Page 224 - Linear Algebra Done Right

Change of Basis
for both vector spaces (after all, the two vector spaces in question are
equal). Thus we usually refer to the matrix of an operator with respect
to a basis, meaning that we are using one basis in two capacities. The 215
next proposition is one of the rare cases where we need to use two
different bases even though we have an operator from a vector space
to itself.
Let’s review how matrix multiplication interacts with multiplication
of linear maps. Suppose that along with V we have two other finite-
dimensional vector spaces, say U and W. Let (u 1 ,...,u p ) be a basis
of U, let (v 1 ,...,v n ) be a basis of V, and let (w 1 ,...,w m ) be a basis
of W.If T ∈L(U, V) and S ∈L(V, W), then ST ∈L(U, W) and

10.1 M ST, (u 1 ,...,u p ), (w 1 ,...,w m ) =

M S, (v 1 ,...,v n ), (w 1 ,...,w m ) M T, (u 1 ,...,u p ), (v 1 ,...,v n ) .
The equation above holds because we defined matrix multiplication to
make it true—see 3.11 and the material following it.
The following proposition deals with the matrix of the identity op-
erator when we use two different bases. Note that the k th column of

M I, (u 1 ,...,u n ), (v 1 ,...,v n ) consists of the scalars needed to write
u k as a linear combination of the v’s. As an example of the proposi-

2
tion below, consider the bases (4, 2), (5, 3) and (1, 0), (0, 1) of F .
Obviously

4 5

M I, (4, 2), (5, 3) , (1, 0), (0, 1) = .
2 3

The inverse of the matrix above is 3/2 −5/2 , as you should verify. Thus
2
−1
the proposition below implies that

3/2 −5/2
M I, (1, 0), (0, 1) , (4, 2), (5, 3) = .
−1 2
10.2 Proposition: If (u 1 ,...,u n ) and (v 1 ,...,v n ) are bases of V,

then M I, (u 1 ,...,u n ), (v 1 ,...,v n ) is invertible and
−1
M I, (u 1 ,...,u n ), (v 1 ,...,v n ) =M I, (v 1 ,...,v n ), (u 1 ,...,u n ) .

Proof: In 10.1, replace U and W with V, replace w j with u j , and
replace S and T with I, getting