Pr

od
uc
ed
6.12 Linear Alg b
era
w
ith
a (b) Show that the mapping T: .Mnn .Mnn given by T(A) =A - AT is a linear operator
on .Mnn· .
Tr
ia
lV
5. Let P be a fixed non-singular matrix in .Mnn· Show that the mapping T: .Mnn .Mnn given
er
si
by T(A) = P- 1 AP is a linear operator.
on
of
6,
PD
Let V and W be vector spaces. Show that a function T : V W is a linear transfonnation if
F
, I
and only if T( av1 + pv2) = a T(v 1) + /JT(v 2), for all vi' v2 E Vandall a, Pe R
An
' 11 no
ta
7• Let T1, T2 : V W be linear transformations. Define
to
11 r-
I I
T1 + T2 : V W by (T1 + T2)(v) = T1(v) + Ti(v), v e V
w
w

I
w
Also, de.fine
.P
D
FA
c T1 : V W by (cT1)(v) = c(T1(v)), v e V ·
(
nn
ot
Show that T1 + T2 and cT are linear transfonnations. at
or
.c
om

1. (a) Yes (b) No (c) No (d) No (e) Yes (f) No (g) No

6.2 THE MATRIX OF A LINEAR TRANSFORMATION

In this section we will show that a linear transformation between finite-dimensional vector spaces
is uniquely determined if we know its action on an ordered basis for the domain. We will also show
that every linear transforma_tion oetween finite-dimensional vector spaces has a unique matrix Ase
with respect to the ordered bases B and C chosen for the domain and codomain, respectively.

A Linear Transformation is Determined by its Action on a Basis

One of the most useful properties of linear transformations is that, if we know how a linear map
T:V W acts on a basis of V, then we know how it acts on the whole of V.
'l,;,~::t /:; ;,i,\:;' 1?·, -.'.t- ,{ . ' ~- ~3/.::h ~0
0- ,:",<. t
.,v~h;~,ec~ri ,orde.re~t ~asisW~r-''."~/ ·' '
<). ' '.;:"\.; " \,; .

· · · · 11<?t v¢p~ssaiHY iJis,(:

\.<,,.,J)/,;,r:/:,'/,s':,):;--:-'. i,t"'• _,· .. ,'-:;,,1::_,l _'··:·~.',~;

•'.~~:~~i:tctx;:
er:niin~4"by'tt§''.a'.cti.o .:
,, -i ., .• ' ' . ' , , , •.. "' ,•-:·f., ~,

Proof Let v be any vector in V. Since B = {vi' v2, ..., vn} is an ordered basis for V, there exist
unique scalars al' a2, •• •, an in JR such that v = a1 v1 + a2 v2 + ... +an vn.
Define a function T: V W by
T(v) = a1 w1 + a2 w2 + ... +an wn
Since the scalars aI.'s are unique, Tis well-defined. We will show that Tis a linear transformation.
Let x and y be two vectors in V. Then
X = b1 v1 + b2 v2 + ... + bn vn
 ,, l "1,'
Pr
od
uc
6.13
ed
£inear Transformations
w
ith
a0d y = c 1 vt + c2 v2 + ... + c
v
a
Tr
for 5001e unique b/ s and c/ s in JR. Then' bYdefitmtion
. n of T we have
ia
lV
er
T(x)-b
. - I wt + b2 w 2 + ... + b W
si
on '
Tiu) - n ,i
v - cl wt+ c2 w 2 + ·•· + C W
of
PD
T(x) + T(v) =
_ (bl + + b,,- w2 + ... + • w.)
b WI F
An n + (c 1 w1 + c2 w, + ... + c, w.)
- ( Ic1)wt + (b2 + c2)w2 + ... + (b + c )w
no
+ y = (bl vi + b2 V2 + •·· + bn v,,) +'(c v n·+ c2nv2 n+ ... +en v,,)
ta
X to
r- 1 1
I I
-(b 1 + cl )vi + (b 2 + c2)v2 + ... + (b + c )v
w
- w I

'I
w
T(x + Y) -- (b 1 + c1)W1 + (b + c )w + ... + (bn + nc )w,,,
.P n . I .
2 2 2
D
FA
ain by definition of T. Hence ' T(x + y) = T'x)
1,: + T(y) • Next, 1or n sea1ar c e m
e: n any ~,
nn I
I I
ag _ex = c(h •• + h, , + ... + h, •.)=(ch,)•, +(ch,)•,+.,.+ (ch,)•.
ot I
at
1 2
or
I,
.c
Ttcx) = (cb1)w1 + (cb2)w2 + .... + (cb n)wn I
om
I

= c(b1 W1) + c(b2 W2) + ... +c(bn wn) I

= c(b\ WI+ b2 W2 + ... + bn wn)
= cT(x)
Hence Tis a linear transformation.
To prove the uniqueness, let L : V W be another linear transfonnation satisfying

L(vt) = w., L(v2) = Wv ... , L(v,,) = w,,

Ifv E V, then•= a •• + a2 • 2 + ... +a,••• for unique scalars a 1, a2, ••. , a, E JR. But then
1
L(v) = L(a 1 Vi+ a2 v2 + ... +an vn)
= a L(vi) + a2 L(v2) + ... + a 11
L(v,,) ( ·: Lis a L.T.)
1
= al Wi + az W2 + ... +an wn = T(v)
L = .T and hence Tis uniquely determined.

EXAMPLE 14 Suppose L: JR 3 JR 2 is a linear transformation wit~

L([l, -1, OJ) = [2, I], L([O, I, -1]) = [-1, 3] and L([O, I, OJ)= [O, I].
3
FindL([-1, I, 2]). Also, give a formula f~r L([x, y, z]), for any [x, y, z] [Delhi
E IR . Univ. GE-2, 2017]

SOLUTION To find L([-1, I, 2]), we need to express the vector•= [-1, I, 2] as a linear
combination of vectors,,= [I, -1, OJ, v2 = [O, I, -I] and • 3 = [O, I, O]. That is, we need to find
constants a1, a 2 and a3 such that
V = al Vi + a2 V2 + a3 V3,
which leads to the linear system whose augmented matrix is
Pr
od
uc
ed
6.14
w
ith Linear Algebra
a

[_: -:J
Tr
ia
lV
er
0 -1 0 2
si
on
We transform this matrix to reduced row echelon form :
of
PD

1 0I 01 -lJ1 0-1]
F
An
1 0
[0
no
-1 ta
0 2

-~]
to
-1 0 2 r-
w
w
w
.P
D
FA
0 1 2
nn
ot
at

ill0 O0-2-li
or
.c
om
;
0 ill 2 I

This gives a 1 = -1, a2 = -2, and a3 = 2. So,

v =-v1 -2v2 +2v . 3
L(v) = L(-v1 - 2v2 + 2v3)
= L(-v1) - 2L(v2) + 2L(v3 )
= -[2, 1] - 2[-1, 3] + 2[0, I]= [O, -5]
i.e., L([-1, I, 2]) = [O, -5]
To find L([x, y, z]) for any [x, y, z] e JR3, we row reduce

_: ;]
[ 0 -1 0 z
to obtain r~O O 1 x+ y+z
-:]
Thus, [x,y,z] =xv1 -zv2 +(x+y+z)v3
L([x, y, z]) =L(xv1 -zv2 +(x+y+z)v3)
. = xL(v1) - zL(v2) + (x + y + z)L(v3)
= x[2, 1] - z[-1, 3] + (x + y + z) [O, 1]
= [2x + z, 2x + y- 2z].
EXAMPLE 15 Suppose L : R 2 R 2 is a linear operator and L([i, 1]) = [1, -3] and
L([-2, 3]) = [-4, 2]. Express L([l, O]) and L([O, 1]) as linear combinations of the vectors
[I, OJ and [O, I]. [Delhi Univ.- GE~2, 2019]

SOLUTION To find L([l, O]) and L([O, I]), we first express the -vectors v1 = [l, O] and
v = [O, I] as linear combinations of vectors w1 = [I, I] and w2 = [-2, 3]. To do this, we row
2
reduce the augmented matrix
Pr
od
Transformations
uc
tifl~ar 6.15
ed
w .
ith
a
Tr
ia
lV
s we row reduce er

[: -! ~]
fill.I'
si
on
of
PD R2 R~ - RI
F
An
l
-211 -1/ 15 1/5OJ
no
R2
5
ta
to 2
r-
w

013/5 2/5]
w
w
.P I I
D
I I
1 -1/5 1/5
FA
nn \ I
ot
3
at
l
V -
and = -25 w1 + -15 W2
or
1 - - wt - -'- W v
5· 5
.c
2 2
om

This gives L(v 1) = I5 i(w)1 _ .!.'5 L(W2, )

= ¾L([l, l]) -"i Lff-2, 3]) · . I

3
= 5[1, -3] - 5[-4,.2] =
f' · [Ts'
2 1
and L(v2) = 5 L(w-1) + 5l,(w'l)
1
2 1
=
5 L([l, 1]) + 5L([-2, 3])
2
== -[1 -3] + -[-4 2] =
5 ' 5' - ' .· ·
1 [-2
·-. ,
5
-=-~] = - 2 (1, O] - 54(0, 1].
5
The:·_Matrix -of a Linear Transformation
We now show that any linear transformation on a fmite-diniensi0nal vector spac.e.c~be e.ipress.ed
as a matrix multiplication. This will enable us to find the effeot ..of any linear-.tranS:fonna1ion•·by
simply using matrix multiplication. .
Le~ Vand Wbe n~n-~vi~l vect~r s~a,c~~' W:ith_dim V= _n ~i~ W-= ."!•,~et:R = {v1~ v2, -~·, ~n}
and C = {wl' w , ..., wm} b~ ordered bases for Vand W, respectively: L.e.tT': V ,W:btfaJ inear
2
transforma~on. Fo:r e~ch- v in V, the coordinate vectors for v and T(v) with respect to oniered
bases Band Care [v]B and [T(v)]c,trespectively. Our•goal is to find .an m x n ma~ A-= (ai)
(1 $ i m ; 1 j ) su:ch-that - · · · · · ·
A{v]B = [T(v)]c ... (1)
 6.16 Linear Algera
b
~ol4s for all. vectors v in V.. Since..Equ_ation (1) must hold for all vectors in V, it must hold .
particular, for the basis vectors in B, that is, . . ' lil
A [v1JB = [Ttv1)Jc, A[vz]B = [Ttvz)]c, .... , A[vn]B = [T(vn)]c ... (2 )

0
1 0
But · ' .... , [vn]B = :'
0 l .,
0 11' 0 12 a1n 1
0 21 °22 02n 0
A[v1J.l! =_ =

Omn 0 Omt

0 1n 0
0 2n I
=

Omn 0 Om2

a1n 0

=
amn I
Substituting these results into (2), we obtain
0 1n
a11 a12
0 22 0 2n
a21
=fT(v1)Jc, = [T(v2)]c, ...... , = [T(vn)Jc,
l!inn
This shows that the successive columns of A are the coordinate vectors of T(v 1), T(v2), .•. , T(vn)
with respect to the ordered basis C. Thus, the matrix A is given by

'
We will- ca:11 this matrix as the matrix of T relative to the bases B and C and will denote it by the
· :1Bc or [T]8c- Thus, · ·
Alic = .[[T(v1)]c [T(v2)Jc ·:· T(vn)]c] ·

From (1 ), the matrix ABC satisfies the P!operty

ABC [v]B· == [T(v)]c for all v E V.

We have thus proved :

· Transformations
v,,•,1e{)r 6.17

.,~. ll,tPLE 16 p·Letd T: R 3 . R3 be the 1mear

.
JP;,r,...
+ .t 'X1 - X3]- 1D th e matrix for Twith , operator given
. by T([xp. X2, x3]) = [3x1 + X2,
t 3 . respect to the standard basis for JR3 •
1
soLUTION The standard basis for R 3 is B = =
bstituting each standard basis vector int th . {et [l, 0, O], e2 = [O, 1, 0), e3 = [O, 0, l]}.
Su O e given formula for T shows that
. T(e1) = [3, 1, 1], T(ez) = [1, 0, 0], T(e3) = [0, 1, -1]
. ce the coordmate vector of any element [Xi, x , x ] m
1n . R 3 with
. respect
· to the standard basis

l
S 2 3

{e1, e2, e3) is [ :: we have

Thus, the matrix A BB for T with respect to the standard basis is :

ABB = [[T(e1)le [T(e2)lB [T(e3)Ja]. = Fi _:}

3
EXAMPLE 17 Let T: P IR 3 be the linear transformation given by T(ax + bx2 +ex+ d)
3
= [4a- b + 3c + 3d, a+ 3b- c + 5d, -2a- 7b + 5c-d]. Find the matrix for Twith respect to the
standard bases B = {x3, x2, x, l} for P3 and C = {e1, e2, e3} for ~.3.

SOLUTION Substituting each standard basis vector in B into the given formula for Tshowsthat
T(x3) = [4, 1, -2], T(x2) = [-1, 3, -7], T(x) = [3, - 1, 5] and T(l) = [3, 5, -1]. Since we are using
the standard basis C for R 3,

Thus, the matrix of T with respect to the bases B~and C is: ·

ABc . [[T{x3)1c [T(x2)lc [T(x}lc [T(l}]cl = l -~

-2 -7
3
-1
5 -1
3~
5 .
 Linear Algebra
6.18 . . · b Tt") - p' ·where ·
. .. . be•the linear transfonnation given y "'. - , . . p E--P3, \\
EXAMPLE 18 Let T. 'P3 1'2 . . <lard bases for 'P and -p2. Use this matrtx to calculate
Find the matrix for T with respect to ~e- sta~ 3
T(4x3 - sx2 + 6x - 7) by matrix multtphcatton. . . .
, · .· • B - {x3 xz x I } and the standard basis for p is
·• •d d b ·s for 'P 1s - , , , ' 2
SOLUTION The s~ ar as1_ . 3f each polynomial in the standard·bases B for 'P3 shows
C = {x2, x, l}. Computing ~e denvattve o
that T(x3) = 3xz, T(x2) = 2x, . T(x) = ~' a~d- T(l) = 0 . .
We conv.:ert these resulting polyno_mi_als in 'P2 to vectors lrt JR :
3x2 [3, 0, OJ; 2l" fO; 2, O]; 1 [O, 0, 1]; and O [O, 0, O] .
Using each of these vectors as columns yields

BC
.
= r~ ~].
0 0 1 0
We will compute T(4x 3 - Sx2 + 6x- 7) using this matrix. Now,
4
-5
[4x3 - Sx2 + 6x - 7]B = 6
. -7

Hence,

o ·o
2 0
[1t4x3 - 5x2 + 6x - 7)Jc = A8 ~[4x3 - 5x2 + 6x-:- 7]8 = [~
. 0 0 1

Converting-·back from ~-coordinates•to-polynof!Uals gives

T(4x 3 - S.x2 + 6x- 7)-= 12.x2 - lOx + 6.
E~liE l9 Let T: JR 3 R2 be the.linear transformation given by T([x 1, x2, x3]) = [-2x1 3x3, +
x1 + 2x2 -x3]. Findthe matrix.for,Twith respect to-the ordered bases B = {[l, -=--3, 2], [-4, 13, -3],
[2, -3,"20n for JR 3'·and C = {[-2, -1], [5, 3]} for R 2 . [Delhi Univ;c6E~2., 2019(Modijied)}
. - .
SOLUTION -By definition, the matrix ABc of Twith respect to the orde-red bases B and C is given
by A0 c =l[T(v1)]c [T(v2)]c [T(v3)]c], where v1 = [1, -3, 2], v2 = [-4, 13, -3], and v3 = [2, -3, 201
.
are the basis vectors -in B. Substituting each·b~is vector in B:intothe given•fotmubrfor Tshows that
. -

T(v
1
) = [4, -7], t(v2 ) = [-1 ,' 25], T(v3) = [56, -24]
Next, we must find the coordinate vector of each of these images,in R 2 with ·respectrto -the C
basis. To do -this, we use the Coordinatization Method. Thus, we·must row reduce matrix
[w 1 w2 I T(v 1) T(v 2) T(v3)],
where w1 = [-2, -1], w2 = [5, 3] are the basis vectors in C. Thus, we row reduce
 £iTlear Transformations
6.19
[~2 514
1 3 -7
--1 56]
25 -2 4 to obtain [l O\-47 128 -2881
-47] O l -18 51 -104

Jieuce [f(v1)1c = [ -18 . ' [T(v2)]

c = (12
81
51 ' [T(v 3)] = [-
288
1
- C -104.J

fbe matrix of T with respec t to th~ bases Band C is A =· [-47 ·128 -2881
BC -18 51 -104.

finding the New Matrix for a L"anear Transfor f ·

we now state a theorem (proof • · ma ion After a Change of Basis
Om1tted) which h l
. w h en we change th . computing the matrix for a linear
tra115forroat1on b e ps us m
::p":" e ases for tbe domain and codomain. · .

':f

A.::,:;sc_ _ _ _,._, [T(v))c

[v]B _ _ _ _ _
Matrix for Tusing B, C
Transition Transition
matrixP matrix Q

[v]v-----A~oE;... , _ _ __... [T(v))c

Matrix fQr I using D, E
FIGURE 6;6 Jllustrates the situation -in.'f.heorem 6.6
EXAMPLE 20 Let T: JR 3 JR 3 be the linear operator given by T[(a, b, c)] = [-2a + b,
-b - c, a + 3c].
(a) Find the matrix ABB for T with respect to the standard basis B = {e1 = [l, 0, O], e2 = [O, l, 01,

-e3 = [O, 0, I]} for R 3.

(b) Use part (a) to find the matrix ADE with respect to the standard bases D = {[15, -6, 4],
[2, 0, !], [3, -1, I]} and E = {[I, -3, I], [O, 3, -!], [2, -2, I]}.
SOLUTION (a) We have T(e ) = [-2, 0, I], T(e2) =[!,-I, OJ, T(e3) = [O, -I, 3]}. Using each
1
of these vectors as columns yields the matrix A BB :

A~= n-l -!l

 6.20 Linear Algebra
(b) To find ADE' we make use ofth~ following relationship:
ADE = .QABB p-1, . . .. (1)
where Pis the transition matrix from B to D and Q is the transition matrix from B to E. Since p-i 18
.
the transition matrix from D to Band Bis the standard basis for -IR 3 , it is given by

p -1 = [~ ; -:]
. To fmd ·Q, we first find Q- 1, the transition m~~ix from E to B, which is given by
.. t· r ·-

Q-1 -=~<r:.3
,I O 2J
3 -2
1 -1 1
It can be 1easily checked thal -' ··
.•:_ ft

Q = (Q -1)-1 = r: - ~i'
0 .1

Hence, using Eq. ( 1), we obtain

1 -2
ADE= QABBp-l = 1 -1 -4
-6][-2~ 1 -o][I[
0 -1 -1 -6
5r -32 ' -43]
-23 -31 ,
[
0 1 3 1 0 3 4 14 18
EXAMPLE 21 Let T: 1)3 JR 3 be the ~inear transformation given by T(ax 3 + bx 2 +ex+ d)
= [c+ q, 2b, a - d]. ·
.,, .
(a) Find the matrix ABC for Twith respect to the standard bases B (for 1'3) and C (for R 3).
. ~. i J

(b) Use part (a) to find the matrix ADE for ·T with respect to the standard bases D = {x3 + x2,
x2 + x, x + 1, 1} for ~3 and E = .{[-;-2, 1, - ~], p, ~- ?,.: O], [3, -6,_2]} fot: R\ ·. _
SOLU'fION _(a) To f¥ld the mattjxA9 c for Twith respeotto the standard bases B == {x3, x2, x; l}
for ·1:)3 t nd C= {e1 =-[l, 0, OJ, e2 = [O, 1, Q], e3 = [.O, 0, _1]} for R3, we first need to find T(v) for
each v e B. By definition of T, we have
T(x 3) = [O, 0, 1], T(x 2) = [O, 2, OJ, T(x) = [l, 0, O] and T(l) = [l, 0, -1]
. . . - -

Since we are using the standard basis C for R3, the matrix BC for Tis the matri~ ~~ose columns
are these images. Thus · ·
O O I
A = O 2 0
BC [
1 0 0

(b) To find ADE' we make use of the following relationship :

ADE = QAecp-I ... (1)
 . ar Transformations
1,,ne .. .
rep is the transition matrix from B tO D . 6.21
~be • fr and Q · th . .. ·
tf!ltlsition matnx om B to D, therefore p-1 is th is e ~an~1tion..~~trix C to E. Since Pis the
-• we-need to convert the polmomial 1. D. e transition matnx from D to B. To compute I
(
f ' • 3 b 2 -. - s n · mto v t · 4 . . .
1ynollllal ax + x + ex + d m D to [a -b _ec ors m JR . This is -done by converting each
po •' , c, d]. Thus -
1,0,0]; 1 1 O]· ..
. Bis the standard basis for Trb3 th ' ' . '. ' (x + l) [O, O, 1, 1]; (1) [O, 0, 0, 1]
Stllce .11\\. , e transition tr· 1
each of these vectors as columns : . ma ix (P- ) from D to B is obtained by using
I 0 0\ 0, . ,. J

I I 0 0
p-1 - 0 i 1 0 \
I

0 0 I I
To find Q, we first find Q-1, the tran~·t·
1 .
are the vectors in E. ion matrix from E to C, which is the-matrix whose columns

= [-2
~l
I
Q-1 1 -3
-3 0

Q = (Q-1)-1 = [-2 -3 -6 -
I
-3 0 2
I
Tl r6 -2
= 16 5 _-9
-9 -3
131
5 .
j

[-6 -2 rl -10
~rn~
1 0 0 0

-9]
0 I
I I 0 0 -15
Hence, ADE = QABcp-t = 16 - 5
-9 -3
2 0
0 0
_i] "'- .
0 I 1 .. () . = 1 26 41 · 1 25- .
I 0 0 l 1 _ -:1 -15 -?3 -14

6.3 LINEAR OPERATORS AND SIMILARITY

In this section we will show that any ~o matrices for the same linear clperator ( on ·a finite~im,e~~iclnal
vector space) with respect to different ordered bases are similar. l

Let V be a finite-dimensional vector space with ordered bases B and C, and let T : V v be a
linear operator. Then we can find two matrices, ABB and Ace, for rwith respect to ordered base~'.
Band C, respectively. We will sho,w that 4 BB ~pd 4cc are similar. To prove this, let P denote the
transition matrix (PC+-B) from B to C. Then .
by Theorem 6.6, we have
Ace = P ABB p-• ABB= p-t Ace P
This shows that the matrices ABB and Ace are similar. We have thus p~oved the following:
~~ ~i ,~-,"<•.·
·•n.~~ ~ iJt, .,
'tz,;~ i~JJ!;_
· ,. ,~\,;>:,. - <J
~pe•~ ' h''
~~:-- . p· emtoj;
,, ,, ,,,,,
·'flJ
~~ A-· foi~~~~i•'
cc '; -"' -
_;,f, then AiJ,::·,_:
. '<
 Linear Algebra
6.22
[v]11 _ _ _ _ _A..:.:ss;__---~ [T(v)]a
Matrix for Tusing B
Transition .
Transition matrixP
matrixP-'

_ _ __.:.:A!;!".
cc_--,--.IIII [T(v)]c
[V] e - I
Matrix for Tusing C
FIGURE 6.7 mustrates the situation1in Theorein 6~7
EXAMPLE 22 LetL: R 2 JR 2 be the linear operator given by L([x,y]) = [2.x- ~' :- ~y]. Find
the matrix for L with respect to the basis {[4, -1], [-7, 2]}. using the me~o? of -~mula~ty:
[Dtlhi •Univ., -GE-2 1,20-1'7]
\ I

SOLUTION First, we find the matrix for L with respect to the standard basis B = {e1, e2} for
JR 2• We have
L(e;) = L([O, 1]) = [-1, -3]
L(e,) = L([l, O]) = [2, 1],

Thus, ilie ~trix for L with respect to the basis Bis AB~= [~ =a
Next, we fi~d-the transition matqx j> fp:mi tlie ·s~ndard 9asis B to the basis C {[4, -1), (-7, 2)}.
To fipd P, we first find p- 1, the transition matrix from C to B, which is given by

·/
.
· . [-14-7J2 · .
p-1 = I. l ,

' l

Hence the matrix Ace for L with respec~ to the basis c :is·given by

I.
. cc
· p·1. p-t.. [2·1 7]r.
= BB
2 1il][ 4 -7] -167
4 L1 -:3 -1 2 = l 37
=
-123]
-68.
. !
EXAMPLE 23 L,et L : ~ 2 , 2 be the linear operator·that performs a counterclockwise rotation
2
through an angle 30°. Find the matrix A for L with respect to the standard ordered basis fqr JR. .
H(;'nce or otherwise find the matrix, for L with respect to the basis C = {[4, -3], [3, -2]}, _similar
to A. . . , _ [Delhi Univ. GE-2,. 2016]

SOLUTION We knowithat if L : JR 2 JR 2 is a linear operator representing the counterclockwise

rotation ofvectors in JR 2 through a fixed angle 0, then the matrix for L with respect to the standard
ordered basis B = {e1 = [l, 0], e2 = [O, l]} is given by
c~s 0 -sin 0]·
[ sme cos0
In particular, the:·mat:rix A for'the ·linear operator L that -performs a 'countercloc~ise rotation
2
through an -angle.30° with .respect to the standard ordeted basis B,1 for R is given by·
r Pr
od
uc
Transformations
ed
6.23
ti/1ear
w

l
ith
cos3oo .
j-
a
A= -sm30°
/2 -l/2J
Tr
ia
lV sin 30° cos3oo -
I
· er . 1/2 ·
find the matnX, for L with respect to th b .
si
. \

\
4
B tC = {[ •--_3], [3,-2]), similar to A, we first find
on
fo . . ~•trix p from the standard basis
th' tr81151uon of e as1s
u,iisition matnx from C to B' is given by o the basts C = {[4, -3], [3 • - 2l l . Now p-'' the
PD
F
An '

L-2 -31 \
no
= [ -~ - -~} - •. =
'
p-1 p
\
ta
! I
to
r-
Jlence the matrix Ace for L with respect to the baSis
, C, s1m1lar
. 3 to
w
w
w 4 A, is :
.P
D
13
FA

f - 7f 3
-1/21[-34
nn
2
Ace= PAP-I = ,/3 /2 3] · _ -9
Jl
ot
l
at
4 112
or
.c -2 - 25 +9
2 2
om
• I '

Matri>< for the Composition of Linear Transformations

We conclude. this. section
th with the r esu
· lt wh.1ch states that the matrix for the composition of lmear
·
tranSformaUons 1~ e product of the matrices for the individual linear transfonnations in reverse
order. More specifically, we have the following : .1

EXAMPLE 24 Let T be the linear operator on JR2 representing the counterclockwise 2

rotation of
vectors in JR2 through1 a fixed angle 0, and let T2 be the linear operator on JR representing the
reflection of vCctors in JR2 through the x-axis'. If A1 arid A2 denote, respectively, the matrices for
T and T with respect to the standard basis for JR2. then we_know that
2
1
· [cos0 --:sin0l T1 0\ '·
A1 == sin0 -_cos0 J au
d
A2 = LO -1 J
By Theorem 6.8, the matrix A for the composite linear operator T2oT1 with respect to the standard

basis for R2 is given by

A =A2A1 =l~ o]~~se -sinOl = \ cos0 -sinOl_

-1 Lsm0 cos0 J L-sm0 . -cos0 J

1. Supp~se L : R3 JR.3 is a linear operator and L([l, 0, 01) = [-3, 2, 41,3 L([O, 1, 01)
=[S, -1, 3] and L([0, 0, 1]) = [-4, 0, -2]. Find L([x, y, z]), for any [x, y, z1 E JR. . and also find
L([6, 2, -7]). [Delhi Univ. GE-2, 20181
r
I 6.24
Linear Algebra

2. Let T: ]R 3 JR 2 be the linear transformation given by T([xp Xz, x3]) = [-2:X1 + 3x3 ,
x + az
-x ]. Find the matrix for Twith respect to the ordered bases B = {[l, -3, 2}, [-4, 3, -3],
[2, -3, 2]} for JR3 and C = {[-2, -1], [5, 3]} for JR 2•
1 3
[Delhi Univ. GE-2, 2019]
•
Let T: ]R • I 4
JR 3 be the linear operator given by T([x, y, z]) = [-6x + Y-:-- z, -
2x + 3y- Sz,
3
3. 3
3x - y + 7z]. Find the matrix for Twith respect to the _standard basis for IR .
. '
4. Let T: JR2 JR 3 be the linear transformation given by T([x,y]) = [13x-9y,-x-2y,-:-llx+ 6y].
2
Find the matrix for T with respect to the ordered bases B = {[2, 3], [-3, -4]} for IR and
C = {[-1, 2, 2], [-4, I, 3], [I, -1, -1]} _for JR 3•

5. Let T: .Mi2 JR 2 be the linear transformation given by

T ([: : ]) - [6a - b + 3c - 2d, -2a + 3b - c '; 4dJ

2
(a) Find the matrix A c for Twith respect to the standard bases B(for M 22) and C(for R ).
8

( b) Use part(a) to find the matrix A DE for T with respect to the ordered bases

-n-([i :].[~ ~J.G :].[: :]) and E-{(-2,5], [-1,2]}.

6. Let _T: 1R 2 JR 2 be the l~ear operator that performs a ~ounterclockwise rotation through an
1t
angle radians (69°).
3
(a) Find the matrix for T with respect to the standard basis for JR 2 •
. . ' 1 .

(h) Find the matrix for Twith respect to the standard basis B = {[4, -1], [-7, 2]}.

7. Let T: R. 2 R. 2 be the linear operator given by T([x, y]) = [3x-4y, -x + 2y]. Find the matrix
for Twith respect to .the basis {[4, -3], [3, -2]}; using the method of similarity.

1. L([x; y, z]) = [-3x + _Sy - 4z, 2x -y, 4x + 3y - 2z] ; L([6, 2, -7]) = [20, 10, 44]

2. [-47
· -18
· 28
11
' ~36]
-14
3. r-6
-2
3 -1
4
3 -5-1]
7
4.
[-32
4 -2 -1
5. (a) [_; -1
3 -1
3
-:J (b)
[ 21 7 21 16]
-51 -13 -51 -38

-~i
- I
6. (a)
2 2
(b) 2 -
--53
- 3
2 7. [-1832
Ii -
I
-17 3 1
2 2 2 2