# Thread: Similar Matrices

1. ## Similar Matrices

1.) Given two matrices, A and B, show that
A = [[1,1],[-1,4]], and
B =[[2,1],[1,3]] are similar.

Then, show that the matrix
C = [[3,1],[-6,-2]] and
D = [[-1,2],[1,0]] are NOT similar.

Could you check to see if I am right? The theorem for similar matrices states that if n x n matrices A and B are similar, then they have the same characteristic polynomial and hence the same eigenvalues (with the same multiplicities).

Or: Two n × n matrices A and B are similar, if there exists a non-singular n × n matrix P such that B = P−1AP.

For A,

(1*4) - (-1*1) = 5, so it is non-singular.

And B,

(2*3) - (1*1) = 5, so it is non-singular

(A - LI)x = 0

A = [[1-L, 1],[-1,4-L]]

(1-L)*(4-L) - (-1) = 0

Solving for L, we get the eigenvalues are 5/2+1/2*sqrt(5), 5/2-1/2*sqrt(5)

B = [[2-L, 1],[1, 3-L]]

(2-L)*(3-L) - 1*1 = 0

Solving for L, we get the eigenvalues are the same as above.

Is this sufficient to show that they are similar, since they both have the same eigenvalues?

To show C and D are not similar:

For C:

(3-L)*(-2-L) - (-6) = 0

Solving for L, we get the eigenvalues are 0 and 1.

For D:

(-1-L)*(0-L) - 2 = 0

Solving for L, we get the eigenvalues are 1 and -2.

Is this sufficient to prove that they are not similar, since they don't have the same eigenvalues?

2. Apparently its sufficient enough to show that the eigenvalues are not the same to show the matrices are not similar; however, it is not sufficient enough to show that if the eigenvalues are the same, the matrices are similar. How would I do it with the P^(-1)AP thing?