Finding the minimal polynomial of a $A$, given a diagonal matrix equivalent to $XI_{10}-A$ Let $A\in M_{10}(\mathbb{C})$. We know that $XI_{10}-A $ is equivalent(By finite number of operations on the rows) to the diagonal matrix:
$$C=\begin{pmatrix}
 1&  &  &  &  &  &  &  &  & \\ 
 & 1 &  &  &  &  &  &  &  & \\ 
 &  &x  &  &  &  &  &  &  & \\ 
 &  &  &x+2  &  &  &  &  &  & \\ 
 &  &  &  &x^2  &  &  &  &  & \\ 
 &  &  &  &  &x+2  &  &  &  & \\ 
 &  &  &  &  &  & (x^2+4)x &  &  & \\ 
 &  &  &  &  &  &  & (x+2)^2 &  & \\ 
 &  &  &  &  &  &  &  &1  & \\ 
 &  &  &  &  &  &  &  &  & 1
\end{pmatrix}.$$
(the blank entries are $0$).
I need to find the minimal polynomial and Jordan form of $A$.
So I know know that the characteristic polynomial of A is: $F_A(x)= x^4(x+2)^4(x^2+4)$, and I'm having a hard realize what I can know in addition, to solve the problem. any hints? thanks!
 A: Let $K$ be a field, $X$ an indeterminate, and $V$ a finite dimensional $K[X]$-module. 
Recall that the multiset of elementary divisors of $V$ is the multiset 
$$f_1^{n(1,1)},\dots,f_1^{n(1,k(1))},$$ 
$$\vdots$$
$$f_r^{n(r,1)},\dots,f_r^{n(r,k(r))},$$ 
defined by the following conditions: the $f_i$ are distinct monic irreducible polynomials, the $n(i,j)$ satisfy $$n(i,1)\ge\cdots\ge n(i,k(i)),$$ 
and we have 
$$V\simeq\bigoplus_{i,j}\frac{K[X]}{\big(f_i^{n(i,j)}\big)}\quad.$$ 
The characteristic polynomial is the product of the elementary divisors, whereas the minimal polynomial is the product of the $f_i^{n(i,1)}$. 
If $W$ is also a finite dimensional $K[X]$-module, then the multiset of elementary divisors of $V\oplus W$ is the "disjoint union" (in the obvious sense) of the multisets of elementary divisors of $V$ and $W$. 
Some diagonal entries of Nir's matrix are equal to 1, and the others are 
$$X,\quad X+2,\quad X^2,\quad X+2,\quad (X+2i)(X-2i)X,\quad (X+2)^2.$$ 
The multiset of elementary divisors is thus 
$$\begin{matrix}
(X+2)^2,&X+2,&X+2,\\ \\ 
X^2,&X,&X,\\ \\ 
X+2i,&&\\ \\ 
X-2i.&& 
\end{matrix}$$ 
The characteristic polynomial is $$(X+2)^4X^4(X+2i)(X-2i).$$ The minimal polynomial is $$(X+2)^2X^2(X+2i)(X-2i).$$ 
The Jordan blocks $J(\lambda,n)$ are 
$$\begin{matrix}
J(-2,2),&J(-2,1),&J(-2,1),\\ \\ 
J(0,2),&J(0,1),&J(0,1),\\ \\ 
J(-2i,1),&&\\ \\ 
J(2i,1).&&  
\end{matrix}$$ 
Let's compute the invariant factors (even if it wasn't required). In the array of elementary divisors, there are four blank entries. We put a $1$ in each of them: 
$$\begin{matrix}
(X+2)^2,&X+2,&X+2,\\ \\ 
X^2,&X,&X,\\ \\ 
X+2i,&1,&1,\\ \\ 
X-2i,&1,&1.  
\end{matrix}$$ 
We get the first invariant factor by multiplying together the polynomials in the first row, and so on. Thus the invariant factors are 
$$(X+2)^2X^2(X+2i)(X-2i),\quad (X+2)X,\quad (X+2)X.$$ 
[Reference: Bourbaki, Algèbre, VII.4.8.]
