A First Course in Linear Algebra, 2017a

520 Vector Spaces
We now discuss the main result of this section, that is how to represent a linear transformation with
respect to different bases.
Let V and W be finite dimensional vector spaces, and suppose
•dim(V)=n and B 1 = { ⃗ b 1 , ⃗ b 2 ,..., ⃗ b n } is an ordered basis of V;
•dim(W)=m and B 2 is an ordered basis of W.
Let T : V → W be a linear transformation. If V = R n and W = R m ,thenwecanfindamatrixA so that
T A = T . For arbitrary vector spaces V and W, our goal is to represent T as a matrix., i.e., find a matrix A
so that T A : R n → R m and T A = C B2 TC −1
B 1
.
To find the matrix A:
T A = C B2 TC −1
B 1
implies that T A C B1 = C B2 T ,
and thus for any⃗v ∈ V ,
C B2 [T(⃗v)] = T A [C B1 (⃗v)] = AC B1 (⃗v).
Since C B1 ( ⃗ b j )=⃗e j for each ⃗ b j ∈ B 1 , AC B1 ( ⃗ b j )=A⃗e j , which is simply the j th column of A. Therefore,
the j th column of A is equal to C B2 [T (⃗b j )].
The matrix of T corresponding to the ordered bases B 1 and B 2 is denoted M B2 B 1
(T ) and is given by
This result is given in the following theorem.
M B2 B 1
(T )= [ C B2 [T ( ⃗ b 1 )] C B2 [T( ⃗ b 2 )] ··· C B2 [T( ⃗ b n )] ] .
Theorem 9.90:
Let V and W be vectors spaces of dimension n and m respectively, with B 1 = { ⃗ b 1 , ⃗ b 2 ,..., ⃗ b n } an
ordered basis of V and B 2 an ordered basis of W. Suppose T : V → W is a linear transformation.
Then the unique matrix M B2 B 1
(T) of T corresponding to B 1 and B 2 is given by
M B2 B 1
(T )= [ C B2 [T (⃗b 1 )] C B2 [T (⃗b 2 )] ··· C B2 [T(⃗b n )] ] .
This matrix satisfies C B2 [T (⃗v)] = M B2 B 1
(T )C B1 (⃗v) for all⃗v ∈ V.
We demonstrate this content in the following examples.