When is $\ker AB = \ker A + \ker B$? 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.
 A: Here is a proof. First notice that the space of all matrices of size $n$ is finite dimensional. Therefore matrices $A,...,A^{m}$ are linearly dependent for a sufficiently large $m$ (this is instead of Hamilton-Cayley). Hence there exists a polynomial $f(x)$ such that $f(A)=0$, $f(0)=0$. Represent $f(x)$ as $x^kg(x)$ where $g(0)\ne 0$, $k\ge 1$. Dividing by $g(0)$ we can assume that $g(0)=1$. We have $A^kg(A)=0$. Hence $Ag(A)=0$ (since $Ker A= Ker A^2$). Now let $ABx=0$. We have that $(g(A)-1)x=A*h(A)x$ (for some $h$), so $(g(A)-1)x$ is   in $Ker B$. Since $g(A)x$ is in the kernel of $A$, we have the decomposition $x=g(A)x-(g(A)-1)x$ where the first summand in $Ker A$, the second in $Ker B$.   
A: Here's another statement of more or less the same result. The ideas in the proof are from this proof in German that Martin Brandenburg linked to in his comment. 
Claim:  Let $n$ be a non-negative integer. Let $A$, $B$ be two $n×n$ square matrices over the complex numbers. If $AB=BA$ and $\ker A \cap \ker B = \{0\}$ then $\ker AB=\ker A \bigoplus \ker B$.
Note: Assuming that $\ker A\cap\ker B=\{0\}$ is not a big restriction, since we can always quotient out to eventually reduce to this case.
Proof: Since $A,B$ commute, it is clear that $\ker A \bigoplus \ker B\subseteq \ker AB$. Further,  $\ker AB$ is invariant under $A, B$. Since all the action is taking place within $\ker AB$, we may assume without loss of generality that $\ker AB$ is the entire space, of dimension $n$. This implies $\operatorname{im} B\subseteq \ker A$.
By the rank-nullity theorem, $\dim\ker B + \dim\operatorname{im} B = n$.  Hence, $\dim\ker A + \dim\ker B\geq n$. Since these spaces intersect trivially by assumption, we are done.
A: 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 $x = Ay + z \in \ker B + \ker A$.
