TITLE: $A \in M_{4x4}^C$, $S:M_{4x4}^C \to M_{4x4}^C, S(X) = AXA$, $A = diag \{J_2, J_2\}$ find Jordan form of $S$
QUESTION [0 upvotes]: Let 
$$
A \in M_{4x4}^C
$$
And let
$$
S:M_{4x4}^C \to M_{4x4}^C, S(X) = AXA
$$
Also: 
$$
A = diag \{J_2, J_2\}
$$
We need to find the jordan form for $S$.

I look at a solution and i dont understand some of the things. 
The minimal polynom of $A$ is $t^2$ therefore:
$$
S^2(X) = S(AXA) = A^2XA^2 = 0
$$
Therefore, the minimal polynom of $S$ divide $t^2$
Now its the problem:
Now assume $S(X) = 0, X = \alpha_{i,j}$
And now i just dont understand how they continue...
Mahybe someone can clarify/show another answer? 
Thank you.

REPLY [1 votes]: I suppose that $J_2$ denotes the $2\times2$ upper triangular nilpotent Jordan block.
The minimal polynomial of $A$ divides $t^2$. Hence all Jordan blocks in the Jordan form of $S$ are $0$ or $J_2$. Since $S(A^T)=A\ne0$ but $S(X)=0$ when
$$
X=\pmatrix{1&0\\ 0&0},\pmatrix{0&1\\ 0&0}\text{ or }\pmatrix{0&0\\ 0&1},
$$
the rank of $S$ is $1$. Therefore the Jordan form of $S$ contains only one copy of $J_2$, i.e. it is
$$
\pmatrix{J_2\\ &0_{2\times2}}.
$$
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