If A and B are symmetric, then AB = (BA)^T. Can you say that A and B are symmetric?

Results 1 to 4 of 4

- Apr 25th 2013, 07:44 AM #1

- Joined
- May 2011
- Posts
- 2

## (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.

- Apr 25th 2013, 12:59 PM #2

- Joined
- Apr 2013
- From
- Green Bay
- Posts
- 68
- Thanks
- 16

- Apr 25th 2013, 02:42 PM #3

- Joined
- Mar 2013
- From
- BC, Canada
- Posts
- 104
- Thanks
- 17

- Apr 25th 2013, 10:25 PM #4

- Joined
- May 2011
- Posts
- 2

## 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 and , not and , 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 and as described above. Turns out I can say they're skew-symmetric.