Matrix similarity transformations introduce small imaginary eigenvalue components

So I'm having an issue with numerical precision. To boil my problem down...

Let A = eye(3) + 1e-6*rand(3) (1)

and let eig(A) return all real eigenvalues (empirically, this happens about 80% the time when I run (1)). I then apply a unitary similarity transformation, B = QAQ*. Now, eig(B) has a small imaginary component, which is a problem for my application.

Is it preferable to symmetrize A prior to the similarity transformation, or to symmetrize B after the similarity transformation...?

I'm not familiar with any theorems that relate the real part of a matrix's eigenvalues to the eigenvalues of its symmetrized form.

Any thoughts?

Thanks!

[David Goodmanson]

Hi Anish, if the unitary matrix Q is known, can you just skip the eigenvalue calculation on B? The eigenvalues of A and B are identical, and if

[V_A lambda] = eig(A)

then the similar eigenvector matrix for B is

   V_B = Q*V_A

(in comparing results, eig(A) and eig(B) may have eigenvalues in different order and their eigenvector matrix columns in the same different order, but that is a bookkeeping detail).

[Anish Potnis]

Hey, thanks for the response. I understood that eigenvalues are preserved under similarity transformations, I just sort of boiled my problem down to help me understand conceptually what exactly the issue is with the numerical side of things. I'm doing a bunch of other kinds of transformations in my code, but I wanted to understand a simplified example first.

Answer 1

[Christine Tobler]

If you know the matrix A is supposed to be symmetric, I'd say symmetrize A, and then symmetrize B again, because Q*A*Q' will introduce a small numerical error and not be exactly symmetrical.

Although if you can directly transform the eigenvalues as David suggest, that would probably be best. It depends on whether the matrix A is meant to be symmetric, I'd say.