1

For a research problem, I found that two real matrices $A$ and $B$ that I generated in due to some procedure always had the property that $AB = B^T A$. Is there any property of matrices $A$ and $B$ due to which this condition holds? I tried doing some matrix algebra, but noticed in my research problem, the matrices $A$ and $B$ are also both singular, so I don't really know how I would proceed on finding any such properties of $A$ and $B$.

Edit: I also verified $B$ is not symmetric and $A$ is symmetric.

Rakesh Sinha
  • 43

1 Answers1

2

just a start, you should be aware that a real matrix is similar, over the reals to its transpose. There is an invertible matrix $C$ such that $$ C B C^{-1} = B^T \; , $$ $$ CB = B^T C $$ Neither one is symmetric.

This begins with the fact that a Jordan block is similar to its transpose, then to keep everything in real numbers you need the real Jordan form.

Will Jagy
  • 139,541
  • Yes, I saw that in the link of a comment above (or a duplicate question to it). What I am looking for here is possibly a sufficient condition for $AB = B^TA$ to hold when both are singular. – Rakesh Sinha Apr 27 '18 at 02:03
  • @RakeshSinha I understand that. How are your matrices being produced? – Will Jagy Apr 27 '18 at 02:04