Jordan canonical form of a matrix $A$ Can you help me to find solution to this problem? 
I have matrix $ A \in M_n(\mathbb{C}): A^2 + A^4 = 0$ and I know, that maximal number of linear independent eigenvectors is $4$. Also known, that minimal polynomial not equal to characteristic polynomial. 
What we can say about Jordan canonical form of this matrix $ A$?

Here is my second answer that I believe is a sufficient guide to a full classification of all Jordan Canonical forms for arbitrary dimension $n$.
So first we need to find the possible minimal polynomials, which are:
\begin{align*}
x, \ x+i, \ x−i, \ x^2, \ x^2+1, \ x(x+i), \ x(x−i), & \\ 
x(x^2+1), \ x^2(x−i), \ x^2(x+i), x^2(x^2+1). & \quad (1)
\end{align*}
Next, note that for $n<4$, the minimal polynomial must have a degree equal or less than $n$. So for example, $n=0$ is trivial, and for $n=2$, we can only have $x, x+i, x−i, x^2, x^2+1, x(x+i), x(x−i)$.
For the case $n \geq 4$, all the possible minimal polynomials listed in $(1)$ are included. To determine the possibilities based on a given minimal polynomial for a dimension $n$, we include at least one block of each root of the multiplicity of the minimal polynomial, finding the rest of the possibilities is then a combinatorial exercise of finding all the possible combinations of extra Jordan blocks such that the sum of lengths of each Jordan block adds up to $n$.
As in my previous answer, I'll do the cases for $x(x^2+1)$ and $x^2(x-i)$, when $n \geq 3$.
For $x(x^2+1)$, we find that there must be at least one $0$-block, $i$-block, and $−i$-block of length one. The rest of the possible blocks can be any $0$-block $i$-block, $-i$-block of length one. Each of the possibilities are diagonal matrices of the form:
$$\begin{pmatrix}
0 & & & & \\
 & \underbrace{\ddots}_{n_1} & & & \\
 & & 0 & & \\
 & & & i & \\
 & & & & \underbrace{\ddots}_{n_2} \\
 & & & & & i \\
 & & & & & & -i \\
 & & & & & & & \underbrace{\ddots}_{n_3} \\
 & & & & & & & & -i
\end{pmatrix},$$
with $n_1, n_2, n_3 \geq 1$ $n_1+n_2+n_3=n$.
For $x^2(x-i)$, we find that there must be at least one $0$-block of length $2$, and one $i$-block of length $1$. And the rest can be any arrangements of $0$-blocks up to length $2$ and $i$-blocks up to length $1$. What we get is
$$\begin{pmatrix}
0 & & & & \\
 & \underbrace{\ddots}_{n_1} & & & \\
 & & 0 & & \\
 & & & \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} & \\
 & & & & \underbrace{\ddots}_{n_2} \\
 & & & & & \begin{pmatrix}
0 & 1 \\
0 & 0
\end{pmatrix} \\
 & & & & & & i \\
 & & & & & & & \underbrace{\ddots}_{n_3} \\
 & & & & & & & & i
\end{pmatrix},$$
with $n_1 \geq 0$ and $n_2, n_3 \geq 1$ such that $n_1+2n_2+n_3=n$.
Hope I didn't make any clumsy misreads, and I hope you can see how to replicate the other cases (which are of similar nature to the ones I did).
