Similar matrices

The relation between a square matrix A and its diagonalized form (when there is a diagonalized form, that is) is a special case of a mathematical relation called similarity.

A square matrix A is similar to another square matrix B if there is an invertible square matrix P with B = P–1AP.

Properties of similar matrices

For any n x n matrices A, B and C

A is similar to itself.

Proof. A = I–1AI

If A is similar to B, then B is similar to A.

Proof. If B = P–1AP, then PBP–1 = PP–1APP–1, i.e. (P–1)–1B(P–1) = A. Then B is similar to A.

If A is similar to B and B is similar to C, then A is similar to C.

Proof. If A is similar to B, then B = P–1AP for some matrix P. If B is similar to C, then C = Q–1BQ for some matrix Q.

Then C = Q–1P–1APQ = (PQ)–1A(PQ), so A is similar to C.

If A and B are similar and invertible, then A–1 and B–1 are similar.

Proof. If A and B are similar, then B = P–1AP. Since all the matrices are invertible, we can take the inverse of both sides: B–1 = (P–1AP)–1 = P–1A–1(P–1)–1 = P–1A–1P, so A–1 and B–1 are similar.

If A and B are similar, so are Ak and Bk for any k = 1, 2, ... .

Proof. If B = P–1AP, then

Bk	= (P–1AP)k = (P–1AP)(P–1AP) ... (P–1AP) (k times) = P–1A(PP–1)A(PP–1) ... (PP–1)AP = P–1AIAI...IAP = P–1AkP

Then Ak and Bk are similar.

If A and B are similar and q is any polynomial, then q(A) and q(B) are similar.

Proof. Suppose A and B are similar, i.e. B = P–1AP for some matrix P.

Suppose that q(t) = c0 + c1t + c2t2 + ... + cktk. Then

p(A) = c0I + c1A + c2A2 + ... + ckAk

P–1q(A)P

= P–1(c0 + c1A + c2A2 + ... + ckAk)P
= c0P–1P + c1P–1AP + c2P-1A2P + ... + ckP–1AkP
= c0I + c1B + c2B2 + ... + ckBk (from the previous proof)
= q(B)

So q(A) and q(B) are similar.

If A and B are similar, so are AT and BT.

Proof. If B = P–1AP, then BT = (P–1AP)T = PTAT(P–1)T = [(P–1)T]–1AT[(P–1)T]. So for Q = [(P–1)T], BT = Q–1ATQ. That means that AT and BT are similar.

Similar matrices share many similarity invariants.

Similar matrices have the same

a) determinant and invertibility

b) characteristic equation and eigenvalues

c) eigenspace dimension corresponding to each common eigenvalue

d) trace

e) rank and nullity

Proof (of the first two only). Suppose that A and B are similar, i.e. that B = P–1AP for some matrix P.

a) det(B)

= det(P–1AP)
= det(P–1)det(A)det(P)
= [det(P)]–1det(A)det(P)
= det(A)

so the matrices have the same determinant, and one is invertible if the other is.

b) det(B – λI)

= det(P–1AP – λP–1P)
= det[P–1(A – λI)P]
= det(P–1)det(A – λI)det(P)
= det(A – λI)

so the matrices have the same characteristic polynomial and hence the same eigenvalues.

Note that similar matrices will not generally have the same eigenvectors.