Let $A\in \mathbb{C}^{4×4}$. We are given the equation
$$A^{4}+12A^{2}=6A^{3}+8A$$
and
$$\textrm{rank}(A)=2\cdot\textrm{rank}(A-2I_{4})=4$$
I already found out that $2$ is an Eigenvalue, but how do I determine the other ones ? I'm mainly looking for hints.
The polynomial $x^4-6x^3+12x^2-8x = x(x-2)^3$ annihilates the matrix $A$ so $\sigma(A) \subseteq \{0,2\}$.
If $\sigma(A) = \{0\}$ then $\operatorname{rank}(A-2I) = 4$ so $\operatorname{rank}(A) = 2\operatorname{rank}(A-2I) = 8 > 4$ which is a contradiction.
Assume $\sigma(A) = \{0,2\}$.
If $\operatorname{rank}(A) = 0$ then $A = 0$ so $\sigma(A) = \{0\}$, a contradiction.
If $\operatorname{rank}(A) = 2$ then $\operatorname{rank}(A-2I) = 1$ so $$\operatorname{null}(A) + \operatorname{null}(A - 2I) = 2 + 3 = 5 > 4$$
which is a contradiction.
If $\operatorname{rank}(A) = 4$, then $0 \notin \sigma(A)$, a contradiction.
Therefore, the only option is $\sigma(A) = \{2\}$, which implies $\operatorname{rank}(A) = 4$ and $\operatorname{rank}(A-2I) = 2$. An example of such a matrix is
$$A = \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 1 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \\ \end{pmatrix}$$