Prove/ Disprove: Let $n$ be a positive integer. Let $A$, $B$ be two $n \times n$ square matrices over the complex numbers. If $AB = BA$ and $\ker A = \ker A^2$ and $\ker B = \ker B^2$ 
then $\ker AB = \ker A + \ker B$.


(Recall that $\ker A$ is the set of all vectors $v$ such that $Av = 0$.)


Background: I am teaching linear algebra this semester. I did not like the standard proof of the Jordan canonical form I found in the textbooks, and thought I could prove it differently, directly from the axioms for a vector space, without using either the determinant, or the classification theorem for finite abelian groups. If the statement above is true, I believe I have a proof for the Jordan canonical form for $T$ by setting $A = (T-\lambda\_1I)^{n\_1}$ and $B=(T-\lambda\_2I)^{n\_2}$ for appropriate $n\_1$ and $n\_2$.


Note 1: If $A = B = \left( \begin{array}{cc}
0 & 1\\\0 & 0 \end{array} \right)$ then $AB = BA$ but $\ker AB \neq \ker A + \ker B$.


Note 2: It is easy to find $A, B$ such that $\ker A = \ker A^2$ and $\ker B = \ker B^2$ and $B$ maps a vector outside $\ker A + \ker B$ to $\ker A$, so that $\ker AB \neq \ker A + \ker B$.


Hence, both conditions are necessary.



When is $\ker AB = \ker A + \ker B$?
Since $\ker A = \ker A^2$, the map $\bar{A} : V/\ker A \to V/\ker A$ is injective. Since $V/\ker A$ is finite dimensional, this map is surjective. So for any $x \in V$ we can find $y \in V$ and $z \in \ker A$ such that $x = Ay + z$.


Now suppose $ABx = 0$ and let $x = Ay + z$ as above. Then $0 = ABx = ABAy + ABz = A^2By + BAz = A^2By$ because $AB = BA$ and $Az = 0$. 


So $By \in \ker A^2 = \ker A$ whence $ABy = 0$. So $B(Ay) = 0$ and $Ay \in \ker B$.


Hence $x = Ay + z \in \ker B + \ker A$.


