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Let $A$ $\in$ $\Bbb R^{n, n}$, $AA^T$ $=$ $A^TA$. Then all complex eigenvalues of $A$ are real.

Is this true or false? How do I show with a short justification?

Please can someone lend a hand?

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closed as off-topic by Namaste, user91500, Claude Leibovici, Shailesh, zoli Apr 10 '17 at 15:42

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    $\begingroup$ All complex eigenvalues are real? What does that mean? $\endgroup$ – Doug M Mar 17 '17 at 21:03
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    $\begingroup$ @DougM Real numbers are in the complex field, but not every complex number is real. The OP is referring to the possibly complex eigenvalues. $\endgroup$ – egreg Mar 17 '17 at 21:05
  • $\begingroup$ @egreg why wouldn't you say "If $A^TA = AA^T$ then $A$ has real eigenvalues."? $\endgroup$ – Doug M Mar 17 '17 at 21:07
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    $\begingroup$ @DougM Possibly the OP's instructor uses that terminology. $\endgroup$ – egreg Mar 17 '17 at 21:12
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The condition means that $A$ is normal. A normal real matrix has only real eigenvalues if and only if it is symmetric. But antisymmetric and orthogonal matrices are normal.

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$$ A = \left( \begin{array}{rr} 0 & 1 \\ -1 & 0 \end{array} \right) $$

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The eigenvalues in your case are just a positive real number and I think the eigenvalues are equal for both cases $AA^{T}$ $A^{T}A$. I mean if A is a unitary matrix.

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