Show that $\phi,\psi$ are diagonalisable and calculate the eigenvalues of $\phi(A)= A^T$ and $\psi(A) = A^T + \sqrt{2}A$

Question

Task: Let $V=\mathbb{R}^{n \times n}$ and $\phi, \psi \in \operatorname{End}_K(V)$ defined by

\begin{align} \phi(A)&= A^T \qquad \quad \operatorname{for} A\in V \\ \psi(A) &= A^T + \sqrt{2}A \quad \operatorname{for}A \in V \end{align}

Show that $\phi,\psi$ are diagonalisable and calculate the eigenvalues of $\phi$ and $\psi$.

Question: Is my attempt valid? How to proceed for $\psi$?

My work so far:

Since $\phi(A)=A^T \Rightarrow \phi(A)^2=A$. So the minimal polynomial is $\mu_A = X^2 - X=X(X-1)$ that can be rewritten as a multiplication of linear factors and thus $\phi$ is diagonalisable.

Answer 1

Your attempt isn't quite right; as a quick check, note that $0$ cannot be an eigenvalue for $\phi$, because that would mean for some matrix $A\neq 0$, $\phi(A)=A^T=0$, but clearly $A=0$ if and only if $A^T=0$. What you have shown is that $\phi^2=I$, as $\phi^2(A)=(A^T)^T=A$ for any matrix $A$. This tells you that the minimal polynomial of $\phi$ divides $x^2-1$.

Hint for both diagonalizability and eigenvalues: think about symmetric and skew-symmetric matrices. What happens to these matrices under these maps? Do these matrices span $V$? What does this imply about diagonalizability and the eigenvalues?