Calculate the eigenvalues of the symmetric part of $A$ form the eigenvalues of $A$ and $A^T$ without calculating the symmetric part of $A$?

Question

We have the matrix $A$, $$\begin{pmatrix} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{pmatrix} $$ Clearly $A$ is orthogonal and has eigenvalues $1, e^{2i\pi/3}, e^{-2i\pi/3}$, which are the same for $A^T$. I have a question which wants the eigenvalues of the symmetric part of $A$ from the eigenvalues of $A$ and $A^T$ without explicitly calculating the symmetric part of $A$. It seems like it should be simple, but I'm struggling to get this.

Since $Tr A = \sum_{i=0}^3 \lambda_i$ the sum of the eigenvalues of the symmetric part is $0$, but I'm not sure where else to go. Any hints?

Answer 1

$A+A^T=A+A^{-1}$ has the eigenvalues $\lambda+\frac{1}{\lambda}$. Since $|\lambda|=1$ it is the same as $\lambda+\bar\lambda=2 \mathrm{Re}\,\lambda$.

Answer 2

Hint (?): Notice that every eigenvector of $A$ is also an eigenvector of $A^T$.