7. The Elements of Vector Spaces
11.1 Representing a Linear Transformation by a Matrix
between the finite-dimensional vector spaces 'l!and tW. Wefix anorã
dered basis {Al , ... , An} for 'l!and an ordered basis {Bl , ... , Bm } for tW. It is then possible to find unique numbers ai,j, where i =
1, ... , n,j=1, ... , m,such that
T(Al )=al,lBl +a2,lB2+ +am,lBm, T(A2) =al,2 B l+a2,2 B 2+ +am,2Bm,
which, we have seen may be written more compactly using the E- notation as
m
T(A)= L ai,jBi,
i=l
159
j =1,2, ... , n.
L. Smith, Linear Algebra
© Springer Science+Business Media New York 1998
160 11. Representing Linear Transformations by Matrices The m xn matrix A=(ai,j) is called the matrix of T relative to the ordered bases{AI, ... , An}, {BI , ... , B m}. Note that just as in the case of'l!=IR3='W,that thecolumns of A are the coordinates of the vectors T(Aj ) relative to the basis {BI , ... , Bm }. Note also that in saying "A is the matrix of T" youmust also specify relative to which pair of ordered bases. Here are two examples that illustrate what we mean.
EXAMPLE 1: Let
T: IR2~ IR2 be the linear transformation given by
T(x, y)=(y,x).
Calculate the matrix ofT relative to the standard basis ofIR2•
SOLUTION: By the "standard basis of IR2" we mean the basis EI =(1,0), E2 =(0,1), and that we are to use this ordered basis in both the domain and range of T. We have
T(1, 0)=(0, 1) so the matrix we seek is
T(O, 1)=(1, 0),
EXAMPLE 2: With T as in Example 1 find the matrix ofT relative to the pair of ordered bases {EI,E2 } and {FI, F2 },where FI=(0,1), F2= (1,0). (Here {EI,E2 } is the basis in the domain ofT and {FI,F2 } the basis for the range ofT.)
SOLUTION:We still have
T(1, 0)=(0, 1), T(O, 1)=(1, 0), butnow we must write these equations as
so that
B= [~ ~]
is the matrix that we seek.
The lesson to be learned from the preceding example is that appearance can be deceiving! Italso suggests that we might profitably inquire into when two matrices represent the same linear transformation relative to different ordered bases. Before doing so, let us see a few more examples.
EXAMPLE3: Calculate the matrix of the differentiation operator
relative to the usual basis { 1,X,x2, ••• , xn} forPnOR).
SOLUTION: We have form = 1, 2, ...
Dxm =mxm-1=0+Ox+ ... +mxm-1+Oxm + ... +Oxn,
and of course
D(l)=O.
Thus the matrix that we seek is
o 1 0 0
o 0 2 0
N= o 0 0 n
o 0 0 0
For example,
N= [~ ~] whenn = 1,
N= [~ ~~]000 whenn=2,
N= [H H] o whenn=3,
0 0 0
etc. Notice that all the nonzero entries are along thesuperdiagonal1 of the matrix. So far, the examples have allIed to square matrices, but this is an artificial restriction due to the examples having been chosen as simple as possible. For an example leading to a nonsquare matrix we can simply regard the differential operator of the previous example as a linear transfonnation
and we can ask for its matrix relative to the standard bases ofPnOR) and
Pn-l(IR). (What size is this matrix? Answer: nx(n +1).) Computing
1The superdiagonal of a square matrix A=(ai,) consists of the entries
al,2,a2,3,"" an-l,nof A.
162 11. Representing Linear Transformations by Matrices as before, we see, that the required matrix is
, [~ 01 02 00 n
N = .
0 0 0 0
For example
N' =[0 1] whenn =1,
N' = [~ 01 ~] whenn =2,
N'. [~ 0 20 01 0 ~] whenn =3,
etc.
For a related example, try to calculate the matrix of the differential operator when we consider it as a linear transformation
(n +1 not n - 1). Before you write anything down decide what size it should be.
EXAMPLE 4: Calculate the matrix of the linear transformation T : IR4---+ PI(IR)
given by
T(al, a2, a3, a4)=(al +a3)+(a2+a4)x
relative to the ordered bases
Al =(1,1, 1, 1),
A2=(1,1, 1, 0),
~ =(1,1, 0, 0),
~ =(1,0, 0, 0), for IR4 and
B1=1+x, B2=I-x,
forPl(IR).
SOLUTION: We have (note1=~(Bl+B2),x =~(Bl- B2ằ
T(A1)=T(1,1, 1, 1) =2+2x=2B1+OB2,
3 1
T(A2)=T(1,1, 1,0) =2+x = 2B1+2B2, T(Aa)=T(1,1,0,0) = 1+x =B 1+OB2,
1 1
T(~)=T(1,0, 0, 0) = 1 = 2B1+2B2, so the matrix sought is
[2o i 1 ~]
! 0 ! .
2 2
ExAMPLE5: Let the linear transfonnation T : ]R3---+ ]R2 be given by the fonnula
T(x, y,z)=(x+y, y +z).
Calculate the matrix of T relative to:
(a) The standard bases of]R3 and ]R2.
(b) The bases
Al = (1,0,0), A2 = (0, 0,1), Aa = (1, -1,1) and
E1=(1,0), E2=(0,1).
SOLUTION:We have
T(1,O,0)=(1,0),
T(O, 1, 0)=(1, 1), T(O,O, 1)=(0,1),
so that
A=[~ ~ ~]
is the matrix for part (a). On the other hand, T(1,-1,1)=(0, 0), so that
8=[100]o 1 0
is the matrix for part (b).
164 11. Representing Linear Transformations by Matrices EXAMPLE6: Let T :P3(m)---> ps(m)be the linear transformation given by the formula
T(p(xằ =(1+x - x2).p(x).
Find the matrix ofT relativetothe standard bases ofP3(m)and Ps(m).
SOLUTION:Wecompu~
T(1)=1+x-x2,
T(x)=x+x2- x3,
T(x2 ) =x2+x3_ x4,
T(x3 )=x3+x4_ xS,
and recall that the columns of T are the vectors T(l),T(x),T(x2), and T(x3 )expressed in terms of the basis { 1,x, x2,x3,x4, xS}. So we obtain
1 0 0 0
1 1 0 0
-1 1 1 0
0 -1 1 1
0 0 -1 1
0 0 0 -1
as the required matrix.
As a last example consider:
EXAMPLE7: Let S ={u, v, w}and define E : Funm(S)---> m
by E(f) = f(u)+f(v)+f(w), f EFunm(S).
Determine the matrix of E relative to the ordered bases {Xu, Xv, Xw} for Funm(S),
{l} form.
SOLUTION: We compute
E(Xu)=Xu(u)+Xu(v)+Xu(w)= 1+0+0= 1, E(Xv)=Xv(u)+Xv(v)+Xv(w)= 0 + 1 + 0 = 1, E(Xw)=Xw(u)+Xw(v)+Xw(w) = 0 + 0 + 1 = 1, so the required matrix is
[1 1 1]
forE.
EXAMPLE 8: Let TIR2---> IR2be the linear transformation given by T(x, y)=x+2y, 3x+2y).
The matrix ofT with respect the the standard bases is
On the other hand,
T(2, 3)=(8, 12)=4(2, 3),
T(l, -1)=(-1,1)=-1(1, -1),
and the vectors Al =(2, 3), A2=(1,-1) are a basis for IR2•With respect to this basis (used twice) the matrix ofT takes the much simpler form of the diagonal matrix
Discovering such hidden symetries in a linear transformation is the subject of much of the rest ofthis book.
The preceding examples illustrate(Ihope!) that the matrix of a linear transformation T :'ll---> 'W can be exceedingly complex if some care is not exercised in the choice of the basis relative to which the matrix is computed. Indeed finding a basis relative to which the matrix is as simple as possible should clearly be one's goal ifmatrices are to simplifY our numerical computations with linear transformations.
11.2 Basic Theorems
With a number of numerical examples behind us, we turn to a more careful investigation of the relation between linear transformations and matrices.
THEOREM 11.2.1: Let 'll and 'W be finite-dimensional vector spaces. Suppose that {AI, ... , An} and {BI, ... , Bm } are ordered bases for'll and'W respectively. Then assigningtoeach linear trans- formationT :'ll---> 'Witsmatrixrelativetothese ordered bases defines anisomorphism
M :L('ll,'W)---> Matm ,n
ofthe vector space oflinear transformations from 'll to 'W with the vector space ofmxn matrices.
166 11. Representing Linear Transfonnations by Matrices PROOF: Itis clear that M is a linear transfonnation. Indeed our definition of matrix addition and scalar multiplication was rigged up so that this would be so. To show that Mis an isomorphism we will actually construct a linear map
L :Matm,n - + £('fl,'W) such that
Lã M(T)=T for allTinÊ('fl,'W), Mã L(A)=A for allAinMatm,n.
So suppose that A=(ai,j)is anm xn matrix. Define T :'fl- + 'Wtobe the linear extension of
T(AI)=al,IBI+a2,IB2+ +am,IBm, T(A2)=al,2B I+a2,2B 2+ +am,2Bm,
Thus, if
C =clAI+c2A2+ ... +cnAn
is an arbitrary vector of'fl,we have the horrendous fonnula T(C)=(al,lCI+al,2C2+ +al,ncn)BI + ...
+(am,lci+a m,2 C2+ +am,ncn)Bm
Itis immediately clear that the assignment ofthe linear transfonnation T to the matrix A defines a linear transfonnation
L :Matm,n- + £('fl, 'W).
The matrix ofL(A) with respect to the basis pair {AI,"" An} and {BI , ... , Bm } is by definition nothing but A. That is,
ML(A) =A V A EMatm,n.
On the other hand, suppose that T :'fl- + 'Wis in£(0/,'W). Then T and LM(T)are given on the basisAI, ... , An by one and the same fonnula, towit:
m
T(Aj ) =I:ai,jBi =LM(T),
i=l
j =1, ... , n.
Since At, ... , An is a basis for'll, one has for any vector a unique expression
n
C= LCjAj ,
j=t
so we have
T(C)=t, (t, a,JCj) B, =(LMXTXC),
and henceT=LM(T);that is,
LM(T)=T 'ifT EL('ll, 'W).
Therefore,Mis an isomorphism with inverseL. D
Before examining several interesting consequences of Theorem 11.2.1, we should note that the proof of this theorem is of interest in its own right. Namely, given the following data,
(1) a finite-dimensional vector space'llwith basis At, , An, (2) a finite-dimensional vector space'Wwith basisBt , , Bm ,
(3) an m xn matrixA =(ai,j),
we may manufacture a linear transformation
L(A)=T :'ll---> 'W
by the formula
whereC=2:cjAj ,or, what is the same thing,by requiring that the ma- trix ofTrelative to the ordered bases {At, ... , An} and{Bt , ... , Bm }
beA.
EXAMPLE 1: Find the value of the linear transformation T:
IRs---> P2(IR) whose matrix relative to the bases {Et ,E2,E s} and
{ 1,x, x2} is
A = [~ ~ =iJ
on the vector C =(1,1, -1)ofIRs.
168 11. Representing Linear Transformations by Matrices
SOLUTION:We have
T(C)=T(lEI+1E2-lE3 )
=T(EI )+T(E2 ) - T(E3 )
=(1+2x+3x2)+(4x)- (-1-3x+2x2)
= 1+1+2x+4x+3x+3x2- 2x2
=2+9x +x2•
Reflection on the above example and equation (*) will show that the coordinates ofT(C) relative to the ordered basis {BI , ... , Bm } appear as the entries in thecolumnvector
[ CI] [al,l al,n] [CI]
Aã : = : . : : .
c'n a~,l' .. ~m,n c~
Thus we may solve problems such as example1by matrix multiplication.
EXAMPLE 2: Let T :m3 ---...m4be the linear transformation whose matrix relative to the standard bases ofm3---... m4is
A= [-~ -i -i]
Calculate the value of T on (1, 2,--4).
SOLUTION:We have
Therefore, T(1, 2, 4)= (2, 7, -7, 6).
While it may appear strange that we calculate the value ofT(C) by using the coordinates of C to make a column matrix and that our answer appears as a column of numbers instead of a row, let us just remark that we could very well have agreed to write vectors in mn
as columns of numbers instead of rows. The reason for not doing so is that English reads from left to right. The interchange of rows and columns in the mathematical formalism is an unfortunate, though not accidental, occurrence, and is tied up with the difference between covariance and contravariance in physics. In Chinese, column vectors would be perfectly natural.
EXAMPLE 3: Let T :1R3---+IRbe the linear transformation whose matrix is
[1, -2, 1]
relative to the standard bases. Find 1(6, -4, 9).
SOLUTION: We have
T(6,-4,9)=(1, -2, 1). [~] =[6+8+9]=[23],
EXAMPLE 4: Let T :1R3---+1R4be the linear transformation with matrix
[! i ~]
relative to the standard bases of1R3and1R4• Find bases for the kernel and image of T.
SOLUTION: If(x, y,z) E1R3 , then by matrix multiplication we can find a formula for1(x, y, z):
So
T(x,y, z)=(x, y +Z,x+y, z),
and
(x, y, z) Eker(T) if and only if 0=(x, y +Z,x+y, z), so
O=x }x=o and y=O,
O=x+y o=y+z}
z =0 and y = O.
O=z
Therefore, ker(T)= {o}. To find a basis for the image of T, note that since ker(T)={O}, the image has dimension 3. Therefore, the vectors
T(1, 0, 0), 1(0, 1,0), T(0,0,1)
170 11. Representing Linear Transformations by Matrices are a basis for the image. By the definition of the matrix of a linear transformation, the components of these vectors are the columns ofthe matrix, so
(1,0, 1,0), (0, 1, 1,0), (0,1,0,1) is a basis for Im(T).
Let us return now to the consequences of Theorem 11.2.1 we hinted at previously. First we have:
COROLLARY 11.2.2: Let 'lI and 'W be finite-dimensional vec- tor spaces. Then L('lI,'W) is also finite-dimensional, and moreover, dim(L('lI,'Wằ =dim('lI). dim('W).
PROOF: This is immediate from the fact thatL('lI,'W)andMatm,n are isomorphic, wherem =dim('lI) andn =dim('W). 0
Before we can state further corollaries, we need the following very important result.
PROPOSITION 11.2.3: Letv.,'lI, and'Wbefinite-dimensional vector spaces with bases {CI , ... , Cp }, {AI, ... , An}, and{BI , ... , Bm }. Suppose that
S :V.---+ 'lI,
arelinear transformations. LetB =(bi,j) be the matrix ofS relative to the bases {CI, ... ,Cp } and{AI,' .. , An},andA=(aj,k) the matrix ofT relative to the bases{AI," ., An} and {BI, ... ,Bm }. Then the matrixofTã S relativetothe bases{CI , ... , Cp } and{BI , ... , Bm } is the matrix productAB.
PROOF: This is an immediate consequence of the definition of the
product of two matrices. 0 .
EXAMPLE 5: Let
S :m3---+m4, T:m4---+ m2
be the linear transformations whose matrices relative to the standard bases are
[ 1 0-1]
= -2 -1 -2
B 0 2 5 '
4 3 0
A = [-1 03-1]2 4 1 2'
Find the matrix of the transformation T . S :m3---+m2relative to the standard bases.
SOLUTION:According to proposition11.2.3we need only calculate the matrix productAB,which is
[-~ ~ ~ -~]. [-~ -~ =~] = [~ : ~],
and so the matrix that we seek is
[-~: ~~].
COROLLARY 11.2.4: A linear transformationT :0/---+ 0/isaniso- morphism ifand only ifits matrixA is invertible. (The matrixA may be computed relative toany pair{AI, .. " An} and {B1 , ... , Bn } of ordered bases for0/.)
PROOF: Suppose that T is an isomorphism. Let 8 :0/---+ 0/ be the linear transformation inverse to T. LetB be the matrix of 8 relative to the basis pair {B1 , ... , Bn } and {AI,' .. , An}. (N.B. We have inter- changed the roles of the bases {B1 , .•. , Bn } and {AI, ... , An}. Thus if B=(bi,j), then 8(B)=L:bi,jAi .) According to Proposition 11.2.3 the matrix productABis the matrix ofthe linear transformation T . 8 : 0/---+ 0/ relative to the basis pair {B1 , ..• , Bn } and {B1 , ••. , Bn }. But Tã 8(C)=C for all C E0/, since T and 8 are inverse isomorphisms. In particular,
Tã 8(Bj )=OBI+OB2+ ... +OBj_1+IBj+OBj+1+ +OBn
and hence the matrix ofTãS relative to the bases {B1 , , Bn } and {B1 , ... , Bn } is
1=
1
o
1 o
1
Therefore,AB =I. Likewise, according to Proposition11.2.3,the matrix product BA is the matrix of the linear transformation ST :0/---+ 0/
relative to the basis pair {AI, ... , An}and {AI,' .. , An}. But 8T(C)=
C for all C in 0/ because 8 and T are inverse isomorphisms, and hence as before, we find that BA=I. This shows that ifT :0/---+ 0/ is an isomorphism, then any matrix A representing T is invertible.
To prove the converse, we suppose that the matrix A of T is invert- ible. Let B be a matrix such that AB=1=BA. Let 8 :0/---+ 0/ be
172 11. Representing Linear Transformations by Matrices the linear transformation whose matrix relative to the ordered bases
{BI , ... , Bn } and {AI, , An} is B. (N.B. We have again inter-
changed the roles of{BI, , Bn } and {AI,"" An}.) Then the ma- trix ofS . T :0/- + 0/ relative tothe ordered bases {AI,"" An} and {AI,"" An} is
1=
1
o
1 o
1
Therefore, Sã T and 1have the same matrix relative to the bases {AI, ... , An} and {AI, ... , An},so that by Proposition 11.2.3 S . T=I;
that is,
Likewise, we see that
Sã T(C)=C
Tã S(C)=C
'ItC E0/.
'ItC E0/, so that S and T are inverse isomorphisms. D
Note that in the proof of Corollary 11.2.4 we used the fact that the ma- trix ofthe identity transformation1:0/- + 0/, which is defined by HC)= C for all C E0/, relative to the bases{AI, ... , An}and{AI, ... ,An} is the identity matrix. This is not the case if we have twodifferentbases (or even different orderings on the same basis) in0/.
EXAMPLE 6: Find the matrix of the identity linear transformation 1:m3- + m3relative to the ordered bases
and EI =(1,0,0), E2=(0, 1,0), E3=(0, 0, 1) FI =(1, 1, 1), F2 =(1, 1,0), F3 =(1, 0, 0) ofm3.
SOLUTION: We have
I(EI )=(1, 0, 0)=(1,0,0)=OFI+OF2+IF3 ,
I(E2)=(0,1,0)=(0, 1,0)=OFI+ IF2+(-I)F3 , I(E3 )=(0, 0,1)=(0, 0,1)=IFI+(-I)F2+OF3 •
So the matrix we are looking for is
The moral of the example is that appearances arereallydeceiving.
In view of Example 6 it is reasonable to expect that when we cal- culate with matrices of transfonnations T : 0/---+ 0/, we insist upon using the same ordered basis {AI, ... , An} twice to do the calcula- tion, rather than work with distinct ordered bases {AI, ... , An} and {BI, ... , Bn }. Other reasons for insisting on using only one ordered basis{AI, ... , An} when we study a transfonnation T : 0/---+ 0/from the same space to itself will appear in Chapter 14.
We turn next to another consequence of Proposition 11.2.3 which sub- stantiates a remark we made in the last chapter concerning inverses of matrices.
COROLLARY 11.2.5: Suppose thatAandBare squarematricesof size nand AB=I. Then BA=I. (Thus, tocheck thata squarematrix Bisthe inverse ofa square matrix A, we need only check oneofthe two conditions AB =1and BA =1.J
PROOF: Let T, S :mn---+ mn be the linear transfonnations whose matrices with respect to the standard basis ofmn are AandBrespec- tively. That is,
L(A)=T, L(B)=S, or, what is the same thing,
M(T)=A, M(S)=B,
where Land M are as defined in Theorem 11.2.1. Then by Proposition 11.2.3,
M(Tã S)=M(T)ã M(S)=AB= 1=M(I).
Therefore, since M is an isomorphism, TãS=I.
From this we may conclude that S :mn---+mn is an isomorphism as follows. Suppose C Eker(S). Then S(C)=0, so
o = nO) =T(S(Cằ =(T . S)(C)=I(C)=C.
Therefore S has kernel {O} so by proposition 8.4.2 we have n =dim(mn) =dim(lm(Sằ +dim(ker(Sằ =dim(lm(Sằ.
This implies that Im(S)= mn by 6.3.6 and hence S is an isomorphism by Proposition 8.6.1. Therefore there is a transfonnation
f: mn---+ mn
such that
sf=l.
174 11.Representing Linear Transformations by Matrices
Then T . (Sf)=T(I)=T
(TS) .f =If=f so that T=f;that is, ST=I .Hence
I=M(I)=M . (ST)=M(S) . M(T)=B . A as required. D
While the proof of Corollary 11.2.5 may seem complicated, we invite the reader to try to prove the result using only matrices. Even in the 3x3 case such a proof will be most painful! Corollaries 11.2.3 and 11.2.5 illustrate an important point, namely, that some results concerning linear transformations are best proved using matrices, and conversely, some results concerning matrices are best proved using linear transformations.
11.3 Change of Bases
In the study of linear transformations we will very often make use of matrix representations to prove theorems and make computations. The matrix representation for a linear transformation T :'l!---+ W depends on a choice ofbases for'l!and W. Our initial choice of these bases may be bad and unsuited to the problem at hand in that we obtain a matrix that does not convey the information we seek. Itis therefore quite natu- ral to change the bases to obtain a "better" representation for T, because to change the bases does nothing to the linear transformation, but it does change the matrix representation. Several natural and important questions therefore arise. First ofall, what is the relation (numerically, that is) between two matrix representations ofT computed with respect to different pairs of bases in 'l!and 'W? Secondly, if we are given two mxn matrices A and B, when (that is, what numerical relations must hold between them) can we conclude that they represent one and the same linear transformation T, but relative to different bases? As it turns out, both questions may be answered simultaneously. However, before doing so, we consider one more example ofthe phenomena under discussion.
EXAMPLE 1: Let T : m3---+m3be the linear transformation given by the formula
T(x, y, z)=(y+Z,x+Z,Y+x).
Calculate the matrix of T relative to
(a) the standard basis ofm3 used twice, and
(b) the basis {(1, 1, 1),(1,-1, 0),(1,1, -2)} used twice.
SOLUTION: To do part (a), we compute as follows:
T(1, 0, 0)=(0, 1, 1), T(O, 1,0)=(1, 0, 1), T(O, 0, 1)=(1, 1, 0).
Thus the desired matrix is
To do the computations for part (b), let us set Fl =(1, 1, 1), F2= (1, -1, 0), Fa=(1, 1, -2). Then we find
T(Fl )=T(1, 1, 1)=(2, 2, 2)=2Fl +OF2+OFa, T(F2)=T(I, -1, 0)=(-1,1,0) =OFl -F2+OFa, T(Fa)=T(1, 1, -2)=(-1, -1, 2)=OFl +OF2 - Fa.
So the matrix for part (b) is
B=[~-~j]
This illustrates that the matrix of a transformation depends heavily on the bases we use to compute it.
THEOREM11.3.1: Let A and B be m x n matrices, 'l! an n- dimensional vector space, and 'W an m -dimensional vector space.
ThenA and Brepresent the same linear transformation T :'l!~ 'W relativeto(perhaps) different pairsofordered bases ifand only ifthere exist nonsingularmatrices PandQsuchthat
(Note thatPisanmxm matrix andQ isannxn matrix.)
PROOF: There are two things we must prove. First, if A and B represent the same linear transformation relative to different bases of 'l!and'W,we must construct invertible matrices P andQsuch that
Second, if we are given invertible matrices P andQsuch that
176 11. Representing Linear Transformations by Matrices we must construct a linear transformation
T:o/-+'W
and pairs of ordered bases for 0/ and 'W such that A represents T relative to one pair and B represents T relative to the other. Consider the first of these. We suppose that we are given bases {AI,"" An}
and {B1 , ... , Bm } such that the matrix of T relative to these bases is A, and bases {C1 , .•. , Cn } and {D1 , ... , Dm } such that the matrix of T relative to these bases is B.
Let P be the matrix of (remember Section 11.2 Example 6) I:'W-+ 'W
relative to the bases {D1 , ... , Dm } and {Bl, ... , Bm }. Then by Corollary 11.2.4, P is invertible. LetQbe the matrix of
1:0/-+0/
relative to the bases {C1 , ... , C n }and {AI,' .. ,An}. Then by Corol- lary 11.2.4, Q is invertible, and by Proposition 11.2.3 Q-I represents the matrix of
1:0/-+0/
relative to the bases{AI,"" An}and {C1 , ... , Cn }.
Therefore by Proposition 11.2.3 p. B is the matrix of T : 0/ -+ 'W relative to the bases {Cl, ... , Cn } and {B1 , ... , Bm }. Ifwe apply Proposition 11.2.3 again we see that P . B .Q-Iis the matrix of T rel- ative to the bases {AI,.' ., An} and {B1 , ..• , Bm }. But A is also the matrix of T relative to the bases {AI,"" An} and {B1 , .•. , Bm }, so that by Theorem 11.2.1,
A = p.B.Q-I as required.
Toprove the converse, suppose we are given invertible matrices P and Q such that A=PBQ-I. Choose bases{AI, ... , An}and {B1 , ... , Bm }
for 0/ and'Wrespectively. Let T : 0/ -+ 'Wbe the linear transformation whose matrix is A relative to these bases. Let
C1=P(A1), , Cn =P(An ),
D1=Q(B1), , Dm =Q(Bm ).
Since P and Q are isomorphisms, the collections {C1 , ... , Cn } and {D1 , .•. , Dm } are bases for 0/ and 'Wrespectively. (See, for example, Propositions 8.5.2 and 8.5.3.) A bruteforce computation now shows that B is the matrix of T relative to the bases {C1 , ... , Cn } and {Dl, ... , Dm }. D
We will return to Theorem 11.3.1 in Chapters 16 and 17 where we will discuss in more detail when A and B represent the same linear transformation relative to distinct bases.
EXAMPLE 2: Let us see how Theorem 11.3.1 applies to Example 1. Recall that we are given the linear transformation T :m,3~ m,3
defined by
T(x,y,z) = (y+z, x+z,Y+x), and
[011]
A= 1 0 1 110
is the matrix of T relative to the standard basis of IR3,while
[ 2 0 0]
B= 0 -1 0 o 0-1
is the matrix of T relative to the basis {(1, 1, 1), (1, -1, 0), (1, 1, -2)} of
m,3.
According to Theorem 11.3.1 there are invertible matrices P and Q such that A= P . B . Q-I. Our task is to compute P and Q-I. The proof of Theorem 11.3.1 tells us how. Namely:
(1) P is the matrix of I :IR3~m,3 relative to the basis pair {(1, 1, 1), (1, -1,0), (1, 1, -2)} and {(1, 0, 0), (0, 1,0), (0, 0,1)}.
(2) Q-I is the matrix ofI :m,3~ IR3relative to the basis pair {(1, 0, 0), (0, 1,0), (0, 0,1)}and {(1, 1, 1), (1, -1, 0), (1, 1,-2)}.
The computation of P is easy and gives us
[
1 1 1]
P= 1 -1 1.
1 0-2
The computation ofQ-I is not hard and depends on the following equa- tions,
1 1 1
(1,0,0)=3(1,1,1)+2(1, -1, 0)+6(1, 1,-2), 1 1 1
(0,1,0)=3(1, 1, 1) - 2(1, -1, 0)+6(1, 1, -2),
1 1
(0,0, 1)= 3(1, 1, 1)+0(1, -1, 0) - 3(1, 1, -2),