I have found similar results here and mathematics stack exchange but they all imposed specific conditions that don't suit this problem in particular. The problem is as follows.


Let A,B be square $n\times n$ matrices such that 


1) $[A,B]=0$


2) $AB\neq0$


3) $A^2=B^2=0$


4) $\mbox{Ker}(A)\cap\mbox{Ker}(B)\neq\{0\}$


Then $\mbox{Ker}(AB)=\mbox{Ker}(A) + \mbox{Ker}(B)$.


It seems to be true for all of the examples I can come up with, and one set inclusion is obvious but I don't know how to prove the non-trivial direction to obtain equality. Any advice or references to where this has been solved?


Thanks!



Kernel of $AB$ if $[A,B]=0$ and $AB\neq0$?
The result does not hold. Here is a counterexample:
take $X=\begin{bmatrix}
1 \\
0
\end{bmatrix}$, $E=\begin{bmatrix}
1 & 0 \\
0 & 0
\end{bmatrix}$, $F=\begin{bmatrix}
0 & 0 \\
0 & 1
\end{bmatrix}$, and consider the matrices
$$A:=\begin{bmatrix}
[0]\_{2 \times 2} & [0]\_{2 \times 2} & I\_2 & [0]\_{2 \times 1} \\
[0]\_{2 \times 2} & [0]\_{2 \times 2} & [0]\_{2 \times 2} & X \\
[0]\_{2 \times 2} & [0]\_{2 \times 2} & [0]\_{2 \times 2} & [0]\_{2 \times 1} \\
[0]\_{1 \times 2} & [0]\_{1 \times 2} & [0]\_{1 \times 2} & 0
\end{bmatrix}$$
and $$B:=\begin{bmatrix}
[0]\_{2 \times 2} & E & [0]\_{2 \times 2} & [0]\_{2 \times 1} \\
[0]\_{2 \times 2} & [0]\_{2 \times 2} & F & [0]\_{2 \times 1} \\
[0]\_{2 \times 2} & [0]\_{2 \times 2} & [0]\_{2 \times 2} & X \\
[0]\_{1 \times 2} & [0]\_{1 \times 2} & [0]\_{1 \times 2} & 0
\end{bmatrix}$$
of $M\_7(\mathbb{F})$. 
One easily checks that $A$ and $B$ all have the required properties. However, 
$Ker(A)+Ker(B)=\mathbb{F}^5 \times \{0\}$ whereas $Ker(AB)=\mathbb{F}^6 \times \{0\}$.


