$A^5 = A$ is diagonalizable, $A - \overline A^{T}$ is diagonalizable Whats the best approach to solve those two tasks?
$A$ is an $n\times n$ Matrix over $\mathbb{C}$


*

*If $A^5 = A$, then $A$ is diagonalizable

*$A - \overline A^{T}$ is diagonalizable

 A: If minimal polynomial of $A$ has distinct roots, then $A$ is diagonalizable. (This is a standard theorem in linear algebra.) So 1 is true since $x^{5} -x = x(x-1)(x+1)(x-i)(x+i)$ has distinct roots and the minimal polynomial of $A$ should divide this.
For 2, you don't need the condition $A^{5} = A$. The matrix $B = A -\bar{A}^{T}$ is skew-Hermitian (which satisfies $\bar{B}^{T} = -B$, so that $C = iB$ is Hermitian, which are always diagonalizable, so is $B = -iC$.

Without mentioning anything about minimal polynomial, let's try to prove it directly. (I think this proof may work for any matrices with distinct eigenvalues)
Let's define $V_{\lambda} = \{x\in \mathbb{C}^{n}\,:\, Ax = \lambda x\}$, eigenspace of $A$ for the eigenvalue $\lambda\in \{0, \pm 1, \pm i\}$. Then diagonalizability of $A$ is equivalent to
$$
\mathbb{C}^{n} = \bigoplus_{\lambda} V_{\lambda}
$$
For distinct $\lambda \neq \lambda'$, it is easy to check that $V_{\lambda} \cap V_{\lambda'} = \{0\}$. Hence we only need to show that $\sum_{\lambda} V_{\lambda} = \mathbb{C}^{n}$.
For any given $v \in \mathbb{C}^{n}$, define
$$
v_{0} = v - A^{4}v\\
v_{1} = Av +A^{2}v + A^{3}v +A^{4}v\\
v_{-1} = Av - A^{2}v + A^{3}v - A^{4}v\\
v_{i} = Av - iA^{2}v - A^{3}v + iA^{4}v \\
v_{-i} = Av + iA^{2}v - A^{3}v - iA^{4}v.
$$
Then $v_{\lambda}\in V_{\lambda}$ by direct computation with $A^{5} = A$, and
$$
v= v_{0}  + \frac{1}{4} (v_{1}- v_{-1} -iv_{i} + iv_{-i})
$$
proves the desired claim.
