$ A^{k}=A^{t} $ I would like to know if there is anything interesting in square matrices with the following property: $ \exists k \in {\displaystyle \mathbb {R} }$ such that $A^{k}=A^{t} $.
The exercise states that in this case $J(A)^{t} = J(A)^{k}$ where $J(A)$ is the canonical Jordan Form of the matrix A.
Now, if I suppose that $A \in$ the orthogonal group i can substitute $A^{-1}$ to the previous relation and work from there but what can I say for a generic A? 
Thanks in advance
 A: We can say something about the spectrum of $A$.
Since $A^t$ commutes with $A$ we have that $A$ is normal. We have
$$A = (A^t)^t = (A^k)^t = (A^t)^k = (A^k)^k = A^{k^2}$$
so the polynomial $x^{k^2} - x = x(x^{k^2-1} -1)$ annihilates $A$. Therefore $$\sigma(A) \subseteq \left\{\text{zeroes of } x(x^{k^2-1} -1)\right\} = \left\{0, 1, \omega_{k^2-1}, \omega_{k^2-1}^2, \ldots, \omega_{k^2-1}^{k^2-2}\right\}$$ where $\omega_{k^2-1} = e^{\frac{2\pi i}{k^2-1}}$.
On the other hand we have $A^{k+1} = A^tA \ge 0$ so $\sigma(A)^{k+1} \subseteq [0, +\infty\rangle$.
Hence $0 \le (\omega_{k^2-1}^r)^{k+1}$ so $(\omega_{k^2-1}^r)^{k+1} = 1$. We conclude that $\sigma(A)^{k+1} \subseteq \{0,1\}$, which together with normality of $A^{k+1}$ imply that $A^{k+1}$ is an orthogonal projection.
Hence $A^{2k+2} - A^{k+1} = 0$ so $$\sigma(A) \subseteq \left\{\text{zeroes of } x^{k+1}(x^{k+1}-1)\right\} = \{0, 1, \omega_{k-1}, \omega_{k-1}^2,\ldots, \omega_{k-1}^{k-2}\} $$
where $\omega_{k-1} = e^{\frac{2\pi i}{k-1}}$.
If we additionally assume that $A$ is invertible, then $\sigma(A) \subseteq S(0,1)$ so $A$ is orthogonal.
