Conditions for "$AA^T=A^TA$ implies $A$ symmetric" to hold.

Question

This claim arose in this question Show that $A$ is symmetric, with $A \in M_n(\mathbb R)$ where it is assumed additionally that $AA^TA$ is symmetric.

I'm considering weakening hypotheses.

Let $A$ be a real matrix such that $AA^T=A^TA$. This implies that $A$ is normal.

According to the spectral theorem (for complex-valued normal matrices), $$A=UDU^*$$ for some complex diagonal matrix $D$ and complex unitary matrix $U$.

I am not sure I can infer from here that $D$ and $U$ must have real entries. Nevertheless, if $A$ has real eigenvalues, then $D$ must have real entries.

• Are real eigenvalues sufficient to prove that $U$ has real entries ?

• Is there a way to circumvent the use of the spectral theorem ?

• Can you think of other hypotheses for the result to hold ?

• Is the claim " $AA^T=A^TA$ implies $A$ symmetric" true for any real $A$ without conditions on eigenvalues ? (EDIT: Answer is no.)

Answer 1

The claim isn't true. Take for example an orthogonal matrix $O$ so we have $$OO^T=O^TO=I_n$$ and $O$ isn't symmetric.