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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 at 0:22
    
Could you sketch out why that is true? –  science404 Nov 7 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 at 17:29

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} $.

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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

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