Let $A,B$ be symmetric real matrices. Is $AB$ similar to a symmetric matrix?
This is a problem in my exam. Not a conjecture :v
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T C
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Oh sorry, it's actually in my exam – T C Dec 11 '16 at 14:13
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In fact AB is symmetric only if A and B commute. – Dec 11 '16 at 14:17
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Yeah. The difficulty here is that they do not – T C Dec 11 '16 at 14:17
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@Rohan as said for example in item 5 of (https://math.vanderbilt.edu/sapirmv/msapir/jan22.html). – Jean Marie Dec 11 '16 at 14:21
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@Rohan ... but the question of the OP is "is AB similar to a symmetric matrix... – Jean Marie Dec 11 '16 at 14:23
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Can people show me a solution and then downvote me later ? – T C Dec 11 '16 at 14:24
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@chítrungchâu What have you tried? – πr8 Dec 11 '16 at 14:53
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Somewhat related (http://mathoverflow.net/q/106191)(http://math.stackexchange.com/q/982797). – Jean Marie Dec 11 '16 at 18:07
1 Answers
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Here is a counter-example of two symmetric matrices $A$, $B$ whose product, besides being non symmetrical, cannot be similar to a symmetric matrix.
Consider matrices
$$A=\pmatrix{1&2\\2&1} \ \ \ \text{and} \ \ \ B=\pmatrix{1&0\\0&-1}.$$
$AB=\pmatrix{1&-2\\2&-1}.$ which is non symmetric.
Moreover, the characteristic polynomial of $AB$ is $\lambda^2+3$: thus, the eigenvalues of $AB$ are $\pm i \sqrt{3}$. If it was similar to a symmetric matrix, it would have the same real eigenvalues.
Jean Marie
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