I need to find all the matrices $A\in M_{4x4}\left(\mathbb{C}\right)\:$ such that:
$$A^4-2A^2+I\:=\:0$$
which means $\left(A^2-I\right)^2=0$
So I see that there is a few groups of which can give this result:
- matrices which $\left(A^2-I\right)=0$
- matrices which $\left(A^2-I\right)!=0$ but $\left(A^2-I\right)^2=0$
But I still don't see how to find them explicitly and generally, although I can write some matrices of the first case which fit in, and for the second group its is harder.
Does anyone see how can I generalize the solution using some knowledge about characteristic polynomial, minimal polynomial, eigenvalues (1,-1) to find all the not similar matrices which fit the formula ?
Hint
You have $p(A) = 0$ where $p(x) = (x-1)^2(x+1)^2$.
According to Jordan normal form or Lemme des noyaux (sorry this in in French...), the matrix $A$ is similar to a matrix
$$\begin{pmatrix} A_1 & 0\\ 0 & A_{-1}\end{pmatrix}$$ where $\dim A_1 = \dim \ker p_1(A)$, $\dim A_{-1} = \dim \ker p_{-1}(A)$, $p_1(A_1) = 0$, $p_{-1}(A_{-1}) = 0$ where $p_1(x) = (x-1)^2$ and $p_{-1}(x)=(x+1)^2$.
Finally, this provides all the matrices similar to one of the matrices
$$\begin{pmatrix} \epsilon_1 & 0 & 0 & 0\\ 0 & \epsilon_2 & 0 & 0\\ 0 & 0 & \epsilon_3 & 0\\ 0 & 0 & 0 & \epsilon_4\end{pmatrix}, \begin{pmatrix} \epsilon_1 & 1 & 0 & 0\\ 0 & \epsilon_1 & 0 & 0\\ 0 & 0 & \epsilon_2 & 0\\ 0 & 0 & 0 & \epsilon_3\end{pmatrix}, \begin{pmatrix} \epsilon_1 & 1 & 0 & 0\\ 0 & \epsilon_1 & 0 & 0\\ 0 & 0 & \epsilon_2 & 1\\ 0 & 0 & 0 & \epsilon_2\end{pmatrix} $$
where $\epsilon_i \in \{-1,1\}$.
If you don't know the theorems mentioned above, you can mimic their demonstration to your special case.