# Similarity

• Oct 17th 2008, 07:10 AM
slevvio
Similarity
Hello, I was hoping someone could help me with this question.

Prove that the matrices

$A = \begin{bmatrix} -14 & -27 & -24 \\ 4 & 9 & 6 \\ 6 & 10 & 11 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 1 & 1\\ 0 & 2 & 1 \\ 0 & 0 & 3\end{bmatrix}$

are similar, and find an invertible matrix $P$ such that $P^{-1}AP = B$.
[Hint: There is no need to work out the characteristic polynomial of A as similar matrices have the same characteristic polynomial. Show that each of the matrices A,B is similar to the same diagonal matrix.]

I worked out that B was similar to diag (1,2,3) by calculating the eigenvectors but I have no idea how to show now that A is similar to diag (1,2,3) without calculating the characteristic polynomial and working out all the eigenvectors. Is it enough that the trace of A and the trace of B, as well as the determinant of A and the determinant of B, are equal?
• Oct 17th 2008, 07:33 AM
Laurent
Quote:

Originally Posted by slevvio
I worked out that B was similar to diag (1,2,3)

Before anything else: the eigenvalues of B are 1 (twice) and 4, so you should first check your computation (or your post, more probably, since the eigenvalues of A are 1,2,3).

That said, the determinant and trace aren't enough: they give you the sum and the product of the three eigenvalues, and you would need a third information to find them.

I don't think you can avoid computing the roots of the characteristic polynomial. To show that the matrices are similar, you don't need to find the eigenvectors because the eigenvalues are different and hence the matrices are diagonalizable. To find the matrix $P$, you'll need them however.
• Oct 17th 2008, 10:25 AM
slevvio
Thanks a lot -- my post is wrong i have an extra 2 in the matrix