IMC 2002 · Day 2 · P11

The direction $\Leftarrow$ is trivial, since if $A = S \overline{S}^{-1}$, then $$A \overline{A} = S \overline{S}^{-1} \cdot \overline{S} S^{-1} = I_n.$$ For the direction $\Rightarrow$, we must prove that there exists an invertible matrix $S$ such that $A \overline{S} = S$.

Let $w$ be an arbitrary complex number which is not 0. Choosing $S = wA + \overline{w} I_n$, we have $A \overline{S} = A (\overline{w}\, \overline{A} + w I_n) = \overline{w} I_n + w A = S$. If $S$ is singular, then $\frac{1}{w} S = A - \left( -\overline{w}/w \right) I_n$ is singular as well, so $-\overline{w}/w$ is an eigenvalue of $A$. Since $A$ has finitely many eigenvalues and $-\overline{w}/w$ can be any complex number on the unit circle, there exist such $w$ that $S$ is invertible.