MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A famous problem in linear algebra is that "A $n\times n$ matrix $A$ over a field $\mathbb{F}$ is similar to its transpose $A^T$."

I know one proof using Smith Normal form. However, I want to know an elementary proof avoiding any concepts related to SNF.

My question is: "Is there an elementary way to prove this?"

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 $$A=PA^T P^{-1}.$$

share|cite|improve this question
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. – Mariano Suárez-Alvarez Feb 19 '13 at 19:58
@Mariano: Yeah, it might not be possible to avoid those to prove it. The matrix P coming from Jordan canonical form proof is already complicated enough. Maybe requiring to provide explicit P is not possible without those. The generalized eigenvectors are involved and an inversion that transforms Jordan block to its transpose. – i707107 Feb 19 '13 at 21:08
There is a proof using rational canonical form, which basically just needs the fact that $A$ and $A^{t}$ have the same minimum polynomial, but I don't think that will be considered explicit .enough – Geoff Robinson Feb 19 '13 at 21:40
In the two by two case $P$ can always be chosen to be symmetric. Is this true in general? – Tom Goodwillie Feb 19 '13 at 23:21
@Geoff: the fact that $A$ and $A^T$ have the same minimal polynomial is obvious: if $p(A)=0$, then $p(A^T)=p(A)^T=0$. – Mark Sapir Feb 20 '13 at 4:39

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.