If $AB=0$ for two commuting linear operators and their kernels intersect trivially, do they span the whole vector space? $\DeclareMathOperator{Ker}{Ker}\DeclareMathOperator{Im}{Im}$
In the finite dimensional case, this is clear by counting dimensions: The image of $B$ must lie completely in the kernel of $A$, and thus we have
$$
V \simeq \Im B \oplus \Ker B\leq \Ker A \oplus \Ker B
$$
And in the finite dimensional case, we can verify because of the trivial intersection that $V$ and $\Ker A +\Ker B$ have the same dimension, thus must be equal.
However, is there a way to conclude this in the infinite dimensional case? Just because $V\simeq U \oplus W$ for two trivially intersecting spaces does not necessarily imply that: Consider for instance $k^\mathbb ℤ$ for some field $k$ and $U$ taking components $(1,\infty)$ and $W$ components $(-\infty,-1)$. The zeroth component is missing in their span.
However, can we fix that in our special case of two commuting operators whose product vanishes on the entire space?
As @Servaes correctly pointed out, it is not true if our operators do not commute. But what if they do?
 A: No; let $k$ be a field and $V=k^{\Bbb{N}}$, and consider the linear maps $A,B:\ V\ \longmapsto\ V$ given by
$$A((x_n)_{n\in\Bbb{N}})_m=\left\{\begin{array}{ll}
x_m&\ \text{ if }\ m\ \text{ is odd}\\
0&\ \text{ if }\ m\ \text{ is even}
\end{array}\right..$$
$$B((x_n)_{n\in\Bbb{N}})_m=\left\{\begin{array}{ll}
0&\ \text{ if }\ m\ \text{ is odd}\\
x_{m/2}&\ \text{ if }\ m\ \text{ is even}
\end{array}\right..$$
Then $\ker B=0$ so the kernels intersect trivially, and $AB=0$. But $A\neq0$ so $\ker A+\ker B\neq V$.
More legibly; the linear maps $A$ and $B$ are defined as
$$A(x_0,x_1,x_2,x_3,x_4,x_5,x_6,\ldots)=(0,x_1,0,x_3,0,x_5,0,\ldots),$$
$$B(x_0,x_1,x_2,x_3,x_4,x_5,x_6,\ldots)=(x_0,0,x_1,0,x_2,0,x_3,\ldots).$$
A: No, it's not true. Let $A, B: k^{\mathbf{N}} \to k^{\mathbf{N}}$ be $\DeclareMathOperator{Ker}{Ker}$
$$A:(x_1,x_2,x_3,\cdots)\mapsto (0,x_1,0,x_2,0,x_4,0,x_6,\cdots),
$$
$$B: (x_1,x_2,x_3,\cdots)\mapsto (0,0,x_1,0,x_3,0,x_5,\cdots).
$$
Then, $AB=BA =0$, but $\Im A +\Im B = \Ker B + \Ker A = \langle e_2,e_3,e_4,\cdots \rangle < k^{\mathbf{N}}.$ 
