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 $n_2$.

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.