# Product of two symmetric banded matrices - real eigenvalues?

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?

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It is true if at least one of the matrices is positive semi-definite. – Algebraic Pavel Jun 12 '14 at 0:22
Could you sketch out why that is true? – science404 Nov 7 '14 at 17:26
If $A$ is positive definite, then $AB$ has the same eigenvalues as $A^{-1/2}(AB)A^{1/2}=A^{1/2}BA^{1/2}=:C$. Since $C$ is symmetric, it has real eigenvalues and so does $AB$. It is still true if $A$ is only semidefinite but the proof is slightly more complicated. AFAIK it can be found in the Horn's and Johnson's Matrix Analysis. – Algebraic Pavel Nov 7 '14 at 17:29

No, $\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix} \begin{pmatrix}0 & 1\\ 1 & 0\end{pmatrix}= \begin{pmatrix}0 & 1\\ -1 & 0\end{pmatrix}$.
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 '14 at 23:18