### Change of Basis

##### 3. Matrix for linear transformation

Let T: VW be a linear transformation. Let B = {b1, b2, ..., bn} and C = {c1, c2, ..., cm} be bases of V and W. Then by isomorphisms [?]B: VRn, [?]C: WRm, T is translated to a linear transformation [?]C T [?]B-1: RnRm. The linear transformation [?]C T [?]B-1 is given by a matrix A, which we call matrix of the linear transformation with respect to the bases B and C.

Since

i-th column of A
= [?]C T [?]B-1(ei) (this formula)
= [?]C T(bi) ([bi]B = ei)
= [T(bi)]C,

we conclude the following formula for A.

The matrix of a linear transformation T: VW with respect to bases B = {b1, b2, ..., bn} and C = {c1, c2, ..., cm} of V and W is

[T(B)]C = [ [T(b1)]C [T(b2)]C ... [T(bn)]C ].

The notation [T(B)]C in introduced to suggest that the matrix is obtained by the following steps:

1. Apply T to vectors in B.
2. Take the coordinates with respect to C.

For an explicit example, in case m = 3, n = 2 and

T(b1) = a11c1 + a12c2 + a13c3,
T(b2) = a21c1 + a22c2 + a23c3,

the matrix of T is

 [ a11 a21 ]. a12 a22 a13 a23

More concrete examples are given here, in which V = W = P2 and B = C = {1, t, t2}, and given in this exercise and this exercise, in which V = W = Pn and B = C = {1, t, t2, ..., tn}.

Example Consider the linear transformation

 T(p(t)) = [ p(0) p(1) ]: P3 → M(2, 2). p'(0) p'(1)

We use the bases

 B = Hermite = {1, t, t2 - 1, t3 - 3t}, C = { [ 1 0 ], [ 0 1 ], [ 0 0 ], [ 0 0 ] } 0 0 0 0 1 0 0 1

for P3 and M(2, 2). Then

 T(1) = [ 1 1 ], T(t) = [ 0 1 ], T(t2 - 1) = [ -1 0 ], T(t3 - 3t) = [ 0 -2 ] 0 0 1 1 0 0 -3 -3

tell us the matrix for T with respect to the bases is

 [T(Hermite)]C = [ [T(1)]C [T(t)]C [T(2t2)]C [T(4t3 - 3t)]C ] = [ 1 0 -1 0 ]. 1 1 0 -2 0 1 0 -3 0 1 0 -3

We may conclude, for example, that rankT = 3 from the matrix.

The following lists some properties of the matrix for linear transformation.

linearity
[(T + S)(B)]C = [T(B)]C + [S(B)]C, [cT(B)]C = c[T(B)]C
transitivity
[T(C)]D [T(B)]C = [T(B)]D
inverse
[T(B)]C-1 = [T-1(C)]B
identity
[id(B)]B = I