Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $A$ and $B$ be real symmetric banded matrices but $AB$ is not symmetric. Are the eigenvalues of $AB$ real?

A more specific case: let $D$ be a real diagonal matrix, $B$ real symmetric and banded, and $DB$ is not symmetric. Are the eigenvalues of $DB$ real?

share|improve this question
It is true if at least one of the matrices is positive semi-definite. –  Algebraic Pavel Jun 12 at 0:22

1 Answer 1

No, $ \begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix} \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}= \begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix} $.

share|improve this answer
Thanks for that counter-example. I actually am interested in banded symmetric matrices, so I will edit the original question. –  science404 Jun 11 at 23:15
I don't know what you mean by banded, tridiagonal? Are my matrices not banded, if they are just $2\times 2$? –  Peter Franek Jun 11 at 23:18
Yes, tridiagonal, pentadiagonal, etc. –  science404 Jun 11 at 23:21
Actually, banded isn't quite right, because I'm considering also block diagonal, block tridiagonal, etc. Basically, the matrix will be sparse and non-zero values concentrated near the main diagonal. The counter-example you gave was anti-diagonal. I'm not consider such matrices (anti-triadiagonal, and so on). –  science404 Jun 11 at 23:24

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.