When does $AA^T$ commute with $A^T$?

Question

Suppose $A \in M_n(\mathbb R)$ is a real matrix. I am wondering what conditions would guarantee $AA^T$ commute with $A^T$, i.e., \begin{align*} AA^T A^T = A^T A A^T. \end{align*} If $A$ is normal, I think the relation holds since then $A^T$ is a polynomial in $A$. On the other hand, the commutativity is exactly $A(AA^T-A^TA) = 0$ which is columns of $(AA^T-A^TA)$ are in $ \text{ker}(A)$. Is this a property of some class of matrices?

Answer 1

$AA^T$ is real symmetric. By a change of orthonormal basis, we may assume that $AA^T=D\oplus0$ for some positive diagonal matrix $D$. The commutativity condition thus implies that $A^T$ is in the form of $X\oplus Y$, where $X$ has the same size as $Y$. But then $D\oplus 0=AA^T=(X^TX)\oplus(Y^TY)$. Therefore $X$ is invertible and $Y=0$. Plug them back into the commutativity condition, we get $X(X^TX)=(X^TX)X$ and in turn $XX^T=X^TX$. Hence $X$ and in turn $A$ are normal. Obviously this is a sufficient condition too.