sometimes be used in surprising ways.
Example 6.79 shcws its application to a tradiťonal calculus problem.
Let 9 be the vcctor space of all differentiable functions, Consider the subspace W'of
9 given by W - span(e3', xe3*, x2e3*7, Since the set B = {eu', xe3r, x2e3'} islinearly
independcnt (why?), it is a basis for Tť.
(a) Show that the differential operator "D maps 1ť into itself,
(b) Find the matrix of D with respectta B,
(c) Compute the derivative of 5e3' + 2xe3* - x2e3* indirectly, using Theorem 6.26,
and veriSz it using part (a).
]
60lutlon (a) Applying D to a general element of W, we see that
' D(Ae3' + bxe3' + cx2e31 : (3a * b)e3' + (3b + zc)xe3' + 3cx2e3*
ffiŇ*6" (check this), which is again in I,ť,
(b) tJsing the ťormula in pert (a), we see that
ID{e
3")
]n :
It follow that
ID(xe
3-}1s
tD] B : [[D(e="} ]ru |D{xe3,)}s lD(xrur,}Ju)
nte3*} * 3e3*, D{xe3*) * e3* + 3xe3*, D(xžr3x) x 2xe3* + 3xže3*
Ll]
It]
L0]Ii]
:t;i;]
Lo 0 ,.
Ip(x
ž
e3*)J n
Idy
le
Section 6.6 The ]vlatrix of a Linear Transformation §§§
(c) For/(x) : 5e3' + 2xe3' - x2e3',we see by inspection that
l- sl
rr,)]u: | ,|
L-,_]
Hence, by Theorem 6.26, wehave
. 'D(f(,ť}}
]g : tn ]r [/(x) }s :
which, ínturn, implies that í'(x), n{f(x))
with the formula in part (a).
t"l:| 4l
L
*3l
3x2 e'*, in esreem-";
*+
t; 1 :l,
5l
Lo 0 ,]L i]
ŤŤ*
W e3* * 4xe3r *
nGmail The point of Example 6.79 is not that this method is easier than direct
differentiation. Indeed, once the formula in part (a) has been estabkshed, there is
little to do, What is significant is that matrix methods can be used at all in what
appears, on the surface, to be a calculus problem, We will explore this idea further in
Example 6.83.
Let V be an n-dimensional vector space and let 1 be the identity transformaťon on
V. What is the matrix of / with re pect to bases B and, C of V tíB = C] (including the
order of the basis vectors)? What if B + C?
Solutbn LetB = {or,...,vn},Then.í(v1) =v1, ,..,I(vn) -v,r, o
[,l
[J{v,)]n :
|
?
l
: i, [í(vz)]r :
L0]
and, iťffi:C,
Il]n : [[í(v,) ]r
: [e, r
T,*ll
t:lltl
Iil* 2, ...) [í(v,,)lr--
Lo]
[.ť(vr) ]n , , I I(v,,) Jr ]
' , n]
Iol
|0l
|:l
Li]
:,',
tlre rr X n identity matrix. (This is what you expected, isďt it?)
In the case B * C, we have
[J]c--r :
|l", 1
c,",, [v-]cj
l C<--B
the change-of-basis matrix from E to C.
t0l1lc8s oíGomnosllc mil lRuerse llno[] Iían$íOmlfiOns
We now generalize Theorems 3.32 and3.33 to get a theorem that will allow us to
easity find the inverse of a linear transformation between finite-dimensional vector
spaces (ifit exists).
Chapter 6 Vector Spaces
itrti|}írlB*]li
u, In words, this theorem says, "The matrix of the composite is the product of
the matrices]'
* Noťce how the "inner subscripts" C must match and appear to cancel each
other out, leaving the'buter subscripts" in the form D <- B.
ilťĚ { We will show that corresponding columns of the matrices [S o T]e*3 and
[ ]z,_c [7]c*6 are the same, Let v; be the ith basis vector in 6. Then the lth column
of [S. T]e*1is
[(s"Txvl)]p: [(T(vJ]e
- rc] ,[T(v)]c
_ ::]"-,
- 1o 1D*g|Tlc*sÍVi]s
by two applications of Theorem 6.26.But [vi]r : e, (why?), so
IS]2*c I T]c* slV,] s : [S]2*6 [ T] g*6e;
is the ith column of the matrix [S]p-c[7]c*6. Therefore, the ith columns of
[S " T]e*r and [S]o-c[T]c*rare the same, as we wished to prove. , , ,. . l,:.l*&
S, and , : #1 -+ ff2 *re defined by
+ b}x end (a + bx]l * ňK + bxI
á', and 8" fwRz, #tl áfld. S2, f pectively, we see thatChoosing
Recal1 that T; R2 --+
la1
'L;]
: a + (a
the standard bases ď,
lT] e,*ť :
Il ?] and [ ]r,*r, *
[I l]
:rix
8" <_<_
0
0
1
tatr
S]g
0
1
0
.;l;i=,,.,.""" (Verify these.) By Theorem 6.27, the me
[(S " I) ]g,,* : [,S
Ic
l
: Il
L,
of S o 7 wit}r respect to á and ť'ís
,lT]
"t*
lr, 0l :t? 3l
]L,
lJ Li ;]
Use matrix methods to compute r' " n[;] for the linear transformations and Tof
Example 6.56.
Thus, by Theorem 6,26,
Section 6.6 The VIatrix of a Linear Transformation
: [(s "
(,
0l l
al
il + b)
T)]r,,* r|;
L,,.nL;]],,
:L]l]l;l:
Consbquently,
xample 6.56.
ax * {a + b}x?, which esr es with the solution to
ln Theorem 6.24,we proved that a linear transformation is invertible if and only
if it is one-to-one and onto (i.e,, ťit is an isomorphism). When the vector spaces involved
are finite-dimensional, we can use the matrix methods we have developed to
find the inverse of such a linear transformation.
(s" ,|;):m
6.28
ť{* Observethatthematricesof Tandr-l 1if Tisinvertible)aren X rr. If Iis
inyertible, then 7*
1
" T : Iv. Applying The orem 6.27 , we have
I': Uv]s:
[;] i,|,',Irr,*,
This shows that|T]gn1is invertible and that (íTlc* -' : 7T"'l}g*c.
Conversely, assume that Á : l,Tlc*ris invertible. To show that Tis invertible, it
]#*"ť"" is enough to show that ker(T) : {0}. (Why?) To this end, let v be in the kernel of T,
Then 7(v) : 0, so
Á[v]g = |T]g*g|v]r = [T(v)]c = [(}ja = Q
which means that [v] 6 is in the null space of the invertible matrix Á. By the Fundamental
Theorem, this implies that [v]6 : 0, which, in turn, implies that v = 0, as required.
In xampte 6,70, the linear transformation r i re? *+ ff1 defined by
+ (a + b)x
wa shown to be one-to-one and onto and hence ínvertíble,Fínd T--l
la\
'lr] - a
Chapter 6 Vector paces
S*!H,fi i'{ In Example 6.8l, we found the matrix of 7 with respect to the standard
bases á and ' íor W2 and g1, respectively, to be
[l 0]
tr],*,: Li l]
By Theorem 6.28, itfollows that the matrix of T- t
with respect to á' and á is
[T-,]e*e, : ([T]í,*,) , :
[i :] :
t-l :]
By Theorďn 6.26,
IT*l(a + + bx] t,
This mean that
T-|(a + bx}: a&t+ {a a)ez : [,
a
l
Lb aJ
(Note that the choice of the standard basis makes this last catrculation vi
irrelevant.)
The next example, a conťnuaťon of Example 6,79, shows t}rat matrices can be
used in certain integration problems in calculus. The specific integral we consider is
usually evaluated in a calculus cour e by means of two applications of integration by
parts. Contrast this approach with our method.
Show that the dif[erential operator, restricted to the subspace W : span(e3', xe3*,
x2e3*) oíg,is invertible, and use this fact to find the integrď
l *'r" d*
J
,i;+íeíll*it In Example 6.79, we found the matrix of D with respect to the basis
B: {e3',xe3',x2e3'} of 1ťtobe
tn]ff :
By Theorcm 6.28, thereťore, l) is inve
[D-']r: (tt:]d*l *
bx) ] * [r--r7r*r,|o
:L-ll]L;]
r a l:l, -)
[: 1 0l
Io 3 2l
Lo CI 3]
le on 1ť, and the rnatrix of D* i is
1 0l ,
[i -i -1 l
3 2l :|; ; :i|
0 3] Lo 0 {J
rtually
A
."+
rtib
Il
|0
L0
Since integrationis antidffirentiation, this is the matrix corresponding to integration
on I4l. We want to integrate the function x2e" whose coordinate vector is
|-ol
í*'r'*fo: | . I
L,]
Consequently, by Theorem 6.26,
I l,
2r3x
d-1 _ tD- '(*re3*)]s
LJ
'u_ t,-,]n[ x,e,*fn
t i -.i *l I-0l í-fri
Lo 0
Sectiorr 6.6 The }Iatrix of a Linear Tiansťormation
It fo110ws that
Í*'uudx
*
*e" - **n'* + }*
Zr3x
(To be fully correct, we need to add a constant of integration. It does not show up here
because we are workingwithlineartransformations, which must send zero vector to
uero vectors, forcing the constant of integration to be z"čro*s we11.)
ltt]nlng In general, differentiation is not an invertible transformation. (See
Exercise 22,) ÍNhatthe preceding example shows is that, suitably restricted, it sometimes
is. Exercises 27-30 explore this idea further.
chango oíBatis and similarity
Suppose T : V --+ V is a linear- transformation and B and C are two different bases for
7. It is natural to wonder how, if at all, the matrices [7]r and |T}gaíe related. It turns
out that the answer to this question is quite satis$ling and relates to some quesťons
we first considered in Chapter 4.
Figure 6,12 suggests one wayto address this problem. Chasing the arrows around
the diagram from the upper left-hand corner to the lower right-hand corner in two
different, but equivalent, ways shows that r o T = T " I, something we already knew,
since both are equal to T. However, if the ",rppeť' version of T is with respect to the
l
l"
} basts {-
)
} nuri, r
)
Chapter 6 Yector Spaccs
basis C and the "lol,yeť' version is with re peď to,6, ťhen T = I " T = T o / is with
respect to C in its domain and with respect to 6 in its codomain. Thus, the matrix of
f in this case is IT]6*a. But
íT]s*c= |I " T|g*g = [I|s*cIT]c*c
and
lTln*c = [To l1s*c* lTls*níIls*c
Therefore, lI1s*clT]c*c - lT7s*all]a*c.
From ExampL 6.8b, we know that íI]Á*c = P1*g, the (invertible) change-ofbasis
matrix from C to B.If we denote this matrix by P, then we also have
P Ir] c**c
: |T} 7-_BP
o t T}cnc r p*' I r] §*_Bp oť [ T]c : p*l t ť]#p
Thus, the matrices LT) nend lrl cffte iínilar, in the terminalagy oť ection 4,4.
$re ummarize the foregaing di*cu síOn as a theorem"
witlr this nntation,
TheO]Gm S.2S Let Vbe afinite-dimensionalvector spacewith bases B andC and let T:V -->V
be a linear transformation. Then
ÍT]c: P-'lTllp
where P is the change-of;basis matrix from Cto B,
nmill As an aid in remembering that P must be the change-of-basis matrix
frorn C to B, and not B w C, it is instrucťve to look at what Theorem 6.29 says when
written in full detail. As shovrn below, the 'inner subscriPs" must be the same (all Bs)
and must appeaí to cancel, leaving the 'buter subscriPsi' which are both Cs.
íT]c* c = Pg *7fTlg*uPs *,
t t___t L} Ť
I Same Same
I
ame
Theorem 6,29 is often used when we are trying to find a basis with respeď to
which the'matrix of a linear transformation is particularly simple. For example, we
can ask whether there is a basis C of V such that the matrix |T] p of T : V -+ V is a
diagonal matrix. Example 6,84 illustrates this application.
Let T: R2 -> R2 be defined by
lxl lx+3yl
'LÁ= |r, * ry1
If possible, find a basis C for R2 such that the matrix of Twith respect to C is diagonal.
§§$§*§§řÝ}iř The matrlx of 7 with respect to the standard basis á is
l7l" = [I 3l
L2 z)
This matrix is diagonalizable, as we saw in Example 4.24.Indeed, if
.: [' :'l and ,:|n o]
Ll -2) L0 -1]
then P- 1
IT1 tP : D, If we |et C be the basis of R2 consisting of the columns oí P, then
P is the change-of-basis matrix P 6*9 from C b e, By Theorem 6,29,
IT]c * p*l [T]gp x p
s* the matrix ať Twith respect to the basis C {
Ll], [
_;] } '-
diagonal,
Section 6.6 The N,latrix of a Linear 'Transformation
n8n0íl$
r It is easy to check that the solution above is correct by computing [7 ] 3 directly,
\,Ve find that
,Ll] Li]
- -Ll] + 0[-;]
Thus, the coordinate vectors that form the columns of [T]g are
rill]]. =
[;]
and
t,[-;]].
=
[-:]L L,
in agreement wit}r our solution above.
o The general procedure for a problem like Example 6.84 is to take the standard
matrtx íT)eand determine whether it is diagonalizable by finding bases for its
eigenspaces, as in Chapter 4. The solution then proceeds exactly as in the preceding
example,
Example 6,84 motivates the following definition.
iinéri tránsťc
It is not hard to show that if B is any basis for % then T is diagonalizable if and
only ť the matrix [ 7 ] r is diagonalízable. This is essentially what we did, for a special
case, in the last example. You are asked to prove this result in general in Exercise 42.
sometimes it is easiest to write down the matrix of a linear transformation with
re pect to a "non tandarď' basis. We can then reverse the process of Example 6.84
to find the standard matrix. We illustrate this idea by revisiting Example 3.59,
Let{bethe1inethroughtheorigininR2withdirectiOnYectord
standard matrix of the projection onto ť.
|rr'r),
Find the
Chapter 6 Yector paces
SOlnlOn Let T denote the projection. There is no harm in assuming that d is
a unit veďor (i,e,, d! + d1 = 1),.since..,any nonzero multiple of d can serve as a
direction vector for t. Let d' : |-i' I so that d and d' are orthogonal. Since
L4)
d' is also a unit vector, the set D : {ďd'} is an orthonormal basis for R2.
As Figure 6.13 shows, ?(d) = d and T(d') = 0. Therefore,
Ir(d) 1n * and I r(d') ]rr :
íl9u18 6.|t
Projection onto ť
[t 0l
lT)p:
L; ;j
The change-of-basis matrix from D ta the standard basis á is
l d, *d,1
Pť*P: l
"
'"Iu
Ldz dr)
so the change-of-basis matrix fram 8 to D is
l d, - drf-, r dl drT
Pp*# q tP*r}*' : l ""' ""'| T:, | "
La, dr) |,- a, d,)
By Theorem 6,2g,then, the standard matrix of T is
IT]E : P* p[ 7]pFr*_g
| ,l, -,l2-1 [ t
l ll
lll
Ld, d, ] L0
I di d,dr\
:
La,a, di ]
which asrees with part (b) of Example 3.59.
0l l- d,, d"1
0JL_;, ;,)
Sectiarr 6,6 The Nlatrix of a Linear Transformation
Let T l .P 2 --+ 9 2be the linear transformation defined by
r(p(l.D:p(zx-I)
(a) Find the matrix of 7 with respect to the basis B : {L * x, 1 - x, x2) of P2.
(b) Show that 7 is diagonalizable and find a basis C for 92 such that [7]a is a diagonal
matrix.
oliltlnfi {a) In Example 6-78, we
standard basis : {1,
",
x2} is
found that the matrix of T with re pect to the
[7]g o
The change-af-basis matrix from B to
[t 1
I
p.-pt*B:
it -1
Lo 0
It ťcllnw that the matrix of T with respcct to ff is
IT] n: P--lIT]gP
rt llz 2
l; }
l) ?,
Lo 0
r10l
L00
and [r]c - D.
[t -1
|o 2
L0 0
Sis
1l
*4l
4]
0l
?]
0l
0l
1j
olIt 1
-ilti
l]L; l +]Lo
_3.1rl
7l
4l
1
*1
0
HHFňF
XT-M
(Checkthis.)
(b) The eigenvalues of I T]6 are I, 2, and 4 (why?), o we know from Theorem 4.25
that I T] g is diagonalizable. Eigenvectors corresponding to these eigenvalues are
ti]ii]ti]]
0 0l
2 0l
0 4]
f-basis matrix from a basis
ctars of c in terms of á, It
vely,
:P-
and
that
an
:ectil
have
S rt
or,vs 1
p
el
to
we
ťtt
fo11
r*
W
C
fo
Chapter S \Iector Spnces
The preceding ideas can be generďized to relate the matrices [T|g*6 and
lTlc,*B of a linear transformation T: V-+ 1{', where BandB'are bases for V andC
andC' are bases for lV. (See Exercise 44.)
We conclude this section by revisiting the Fundamental Theorem of lrrvertible
Matrices and incorporating some results from this chapter.
,$m .80 The Fundamental Theorem of
(}
fr,
h.
i.
j.
k.
l,
11. detÁ * 0
CI. 0 is nCIt an isenvalue of Á.
T is invertible.
T is one-to-one.
T is onto.
, ker(T) * {0}
t, range(T) : W
p"
q.
r.
P1O0í The equivalence (q) <+ (s) is Theorem 6.20, and (r) <+ (t) is the definition of
onto, Since Á is n X n, we must have dim V = dim W = n,From Theorems 6.21 and
6.24,weget{p) + (q) <+ (r),Finally,weconnectthelastfivestatementstotheot]rers
by Theorťm 6.28, which implies that (a) ** (p).
In xercťses 1*12, rtnd the matrix |Tlc*n of the linear
transIarrnationT i V4W withr sp c,t tothebases B andC
ůíV frnd YV, respectively, Verify Theurem 6,26 ío,the vector v
by computing T{v) dirertly enď using the theCIrg?n,
l- r : #r *+ ff, defineď by T(a + bxJ ff b ilx,
ff*(*{1,*}, 1 p(x)-4*2x
2. T:{íl--+? 1 defined by T(a + bx} * b &x,
B: lL * x,L x},C * {t,*},y * p(x) 4 + 2x
3. T:'#z+.Wzdefined by r( Fb)) * p(x + 2},
B - {L, *, x'}, C - {1, x + 2, (x + 2)'),
v ::,
Pk} :*- a + bx + cxT
4, T :Wz-+? zdefined by T(p(*)} * p(x + Z),
B , {1, x + ?, (x + 2)'},C . {1, *, x'l|,
y :": p{ň ,*!?
il + bx * CxL
5, T : ?f z--+ [Q2 defined by T(p(x) ) * ['1o]l,
Lp (t) J'
[p (0)
l
Lp (1) J'
t
[-i] },
| *t1
v - L ,)
["l
La]
ffirc{*r,.tr, 1},ť*
{[l],
y * F{x)*o + bx* cx?
7, T: R2 *+ ffi3 ďefined by
F a fa + 2b1
,L;] :L -;
],fr-
(,{[i][l]Ll]}
8. Repeat xercise 7 with v #
ffi * {1,*, *'},(T"{*,,er},
v:pk)*il,+bx*cxL
6, T , ,$, --* R2 defined by r( F{x})
*
Section 6.6 The \Iatrlx of a Linear Transformation
* _14. Consider the subspace W of T, given by
!A/ * pan (r", e-").
(a) Show that the difrerential operator D map Tť into
itself.
(b) Find the matrix of D with respect to : {u'*, r*'*),
(c) Compute the derivative of í(x) _ e2' * 3e*2*
il*::,*lJ,ff :*ffirT"í-Ť;iliJ,::,o
that it
|* 15. Considerthe sub pace W of$, givenby W - span{r'*,
"*--
e2, Cfis X, e2" sin x).
(a) Find the matrix of D with respect ta B * {r'*,
ť2' o X, 2'
sin x}.
(b)
ffi;T ffiffi:ffi"íHi;j* ;;,-#x
t
verify that it agrees with/'(x) a computeď directlyt
ffi 16. Consider the subspace W of$, given by
-t#
|ff = span(c s Js, sin x, J co X, x sin x),
(a} Find the matrix of D with respect ta ffi * {cos x,
sin x, Jg o x, x sin x}.
(b) Compute the dcrivative of/(x) * cO .x * 2x co J[
indírectly, u*ing Theorem 6.26, and verí$l that it
esrees with/'{x) as omputed directly.
In Exercises 17 anď 18, T : (J + V arud, ; 1l-.+ W are lingar
transfarmatians and B, C, and D are bases for U, V and
ÝV, respeetive[y" Cornpute [ - T]n*n in twa wfl,ys: (a) by
fi.nding, o directly and then omputing its matrix und
ft) by rtnding the matricas o/ and T separately and using
Thgorem 6,27,
17 - T: #i *+ ffiz defined W r{p(x)} s K?+ffiŽ
In xarcíses 19*26, dutermine whether the linear transíormatian
T is invertible by cansidering its mutrix with resPect
ta the standard bases, If T ťs invertible, use Thenrem 6,28
and, the methad af xample 6.S2 fu rt,nď T*I ,
l9. r in Exercise 1
21- r in Exercise 3
20. T in xercise 5
2?", T:Wz+Pzdefined by r{p(x)) : p'(x)
Il]},
{Ll]
g, T: Mzz+ Mzzdefined by r(Á) * Ar, B : C :
{E,, , En, Er,,Err}, v : A*
l: ';
t0. Repeat xercise 9 with ffi , ÍErr, Err, E,,z,E,,} and
C - {E ,r, E rr, E rr, Ě, r}.
I 1. r : Mzz a Mzzďefrrred by r(Á) : ÁB * BA, where
r r _rl
§-: l ] '. l,s:C:{E,,,Erz,Err,Err},
L*l 1],
v- A_ l' 1l
L, d)
\2. T: Mzza Mzzdefined by r(Á) - A * AT', ffi :
{ ,{č,,,En,Err,
Err},v * A*- l'
b]
Lc d)
[íll]],
* ,
definedby
[;]
x
|;;:?r),r*
{1,x},
(=D.-{*,,*r}
1S. r: #r -+ #a ďefined by r(p(x)) * p(x * 1),
, l ff2 -+ Sl defined by, (p(x)) * p(x + 1),
ft - {1,*},f, y f) m {1, *, x*}
lil 13, Consider tlre sub pace Wof S, given by
W * pan(sin x, co J ).
(a) Show that the difrerential operator I) maps 1ťinto
itself.
(b) Find the matrix of .D r,vith respect to
B - {sin x, cos x}.
(c) Compute the derivative ofl{x) : 3 sin x - 5 cos r
indirectly, using Theorem 6,26, and verii, that it
agree with í'(*) as computed directly.
lfl23. T :y}l1?P2 defined by r(p(x)) : p(x) + p'(x)
Chapt*r 6 \iector paces
24,TiMzzaMaz AB, rt here
25, T in xercise 1 1 2§, T in xercíse 1?
defined by T(Á) *
|-l 21
B--
L, l.J
3l. T: R2 -+ R2 defined uu r['l _ t
*4b
l
'|,b) La+5b)
32,. T: R2 *} R2 defined b}, ,[il * l- , ?1'LbJ |a+b)
ffi;ru Exercises 27*3a, use the method af Exampíe 6.B3
to evaluate tlte given inte ral.
27. t (sin x - 3 cos x} dx. (See Exercise 13.) t
28. Í5e-2* dx. (See Exercise 14.)
29, í(r'! c rí* 2e2* sin x) dx. (See Exercise 15,)
30. Í{x cos x * r sin x} dx. (See Exercise 16.)
In Exercises3J *36, a linear transformation T: V *t V is
given, If passiblle, find a basis C fár V such that títematrix
IT]c aíT with respett to čis diagon&l-
33. T: Sr *+ #r defined by T(a + bx) * {4a * 2b} +
(a + 3b)x
34. T:Wz+Wzdefined by r{p(x)) : p{x + 1)
M;s" T:W1 *-* #1 defined by rtp{x)) : p(x) * xp'{x}
36. T: #z *Wzdefined by rtp(x)) : (3x + 2)
37" Let ťbe the line through the origin in ffi2 with direction
vectorfl .- [d'l TTc,*
Lrr
j, lJ * the method of xample 6"85 to
find the standard matrix of a reťlecticn in ť.
38. Let lťbe the plane ínR3 with equation Jc * y *
7z, : ů- [Jse the rnethod of xample 6.85 to find
the standarď matrix oťan orthogonal projection
onto Iď. \furi r thet your &n wer is correct by using
it to ompute the orthogonal projection of v onto tť,
where
Compare your ans\trer with ,xample 5.11.
|Hint: Find an orthoganal decomposition oť R3 as
R3 : 1,'1l + 147t using an orthogonal basis ťbr Ýť. See
Example 5.3.]
39. Let T : V *+ Iťbe a linear transformation between
finite-dimensional vector paces and Let Band Cbe
bases for and lť,respectively. how that the matrix
of T with respect ta B and C is unique, That is, iťÁ is a
matrix such that Á [v] : I r(v) ], fo, all v in Y, then
d x IT]c*n. |Hint; Find values of v that will show
this, one calumn at a time.]
In Exercises 4a*45, íetT: 1/ -+ W be a linear transform *
tian beťween finite-dimensional vector sp c s V anid W
Let B and C be bases for Y and ÝV, respectively, and l*t
A ITJc*n.
40. Show that nullity(r) : nullity(Á).
41. Show that rank(T) : rank(Á).
42.Iť V ,* Iť and ff -. C, show that is di*ganalizable iť
and only if Á is diagonalizable.
43. t]se the results of this section to give a matrixbased
proof oťthe Rank Theorem (Theorem 6.19)"
44. If B' and ť'are also bases for V and trfi respectively,
what is the relationship between Ir] c*ffianď |Tjg,*g,?
Prove your assertion"
45, If dim P: ru and dim W -- m,prwethat tr{U Tť) *
Mrnn.,(See the exercises for Secti*n 6.4.} {Hirct: Let B
and ťbe bases ťor F and 1ď, respectively. how that the
maPpins p(r) * [ ]c*n, for Tin g{V, W}, defines
a linear transťorrnation v 1 ff(\r, W) 4 M**that is an
isomorphism.]
46. Iť F is a vector pace, then the ďual space of is
the vector space Y* * ff(v, re). Ptrove that if ťis
finite-dimensiona"1, then V* x V.
[;]
[xpl ratl0
Tilings, Lattices, and the
Crystallographic Restriction
Repeating patterns are frequently found in nature and in art. The molecular structuie
or cryrtats often exJrŇs repetition, as do the tilings and mosaics found in the
artwork of marry cultures. Tiling (or tessellation) is covering of a plane by shapes that
do not overlap and leave no gaps. The Dutch artist M. C. Escher (1898-1972) produced
many works in which he explored the possibility of tiling a plane using fancifrrl
shapes (Figure 6.14).
v)
a
šť
"c
-C.
cJ
z.
0J
-ť!-"
l
>-
Ě
C}_
ůLJ
-.
L)
U)
til
L)
z,
c..)
_c
F*
arj
(3
s
[-r-1
=
O
e L/)
.
š
q)?
F
Y, ()
;i 6)
uc)
uj
={
H ll c S.l4
T
ťí)
aĎ
C')
Ě
;.
š
;0")
U
q}
a
-c
).
u)
-l3
C
0.)
-
u]
Z
gJ
-fiL
'i
s-
C
a
t{}
(J
-c(J
u1
LlJ
,i
o)
-fr
l-*
O
ť!
A\\Y/
=:i.1_1 ;
,Ě1
F:
lfí
L
o):
:
.i
l
)
uJi
(-) j
;
,ď'"]
-,"',-w,
$ xwr* ffiInvariance
under traRslation
M- C. scher's "Symrn*try ilrawing 103"
,.ůů"L
u
aaaa
ílgute 6.16
A lattice
$ ffi#ffiKnťational
symmetry
M. C, Escher's " ymmetry f}rawing ,103"
ln this exploraťon, we will be interested in patterns such as those in Figure 6.14,
which we as ume to be infinite and repeating in all directions of the plane. Such a
pattern has the property that it can be shifted (or translated) in at least two directions
(corresponding to two linearly independent vectors) so that it appears not to have
been moved at all. We say that the pattern is invariant under translations and has
translatlonal yftrfiretfy in these direcťons, For example, the pattern in Figure 6.14
has translational symmetry in the <iirections shown in Figure 6.15,
If a pattern has translational symmetry in two directions, it has translational symmetry
in infinitel/ many direďiolts.
1. Let the two veďors shown in Figure 6, 15 be denoted by u and v. Show that the
Pattern in Figure ó.14 is invariant under translati onby any iniegerlínear combinaťon
of u and v-that is, by any vector of the form au * v, where a and, b are integers.
For any tr,rro linearly independent vectof u and v in R2, the set of poiilt determined
by all integer linear combinations of u and v is called a httlce. Figure 6.16
shows an example of a lattice.
2, Draw the lattice corresponding to the vectors u and v of Figure 6.15,
Figure 6.14 also exhibits rotational yffiffietTy, That is, it is possible to rotate
the entire pattern about some point and have it appeaí unchanged. We say that it is
invariant under such a rotation. For example, the pattern of Figure 6;14 is irrvariant
underarotaťonof 120oaboutthepointO,asshowninFigure6.17.We callOacenter
of rotational symmeťy (or arotation center),
Note that if a pattern is based on an underlying lattice, then any sym,metries of the
pattern must also be possessed by the lattice,
aaaaaa
3, Explain why, if a po
then36at/0 mustbeaninteger.'(lií36,0/0 = ťt,we sa thepatternorlatticehasn-fulil
rotational symm*ry.)
4. What is t}re smďlest positive angle of rotationa] symmetry for the lattice in
Problem 2? Does the pattern in Figure 6.14 also have rotational ymmetry through
this angle?
5. Take various values of0 suc.h that 0 < 0 < 360o and 360/0 is an integer. Try
to draw a lattice that has rotational symmetry through the angle 0. In parťcular, can
youdraw alattice with eight-fold rotational symmetry?
We will show that values oť0 that are possible angles of rotaťonal qymmetry for a
lattice are severely restricted. The tec.;hnique we will use is to consider rotation transformations
in terms of differerrt bases. Accordingly, let R6 denote a rotation about the
origin through an angle 0 and let á be the standard basis for M. Then the standard
matrix of R9 is
[fig Jr
6. Referring to Problems 2 and 4, take the origin to be at the tails of u and v.
(a) What is the actual (i.e., numerical) value of [RgJg in this case?
(b) Lď 6be the basis {u, v}. Compute the matrix [R6]6.
7. In general, let u and v be any two linearly independent veďor in R2 and
uppo e that the lattice determined by u and v is invariant under a rotation through
an angle 0.IíB * {rr, v}, show that the matrix of RB with respeď to 6 must have the
form
íR,1"-l: t
where a, b, c, and d are integers.
8. In the terminology and notation of Problem 7, Jhow tírat 2 cos 0 must be an
integer. [Hint:IJse Exercise 35 in Section 4,4andTheorem 6.29.]
9. Using Problem 8, make a list of all possible values of d, with 0 < 0 š 360o,
that can be angles of rotational symmeťy of a lattice. Record the corresponding values
of rr, where n = 36a/0, to show that a lattice can have rr-fold rotational sl,rnmetry
ifandonlyif n = I,2,3,4,ot6,T}risresult,knaumasthecrystalhgraphicrcstriction,
was first proved by W. Barlow in 1894.
10, In the library or on the Internet, see whether you can find an Escher tiling
for each of the five possĎle types of rotational symmetry-that is, where the smallest
angle of rotational symmetry of the patteín is one of those specified by the crystallographic
restriction,
Icosď -sin0l:-ll
|sin0 co 6J
rotation c
:
Chapter 6 Vector Spaces
llomo0encous llnmr DlííGíGnti0lluuatlons
In Exercises 69-72 inSection 4.6,we showed that ify = y(t) isa twice-differentiable
function that satisfies the differential equation
y"+fry'+by:0 (1)
then7 is of.the form
N:cÝ^ *cre^^t
if,\1 and )r2 are distincí roots of the associated ďraracteristic equation A2 + a^ +
b : 0. (The case where ,\r = iz was left unresolved,) Example 6.12 and Exercise 20
in this section show that the set of solutioňs to Eguation (1) forms a subspace of S,
the vector space of functions. In tlris seďion, we pursue these ideas furthen paylng
particular attenťon to the role played by vector spaces, bases, and dimension.
To set the stage, w consider a simpler class of examples. A differential equation
of the form
Y'+ay:0
is called a first-order, homogeneous, linear dffirential equation. ("First-ordeť'
refers to the fact that the highest derivative that is involved is a first derivative, and
"homogeneous" means that the right-hand side is zero. Do you see why the equation
is "linear"?) A solution to Equation (2) is a differentiable function 7 = 7(f) that
satisfies Equaťon (2) for all values of f.
It is easy to check that one solution to Equation (2) is y : e*ot, (Do it.) However,
we would like to describe all solutions-and this is where vector spaces come in. We
have the following theorem.
(2)
ilĚ }*íSince the zero function certainly satisfies Equation (2), S is nonempty, Let x
and 2 be two differentiable functions of f that are in S and let c be a scalar. Then
x'*ax:O and y'*ay:g
so, using rules for differentiation, we have
(x + y)',l a(x * !) : y,' * y' * ax * ay : (x' * ax) * (1t' + ay) :0 * 0 : 0
and
(cy)' + a(cy): cy' * c(ay): r(y' * oy): c,0:0
Hence, x * yandc2areďsoinS, soSisasubspaceof?F. .. ,,...,.,.],*ffi
Now we will show that S is a one-dimensional subspace of ?F and that ie-ot} is a
basis, To this end, let x : r( ) be in S. Then, for all t,
x'("t) + x (t} :-: 0 or x' {t)
--; * ex tt}
Sectian 6,7 Ápplications
Define anewfunction z(t) : x(t)eot. Then, bythe Chain Rule for differentiation,
z'(t) : x(t)aen' + x'(t)eot
: ax{t)e"' - ax(t)eot
*0
Since z' is identically zero, e must be a constant function-salr, z(t) : 1r, But this
means that
CI fi (ť) x ke*
ot
, Thereťore,
single solution.y : e*ot, UIe
x(t}eet * z,{t} : k far a1l r
al1 solutions to Equation (2} arescalar multiples of the
have pťOved the ťollowing theorem.
ThOOtem 6.32 If S is the solution space ofy' * ay :0, then dim S : 1 and {e-oí} is a basis for S.
One model for population growth á sumes that the grou,th rate of the population
is proportional to the size of the populaťon. This model works well if there are
few restrictions (such as limited space, food, or the like) on growth. If the size of the
population at time t is p(t), then the growth rate, or rate of change of the population,
is its derivative p'(t). Our assumption that the growth rate of the population is proportional
to its size carr be written as
p'(t) = kp(t)
where k is the proportionality constant. Thus, p satisfies the differential equation
P' - kP = 0, so, byTheorem 6.32,
P(t) : cek'
for some scalar c. The constants c and ft are determined using experimental data.
E, cali is menticlned in Michael
Crichtoďs novel T|te Andromeda
Strain (New York Dell, 1969),
although tlre "villaiď' irr tlrat novel
yras suppo edly an alien virus. In
real life, .E. coli contaminated the
town $rater upply of VValkertCIn,
Ontario, in 2000, resulting in seven
deaths and ca ing hundrecls of
people to become sťriously ill.
The bacterium Escherichia coli (ot E. coli, for short) is commonly found in the
intesťnes of humans and otlrer mammals. It poses severe health riit<s ir it escapes
into the environment. Under laboratory conditions, each cell of the bacterium divides
into two every 20 minutes. If we start with a single E. coli cell,how many will there
be after l day?
$Ohtl0n \Me do not need to use dif[erenťal equations to solve this problem, but we
will, in order to illustrate the basic method.
To determine c and k, we use the data given in the statement of the problem. If
we take 1 unit of time to be 20 minutes, then we are given that p(0) = 1 and p(1) : 2.
Therefore,
c= c.! : cek,o = ! and 2= cek,I : ď
It follows that k : ln 2, so
p(t):etla2-,1n2t*2t
After 1 day, t = 72, so the number of bacteria cells will be p(72) : 272 = 4.72 X 1021
(see Figure 6.18).
Chapter 6 \Iector pace*
5 x 1021
4 x l021
3 x la}l
2x102l
.1 X 1021
Hru]O fi.l
xponential growth
Radioactive substances decay by emitting radiation. It m(t) denotes tlre mass of
the substance at time r, then the'rate of deca| is rr'(í). Physicists have found that the
rate of decay of a substance is proportional to its mass; that is,
m'(t) : 1r*lr7 or m' _ km = 0
where ft is a negative constant. Applyrng Theorem 6.32, we have
m(t) :
"*t
for some con tant c. The ťme required for half of a radioactive substance to decay is
called its half-|ife.
After 5.5 day , a 100 mg sample of radon-222 decayeď to 37 mg.
(a) Find a ťormula for m{t),the ma remaining after f ďays. .
(b) What is the hatf-life of rado n-222?
(c) When will only 10 mg remain?
olutlnn (a)
o
o
olving far k, we find
From m(t) : cekt,we have
100: m(0): cek'a ffi c.l= c
m(t} * tOOe
frf
With time mea ured in days, Ýť are gíven that Ťí(5"5)* 37, Therefnre,
sCI
Therefure, m(t} * 100c*0"18r
1gg,5.
ji; ___
37
*1,5k _ 0.37
5.5k : ln(O ,37)
_ ln(o.rz)
k- -0.18
5.5
10 2a 30 40 50 60 7a
I
I
I
I
l
i
]
ptt)
A73*Y]al
nx(t)
Section 6.7 Applications
íl0utc 0.|s
Radioacťve decay
(b) To find the half-life oírudon-222, we need the value of t for which re(t) = 59.
Solving this equation, we find
1 00e
--0.18f * 50
e*0,18f ff 0.50
-0.18f : h(}) : *ln 2
o
Hence,
ln2
and t=*=3.850.18
Thus, radon-222has a half-life of approximately 3.85 days. (See Figure 6.19.)
(c) We need to determine t}re value of ísuch that rn(t) = 10. That is, we rnust solve
the equation
-J
100e
*0.18í
? 10 or 9*0,18ť "* 0,1
Thking the natural lagerithm of bath sides yietds *0.18f * ln 0.1. Thuu,
ln 0.1
f *-_ * ro L2,79
* CI,18
o 10 ms of the ample will ťemain after approximately L2.7g days.
The soluťon set of the second-order diťferential equation 1" * ay' * bf - 0
is also a subspace of 9 (Exercise 20), and it turns out that the dimension of S is 2.
Part (a) of Theorem 6.33, which extends Theorem 6.32, is implied by Theorem 4.40.
Our approach here is to use the power of veďor pace ; doing so ďlows us to obtain
part (b) of Theorem 6.33 as well, a result that we could not obtain with our previous
methods.
Chapter 6 Vector paces
.ss ;pace oť
y"+fry'+by:0
:e roots of the characteristic equation A2 * ah
h2Á1
lemnlti
r Observe that what the theorem says, in other words, is that the solutions of
y" * ay' * by = 0 are oft}re form
in the first ase and
y : ťrÁ,f + C2e^zt
y : ťr A,f
+ C2te^rt
in the second case.
o _Compare Theorem 6.33 with Theorem 4.38, Linear differential equaťons
and linear recurrence relations have much in common. Although the former belong to
continuous mathematics and the latter to discrete mathematics, ňrere are marry parailás.
PíOOí(a) We first show that {e^,r, rr,r} is contained in . Let,\ be any root of the
characteristic equation and let/r) = y''. th.r,
í'(t) = he^t and f"(t) = l2e^t
from which it follows that
Í' * oí' + bí=
^ze^t
+ a^e^t + be^,
= (A2 -| a}t * b)e^t
:O.eiÍ: 0
Therefore,/is in S, But, since ,\1 and r\2 are root of the characteristic equation, this
mean that e^'íand. et't are in S.
The set {e^'t, g\tY is also linearly independent, since if
then, *etting f : 0, we have
Cre^'t + Cne^,t -= 0
CI + C2:0 0r C2: *CL
Next, we set í= 1 to obtain
cle^t - cre^' = 0 ar cl(e^' * eM) :0
But e^, - e^, * 0, since el, - etr,:0 implies that e^,- g'\,, which is clearlyim-
tl^,,]|t,ť,l*}i!|i,:i:lfl,í20,)
We áeduce that c1 = 0 and, hence, c2 = 0, so
Since dim - 2, {eÁ",
Á'r}
must be a basis for , (b)
}fou ffrs asked to prove this pfoperty in Exercise 21.ílgurc "20
i22
Section 6.7 Applications
Find all solutions of y" - 5y' * 6y : g.
íÝ}iŤ{:ilťiThecharacteristicequationisr\2 - 5^ +6: (r\ -2)(^ - 3):0.Thus,
the roots are 2 and 3, so 1e2t, e3'i is a basis for the solution space. It follows that the
solutions to the given equation are ofthe form
"ffi cpzt*Cz&3'vIJLA
The con tants c1 and c2 cáíl be deterrnined if additional equations, calledFind
the solution of y" * 6y' * 9y : 0 that satisfies y(0) : 1, /'(0) : 0,
S*it! $ť}ti The characteristic equation is r\2 + 6^ + 9 : (,\ + 3)2 :0, so -3 is a
repeated root. Therefore, !e*3t, íe-3Í}is a basis for the solution space, and the general
solution is of the form
/=cf-3'lcrte-3t
The first boundary condition gives
1:.y(0):ct-3'D*O:c1
so y : ,*3t + c2te-3t, Differentiating, we have
yl : -3r-3t + c2(-3te-3t + e-3)
so the second boundary condition gives
g: y'(0) - -3e-3'0 * cr(O * e-''o) - -3 * c2
Or Cz: 3
Therefore, the required solution is
y : e*3t + 3fu-
3t* (r + 3t)elr'
Theorem 6,33 includes the case in which the roots of the characteristic equation
arecomplex. !^: p * 4iisacomplexrootoťtheequationi2 * ah* b: O,thensois
its conjugate
^
= p * qi. (See Appendices C and D.) By Theorem 6.33(a), the solution
space S ofthe diíferential equation/" * oy' * by - 0 has {eií,e^'} as abasis. Now
eht : ub,+ ai)t
- ,?t riGlt} -:- *Pí(co qt + i sin rif)
cXt : e?-qi)t - gPt*i{*qí} glÍ{co qt * i sínqf)
so eplcosqt : a+ and eP'sinq :
+
It follows that lePt cos qt, e\t sin qf} is contained in span (e
^', .I) = S. Since epf cos 4f
and ePt sin qt arc linearly independent (see Exercise 22) and dim S : 2, |ePt cos qt,
ePt sin qt} is also a basis for S, Thus, rvhen its characteristic equation has a complex
rootp + ql,thedifferentialequation )," + ay' * by:0hassolutionsoftheform
t
<+
bounď,-
and
y : cfP' cos qt + c. Ft sin qf
Chapter 6 Yector paces
tl ure .2t
T '\
./
,r"'*\
/\l
\*
?
l
ll
Iml
Find all solutions of y " +4*0.
Sfiún Theclraraďeristicequationis r\2 - 2,\ f 4 : 0withroots 1 + iVÍ.Theforegoing
discussion tells us t}rat the general solution to the given differential equation is
y * t
ťs lňt * {zťt sin ť3ť
A mass is altached to the end of a vertical spring (Figure 6.21). If the mass is pulled
downward and released, it will oscillate up and down. Two laws of physics govern this
situation, The first, Hookďs lar, states that if the spring is stretched (or compressed)
x units, the force F needed to restore it to its original position is proportional to r:
"kx
where ft is a positive constant (called the spring constant). Newtotťs Seconil Law of
Motion states that force equals mass times acceleration. Since x = x(t) repre ents
distance, or displacement, of t]re spring at time í,x' gives its velocity and x} its acceleration.
Thus, we have
mx'| ffi *kX sr X" +
Since both k and m are positive, so is K = klry, and our difíerential equation has the
formx" * Kx= 0,whereKisposiťve.
The characteristic equation is ,\2 * K = 0 with roots +l\fr. Therefore, the general
solution to the differenťal equation of the oscillating spring is
x = c.cos VRr * crsinÝRt
uppose the spring is at rest (x = 0) at time f = 0 seconds and is stretched as far
as possible, to a lengh of 20 cm, before it is released. Then
0 * Jť(0) # cl ť0 0 + cr sin 0 * c1
{il--0
} :20 sin ,iK,
lgure §.22
Section 6.7 Applications
so í= crsinÝ-K' Since t]re maximum value of the sine futrction is 1, we must have
c2 = 20 (occurring for the first time whe n t =rr /2 K), gving us the solution
x: 20 sinlftt
(See Figure 6.22.)
Of course, this is an idealized solution, since it neglects arry form of resistance
and predicts that the spring will oscillate forever, It is possible to take damping effects
(such as fríction) into account, but this simple model has served to introduce an important
applicaťon ofdifferential equatiorrs and the techniques we have developed.
ffi nOnO GllůOut lintfi] niííG]sntitltilllttion
In xercťses }"*12, nd the solution af the dffirential equation
that satis -es th* given baunďary #§nditirn{s}.
!, y' * 3y: Q,y(t) : 2
2,x'* x* 0,x{1} = 1
13. A strain oťbacteria has a gro$ th retť that is propartional
to the *ize of the pnpulatioí}. Initíally, there are
10ff bacteria; after 3 hgurs, there af l6ff0,
(a} Iťp{r) denotes the number oťb*cteria after
f hours, find a formula for p(r}(b}
How long does it take for the population tc double?
(c} When wi}l the population r*ach on million?
14. Thble 6"2 gives the population oťthe t]nited tates at
lO-year intervals for the years 1900*2ůůů.
(a} Asgumin an exponential growth model, use the
data for 19Sů and 1910 to find a ťormula far p{t},
the popu}ation in year t- |Hinf: Let f : S be 1900
and let f * 1 be 1910.] How accurately doe y ur
ť.*rmula calculate the tI. . population in 200ú?
Year
(b) Repeat part (a), but use tlre data far the years I970
and 19S0 to solve for p{t). Does thi epproach give
a betteť áppťoximation for the ye*r 2000?
(c) What can you conclude about tJ"S, population
growth? r
Population
(in millions}
1
:1
1
l_
1
1,/(1) 0
:0
{
*l
1900
19 10
L920
193ff
L94ú
1950
196ů
L97
1980
1990
200CI
76
92
106
L23
131
150
L79
2ů3
227
25a
28L
Source: U. , Bureau of the Census
l5. The hatf-life of radium -226is 1590 year . upposť l,y§
start with a ample of radium -226 whose ma ís50 rns.
(a) Find a formula ťor the mass m{t} remainíns after
t yearc and u e this ťormula to predíct the mfls
remainins aťter 1000 year .
(b} When will only 10 mg remain?
!6, Radiocarbgn datingis a method used by scientistg
to estimate the age of ancient objects that wer onťf,
living matter, such as bone, leather, wood, or paper.
-a
§Hb
Chapter S Yector Spaces
Al1 of these contain carbon} a proportion of w]rich is
carbon- 14, a radioactive isotop* that is continuously
being formed in the pper atmosphere. Since livins
organi ms take up radioactive carbon along with
other carbon atom , the ratio between the two forms
remains constant. }íowever, when an organism dies,
the carbon-14 in its cel}s decays and is not replaced.
Carbon-l4 has a known half-liťe of lrua year , o
by measuring the concentratíon of carbon-l4 ínan
object, scientists can determine its approximate &ge.
One of the most ucce ful applications of radiocarbon
dating has been to determine the age of the
$tonehense monument in ngland (Figure 6.23).
amples taken from the remains of wooden posts
were found to have e ccncentratian oť carbon- 14 that
&ra 45g/o of that found in living material. What is the
estimated age oť the se posts?
lsura .2S
tonehenge
17. A ma is attached ta a spring, as in Example 6.92. Át
time f : 0 sec*nd, the spring is stretched to a length of
10 cm below its positíon &t rest. The spring is released,
and its length lfi secands later is observed to be 5 cm,
Find a formula ťor the len8th of the spring at time
f seconds.
l8. A 50 s rna is attached to a spring, &s in Example
6.92. Iťthe period oťoscillation is 10 seconds,
find the sprins constant.
19. A pendulum consists of a ma , called a bob,that
is affixed to the end oť a string of length I (see
Figure 6.24}. When the bob is moved from its rest
position and released, it swings back and forth. The
time iť takes th* pendulum to swing from its farthe*t
right position to its ťarthest leťt position and back to
its next ťarthest right position is called the periad ať
the pendulum.
T Lň
'0 \r
\i
\. i
\-*-ffi*
-n-q#ffi
tlsute 6.2{
Let 0 * 0(t} be the angle of the pendulum from
the vertical. It can be shown that if ihere is no resist&nce,
then when 8 is small it satisfies the differential
equation
0" + {, -- 0
L
where g is thc constant of acceleration due to gravity,
eppťoximately g,7 mls'. ,rppnse that l : 1 m and
that the pendulum is at rest (i.c., § r 0) at time
f * 0 second. The bnb is then drawn to the right X
en angle of #1 radians and released.
(a) Find the psriod of the pendulum.
{b) Does the period depend on the angle 01 atwhich
the pendulum is released? This question was
posed and an wered by Ga}ileo in 1638. [Galiteo
Galilei (L564-L642) studied medicine a a student
at the t]niversity of pisa but his real interest was
always mathematics. In 15g2, Galileo we eppointed
profes cr of mathematics at the univerity
of Padua in Venice, where he taught primarily
secmetry and astronomy. He was the first to use
a telescope to look at the stars and planets, and in
so doin$, he produced experimental data in support
of the copernican view that the planets re*
volve around the sun and not the earth. For this,
Galileo \,va summoned beťore the Inquisition,
placed under house &rrest, and forbídden to publish
his results. While under house arrest, he was
able to write up his research on falling objects and
pendulum . His notes \ťere smuggled out of ltaly
, and published as .Disc ouť es on Two }ťew Sciences
in 1638.]
20. Show that the solution set of the second-order
differential equation y" * ay' * by: S is a subspace
oť ?F,
2l. Prove Theorem 6.33(b),
22. how that e?t cos qt anď ePÉsinqt are linearly
independent-
E
O
q
_:l
(-)
c
(a
c_)
=_.C
u)
_c
(
c)
c}
i]*
"Ťo)
ť
.
ó__
o
l{ss llofrin titl ls ffIllil ll0nteilt
basis, 446
Basis Theorem, 453
c}range-of-basis matrix, 465
comp*sition of linear
transformations, 477
coordínnte vectar, 449
diagonalizable linear
transformation, il9
dimension, 453
Fundamenťal Thecrem of Invertible
Matrices, 51?
ídentity transformation, 47 4
invertible linear
transťormation, 47B
isomorphism, 493
keinel of a linear
transťormation, 4B2
linear combination of vectors, 433
linear transformation, 472
linearly dependent vector ,
443,446
linearly indep endent vectors,
443,446
matrix oť a linear
transfarmation, 498
nullity oť a linear
transformation, 484
one-ta-Oí}e, 488
snto, 488
rs.nse of a linear
transformation, 482
rank of a linear
transformation, 484
Rank Theorem, 486
pan of a set of vectors, 438
standard basis, 447
subspace, 434
trivial subspace, 437
vector, 429
vector pac , 4ž9
zero subspace, 437
rerCI transformation, 474
fieuinw 0unstions
1. h{ark each oť the followins statements true or ťalse:
(a} If Y = span(yr, . . . , yn), then ev*ry spanning set
for V contaíns at least n vectcr ,
{b} Iť {u, v, w} is a linearly independent set oťyťtor ,
then so is {u t y,y * $I, u * w}.
(c) Mzzhas a basis ccnsisting of ínvertible matrice .
(d) Mr2 has a basis consisting of matrices whose trace
is zero.
{e) The transformation T: [Rn --+ K defined by
T(x) all*
ll is a linear transformation.
(0 Iť T ; 1/ -+ 1ťis a linear transformatian and dim
V * ďim Tť, then T cannot be both one-to*one
anď onto.
k) If r: V -+ Tď ísa linear transformation and
ker(T) * Y, then |tf : {fi}.
(h} If r i Mn a W ais a linear transťormation and
nullity{T) * 4, then T is ontCI,
(i} The vector pace \rr * {p(x) in #n :p(1) * 0} is
isomCIrphic to #E.
(i} IťJ : tr/ * Y is the identity transťarmation, then
the matrix [lJc*r is the identity matrix for any
bases fr and c oť v.
In Questians 2*5,
2,V: ffi2, 14/ -
'_1
,o+ b_ c* d
dl
* a+ c- b+ r\
,4. : s3, Ýť : {p(x) in93: x'p{t/x} - p{x)}
5. V : W,W *
{fi* ? :f{x + n) : frx} for all x}
6. etermine whether {1, co 2x,3 sinzr} is linearly
dependent or independent.
7. LetÁ and B be nonzero n T n mattices such that Á is
syínm tric and B is skew-symmetric. Prove that {Á, B}
is linearly independent.
In Questions 8 and 9, find a basis fo, W and state the
dimension af W.
(la b1 'l
8.|tf * {l*
*,l
,o+d*b+cLtJl vv
tt, aJ I
g. W: lryk} in Wslpt*x) : P(x)}
10. Find the change-of-basis matríces Pcnnand Fr*rwith
respecttothebasesff, : {1, 1*x, 1*x+x'Iand
C: {I * Jí,x * xŽ,L + x'} oťS2.
In Questions ]1-13, determine whather T is a linear
transiormatiott.
11. T : R2 -+ R2 deťined by T(x) * }x!, where y :
{l-3,V* Mrr, }f *
LLc
determine whether W is fr sLtbspace oíV"
í[,,l ,
"2
+ 3v2* 0}
LLl) ' ) [,l
L2]
Chapter 6 \&ctor paces
12. r i Mnn Mnrdefined by r$) * Ált
13. r i W,*W ndeťined by r(p(x)) * p(Zx * 1)
l4. If T :W2-+ Mzzísn linear transformation such that
T(1) - [l |],
rt, + x) :
[l l]
*o
r(1 * x + x2} ffi
[l
-i], find T(5* 3x + 2x?},
1 . Find the nullity of the linear transformation
T: Mno.+ ffi defined by T(Á} * tr(Á). &
16. Let Iťbe the vect*r space of upper triangu|ar 2 X 2
matrices.
{n) Find n linear transformation T i Mzz a Mzesuch
that ker(ť} * W,
(b) Find a línear transťormation ? i Mzz ň Mz,zsuch
that range{T) * W,
17. Find the matrix IT]c**r oťthe linear transformatian
T in Question 14 with respect to the standard bases
ffi : {1, x, x21 *ÉW2 and, ( : {E,r, En, Ezt, Erz] af M22,
l8. Let -- {r,, r , r , vn} be a set of vectors in avector
pece Y with the property that eyery vector in V cfirt
be written a a linear cgmbination of v1, 1 . , , vn in ex*
a tly on }ťay. Prove that ,. is a basis fgr F,
19. If T : t] -+ and : 1/ *r Tť are linear transforrnations
such that range(T) ker( ), what can be
deduced about " T?
20. Let T : Y * V be a linear transformation, and let
{r,, . . t,vn} be a basis far Vsuch that {r(v,), . . .,
T(vr)} ísalso a basis for V, Prove that T íninvertible.
J