math111_logo Change of Basis

2. Computing the matrix for change of coordinates

Let B = {b1, b2, b3} and C = {c1, c2, c3} be two bases of the euclidean space R3 (or a subspace of R3). Then they induce matrices [B] = [b1 b2 b3], [C] = [c1 c2 c3]. As we have seen in the proof of the formula [v]C = [B]C[v]B, the matrix [B]C for change of coordinates fits into

[B] = [C][B]C.

In particular, if we take B to be the standard basis of R3, so that [B] = I and [v]B = v, then the formula [v]C = [B]C[v]B becomes [v]C = [C]-1v. This is the earlier formula for the coordinates with respect to any basis of Rn.

The equality [B] = [C][B]C shows that the columns of the matrix [B]C are the solutions of the system [C]x = bi, where bi are the columns of [B]. The combined augmented matrix of these systems is the matrix [ [C] [B] ]. This suggests that the computation of [B]C is reduced to row operations on [ [C] [B] ].

Example Let

A = [ 1 3 2 0 1 ].
-1 -1 -1 1 0
0 4 2 4 3
1 3 2 -2 0

In this example, we found a basis B = {(1, -1, 0, 1), (3, -1, 4, 3), (0, 1, 4, -2)} of colA. In this example, we found another basis C = {(1, -1, 0, 1), (0, 1, 2, 0), (0, 0, 1, -1)} of colA. To find the matrix [B]C for changing the B-coordinates to C-coordiantes, we use the fact that the i-th column of [B]C is the solution of [C]x = i-th column of [B]. Thus we consider the combined augmented matrix.

[ [C] [B] ] = [ 1 0 0 1 3 0 ]
-1 1 0 -1 -1 1
0 2 1 0 4 4
1 0 -1 1 3 -2

Row operations change this into

[ 1 0 0 1 3 0 ].
0 1 0 0 2 1
0 0 1 0 0 2
0 0 0 0 0 0

Then it is easy to see that the columns in heavily shaded submatrix are the solutions of [C]x = bi, and

[B]C = [ 1 3 0 ].
0 2 1
0 0 2

Now by an earlier computation, we have

[(0, 1, 2, 0)]B = (-3/2, 1/2, 0).

Then

[(0, 1, 2, 0)]C = [ 1 3 0 ] [ -3/2 ] = [ 0 ].
0 2 1 1/2 1
0 0 2 0 0

This is consistent with the fact that the second vector in C is (0, 1, 2, 0).

Example The following are two bases for the vector space P3 .

Chebyshev = {1, t, 2t2, 4t3 - 3t}, Hermite = {1, t, t2 - 1, t3 - 3t}

To find the matrix for changing the Chebyshev coordinates to Hermite coordinates, we use the translation [a + bt + ct2 + dt3] = (a, b, c, d) and the method of the previous example. Thus we form the matrix

[ [Hermite] [Chebychev] ] = [ 1 0 -1 0 1 0 -1 0 ].
0 1 0 -3 0 1 0 -3
0 0 1 0 0 0 2 0
0 0 0 1 0 0 0 4

Row operations give us

[ 1 0 0 0 1 0 2 0 ].
0 1 0 0 0 1 0 9
0 0 1 0 0 0 2 0
0 0 0 1 0 0 0 4

Thus we conclude that the Hermite coordinates may be obtained by multiplying

[Chebychev]Hermite = [ 1 0 2 0 ]
0 1 0 9
0 0 2 0
0 0 0 4

to (the left of) the Chebychev coordinates.


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