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A famous result in linear algebra is the following.

An $n \times n$ matrix $A$ over a field $\mathbb{F}$ is similar to its transpose $A^T$.

I know one proof using the Smith Normal Form (SNF). However, I want to find an elementary proof avoiding any concepts related to the SNF. My question is: is there an elementary way to prove this?

The requirements are:

  • Do NOT use the structure theorem over PID.

  • Do NOT use the Smith Normal form (nor Jordan canonical form).

  • Do NOT use the concept of invariant factors.

  • Provide an explicit invertible matrix $P$ such that $A=PA^T P^{-1}$.

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    $\begingroup$ This is a lot like looking for a proof which does not need the letter a in order to be written down :-) What exactly do you mean by elementary? The Jordan canonical form is basic linear algebra, really. $\endgroup$
    – Mariano Suárez-Álvarez
    Commented Feb 19, 2013 at 19:58
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    $\begingroup$ It's easy to prove using normal forms, but you can reframe it as the statement that if $A:V\to V$ is linear then there is a nondegenerate bilinear form on $V$ such that $\langle Av,w\rangle=\langle v, Aw\rangle$ identically. Is there any chance of a proof of this that is quite different from the normal-form proofs? $\endgroup$
    – Tom Goodwillie
    Commented Feb 19, 2013 at 23:19
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    $\begingroup$ In the two by two case $P$ can always be chosen to be symmetric. Is this true in general? $\endgroup$
    – Tom Goodwillie
    Commented Feb 19, 2013 at 23:21
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    $\begingroup$ @Geoff: the fact that $A$ and $A^T$ have the same minimal polynomial is obvious: if $p(A)=0$, then $p(A^T)=p(A)^T=0$. $\endgroup$
    – user6976
    Commented Feb 20, 2013 at 4:39
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    $\begingroup$ After many years this question was first asked, I cannot find anything meeting all of my requirement. Actually, finding Smith Normal Form can be done through elementary row/column operations. So, it is already elementary enough, and I do not have any reason left for avoiding such a simple and powerful tool. Also, SNF provides not only the existence of $P$, but also an explicit way to write down a such $P$. $\endgroup$
    – Sungjin Kim
    Commented May 30, 2018 at 22:24

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