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I have two $n\times n$ symmetric matrices.

I know that $\operatorname{tr}(AB)=\operatorname{tr}(BA)$. But it does not mean $AB=BA$. I wonder what is the condition for $AB=BA$.

Adrian Keister
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Simon Lee
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    The condition for $AB=BA$ is... $AB=BA$. Irony aside, I mean that there is not a general characterization, according to my knowledge, at least. – Giuseppe Negro Oct 06 '18 at 19:41
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    Symmetric matrices (presumably real) are diagonalizable. They are simultaneously diagonalizable if and only if they commute. But, this fact is not really useful for detecting $AB=BA$. It is very useful for many other purposes though. – Jyrki Lahtonen Oct 06 '18 at 19:43
  • Jyrki: What about the identity and any symmetric matrix that is not diagonal? These commute, but are not simultaneously diagonalizable. I don't think that condition is if and only if. – kholli Oct 06 '18 at 19:56
  • @kholli They are certainly simultaneously diagonalizable! The identity matrix is diagonalized by any change of a basis, so we only need to worry about the other matrix :-) – Jyrki Lahtonen Oct 06 '18 at 20:02
  • @JyrkiLahtonen Oh, yes, being silly. Good point. – kholli Oct 06 '18 at 20:07
  • https://en.wikipedia.org/wiki/Commuting_matrices

    I saw this article, but I can get neither head nor tail of it, as I know whatever is there in the pdf in this link: 1) https://ncert.nic.in/ncerts/l/lemh103.pdf 2) https://ncert.nic.in/ncerts/l/lemh104.pdf but it might be of help to y'all.

     – Harikrishnan M Dec 22 '23 at 10:31

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