Prove that every real-valued matrix is similar to its transpose ( w/o using complex numbers field)

Asked May 18 '17 at 18:20

Active Dec 09 '21 at 19:34

Viewed 1,215 times

As in the title of question, I'd like to ask for a proof of the following statement:

For every real-valued $n\times n$ matrix $A$ there exists an invertible, real-valued $n \times n$ matrix $P$ such, that $A = P^{-1} A^T P$.

My professor told me that it is entirely possible to do it without using whole complex numbers apparatus. However, I'm not sure how to do it since $\mathbb{R}$ is not algebraically closed.

I know that similar topics have already appeared, i.e. here or here. None of them, unfortunately, fully answers my question.

edited May 18 '17 at 19:29

asked May 18 '17 at 18:20

I_Really_Want_To_Heal_Myself

Presumably you meant to write $A = P^{-1}A^T P$? – NickD May 18 '17 at 18:30
Oh yes, thanks for pointing it out. – I_Really_Want_To_Heal_Myself May 18 '17 at 19:28
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Marc's Answer from your second link does not rely on the algebraic closure. Is your problem with it not being familiar with the Smith Normal Form? – Ken Duna May 18 '17 at 21:13
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This was asked here as well and it's been concluded that the Smith Normal form is the way to go @UsernameUnknown. It seems unlikely that there is going to be any simpler method. – Sarvesh Ravichandran Iyer Dec 11 '21 at 12:42

Prove that every real-valued matrix is similar to its transpose ( w/o using complex numbers field)

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