An $n\times n$ matrix $A$ such that $PA^T = AP$ Suppose that I have an $n\times n$ real matrix $A$ such that $PA^T = AP$ for every invertible matrix $P$. Does this imply that $A$ is a multiple of the identity matrix (i.e., $A= c\, I_{n\times n}$ for some $c\in \mathbb{R}$.) ?
 A: The following proof works over any field, not just $\mathbb R$:
Put $P=I$, we see that $A$ is real symmetric. Hence $PA=AP$ for every invertible $P$. Since the set of all invertible matrices spans the whole matrix space, we in turn get $XA=AX$ for every square matrix $X$. Now it is well known (and also an easy exercise) that a matrix that commutes with every other square matrix must be a scalar multiple of $I$.
A: Yes, indeed we have $A^T= P^{-1}AP$ for every invertible matrix $P$. In particular if $P$ is the identity, then $A^T=A$. Hence $A$ is symmetric and thus have a real eigenvector, say $v$, such that $Av = \lambda v$. Now, note that for any invertible matrix $P$, we have
$$ A Pv = PA^T v = PAv = P\lambda v = \lambda Pv,$$
i.e. $Pv$ is also an eigenvector of $A$. Since for every $w\neq 0$, there exists an invertible matrix $P$ such that $Pv = w$, it follows that $Au = \lambda u$ for every $u \neq 0$. Hence $A$ must be a multiple of the identity, i.e. $A=\lambda I$.
A: Here's a very elementary proof.
First, as the other answers already stated, if you set $P=I$, you directly get that $A^T=A$. Therefore in all the following we can just omit the transposition.
Next, take for $P$ the diagonal matrix that has $-1$ as the $k$-th diagonal element and $1$ for the others. Then $PA=AP$ means the matrix that you get when reversing all signs in the $k$-th row is equal to the matrix that you bet by reversing the signs in the $k$-th column. This is only possible if all nondiagonal elements in the $k$-th row and $k$-th column are $0$, and since this is true for all $k$, $A$ must be diagonal.
Finally, by using permutation matrices as $P$ you find that all diagonal elements must be equal. Therefore $A$ is a multiple of the unit matrix.
On the other hand, for any multiple of the unit matrix it is easily verified that $PA^T=AP$ for any $P$, so there are no further conditions on $A$.
