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Is there any manner to prove that an inversible real matrix is conjugate to its transpose using the polar decomposition ? This would be, in this particular case, a much more simpler proof than using invariants in the general case.

Tig la Pomme
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  • What do you mean, "conjugate to its transpose"? Do you mean, $A^\top = \overline{A}$, where $\overline{A}$ is the matrix formed by taking complex conjugates of each entry? If so, this is equivalent to being Hermitian, and is certainly not true for general invertable matrices. – Theo Bendit May 11 '22 at 10:58
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    No, I meant $$P^{-1}AP=A^T$$ for some invertible matrix P – Tig la Pomme May 11 '22 at 11:00
  • Good to know. For what it's worth, I would have used the word "similar" instead. – Theo Bendit May 11 '22 at 11:05
  • @Tig There's no way to make use of the polar decomposition for this proof that I know of or can think of. See this post for a few options for how to go about this proof. Indeed, the fastest approach is to use some kind of invariant or canonical form. – Ben Grossmann May 11 '22 at 13:43

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