3

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.)

Community
  • 1
Gabriel Romon
  • 35,428
  • 5
  • 65
  • 157

1 Answers1

5

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.

  • What about the eigenvalues of $O$ ? Are they real ? – Gabriel Romon Apr 05 '14 at 12:38
  • Not necessary. Take for example this matrix $$\left(\begin{array}\ \cos\theta&-\sin\theta\\sin\theta&\cos\theta\end{array}\right)$$ –  Apr 05 '14 at 12:41
  • For $2*2$ matrices, orthogonal matrices are either symmetric, or they have complex eigenvalues (their imaginary part is $|sin(\theta)|$). Is is true for higher dimensions? Can you disprove the stronger claim that if $A$ has real eigenvalues, then $A$ is symmetric ? – Gabriel Romon Apr 05 '14 at 12:48
  • Short and sweet, Sami! – amWhy Apr 06 '14 at 12:41