### Change of Basis

##### 5. Similar matrix

To build the matrix for a linear transformation T: VV from a vector space to itself, we naturally want to choose one basis B = {b1, b2, ..., bn} for both copies of V. Then by this formula, the matrix for T with respect to B is [T(B)]B.

Let B' be another basis of V. Then by this formula, the matrices of T with respect to B and B' are related by the equality

[T(B')]B' = [B]B' [T(B)]B [B]B'-1.

The relation leads to the following definition.

Two square matrices A and B are said to be similar if there is an invertible P, such that B = PAP-1.

Thus the matrix of a self linear transformation is changed by similarity when the basis is changed.

Example Consider the linear transformation

T(x, y) = (13x - 4y, - 4x + 7y): R2R2

from R2 to itself. Since x and y are the coordinates with respect to the standard basis B = {e1, e2} = {(1, 0), (0, 1)}, the matrix

 A = [T(B)]B = [ 13 -4 ] -4 7

that we usually associate to T is really the matrix of T with respect to the standard basis. If we choose a different basis B' = {u1, u2} = {(1, 2), (2, -1)}. Then

T(u1) = (13×1 - 4×2, - 4×1 + 7×2) = (5, 10) = 5u1
T(u2) = (13×2 - 4×(-1), - 4×2 + 7×(-1)) = (30, -15) = 15u2

tell us the matrix for T with respect to B' is

 B = [T(B')]B' = [ 5 0 ]. 0 15

Denote

 P = [B']B = [u1 u2] = [ 1 2 ]. 2 -1

It is easy to directly verify that A = PBP-1.

We note that the original formula for T (with respect to {e1, e2}) is too complicated for us to have a good understanding of what T is doing on the plane R2. However, if we look through the basis {u1, u2} instead, then we have a very good understanding of the transformation. Basically T rescale the u1 direction by 5 times and the u2 direction by 15 times.

The example shows the importance of similarity. To gain a good undertanding of a self linear transformation, we find a basis (a suitable invertible matrix P) so that the the transformation behaves particularly nice with respect to the basis (the similarity change via P gives a diagonal matrix).

Example Consider the linear transformation

T(p) = (-1/3)(1 - t )2p''' + (- 1 + t)p'' - 2(1 + t)p' + 6p: P3P3.

From

T(1) = 6
T(t) = - 2 + 4t
T(t2) = - 2 - 2t + 2t2
T(t3) = - 2 - 2t - 2t2

the linear transformation has the following matrix with respect to the usual choice {1, t, t2, t3} of basis.

 A = [ 6 -2 -2 -2 ] 0 4 -2 -2 0 0 2 -2 0 0 0 0

To see more clearly what the transformation does to polynomials, we consider a new basis {1, 1 + t, 1 + t + t2, 1 + t + t2 + t3}. From

T(1) = 6 = 6(1)
T(1 + t) = 4(1 + t)
T(1 + t + t2) = 2(1 + t + t2)
T(t3) = 0 = 0(1 + t + t2 + t3)

the matrix with respect to the new basis is

 B = [ 6 0 0 0 ]. 0 4 0 0 0 0 2 0 0 0 0 0

Geometrically, this means that T is rescaling by factors 6, 4, 2, 0 in the new basis directions.

The matrix for changing the coordinates in the new basis to the coordinates in the usual basis is

 Q = [{1, 1 + t, 1 + t + t2, 1 + t + t2 + t3}]{1, t, t2, t3} = [ 1 1 1 1 ]. 0 1 1 1 0 0 1 1 0 0 0 1

Direct verification shows B = Q-1AQ (by [B]B' = [B']B-1, this Q is the inverse of P above).