# (AB) = (BA)^t

• April 25th 2013, 08:44 AM
Wyaltster
(AB) = (BA)^t
Hello,

This may be a long shot, but does anyone know anything that can be said of square matrices A and B with the property above? A and B are block components of a larger matrix that I'm trying to find the determinant of. Simultaneously diagonalizing them would be my dream come true, but any information would be helpful.

Thanks.
• April 25th 2013, 01:59 PM
mathguy25
Re: (AB) = (BA)^t
If A and B are symmetric, then AB = (BA)^T. Can you say that A and B are symmetric?
• April 25th 2013, 03:42 PM
majamin
Re: (AB) = (BA)^t
Quote:

Originally Posted by Wyaltster
Hello,

This may be a long shot, but does anyone know anything that can be said of square matrices A and B with the property above? A and B are block components of a larger matrix that I'm trying to find the determinant of. Simultaneously diagonalizing them would be my dream come true, but any information would be helpful.

Thanks.

Suppose AB is diagonalized. Assuming $A^{-1}$ exists, and AB is diagonalized as $AB = P^{-1}DP$. Now,

$B = A^{-1} P^{-1}DP$
$BA=A^{-1} P^{-1}D(PA)$
$BA=(PA)^{-1}D(PA)$
$(BA)^T=((PA)^{-1}D(PA))^T = (PA)^T D ((PA)^T)^{-1}$

Since $(A^{-1})^T = (A^T)^{-1}$ and $(AB)^T = B^T A^T$.
• April 25th 2013, 11:25 PM
Wyaltster
Re: (AB) = (BA)^t
Thanks, both of you.

mathguy, they're both skew-symmetric, so bingo :)
I've been seeing this a lot because I've been dealing with a lot of skew-symmetric matrices. Feel like a bonehead for not seeing that.

majamin, thanks, that's interesting. I might have been confusing when I said simultaneously diagonalizable. I'd want to simultaneously diagonalize $A$ and $B$, not $AB$ and $(BA)^t$, but I think what I want to do can't be done. There's a theorem that says that a set of diagonalize matrices commute if and only if they can be diagonalized by a common matrix. That's not applicable to my case, but I was hoping maybe something could be said about $A$ and $B$ as described above. Turns out I can say they're skew-symmetric.