I am trying to understand how intersections of images and kernals under $k$-th powered linear mappings can be non-trival for some values of $k$ rather than others.
Let $Y$ be a 6 dimensional vector space over some field $\mathbb F$ and $T\in L(Y, Y)$.
Suppose there are two vectors $\mathbf{u}$, $\mathbf{v}$ such that $$T^3(\mathbf{u}) = T^3(\mathbf{v}) = \mathbf{0} \quad \quad \textrm{ where } T^2(\mathbf{u}) \textrm{ and } T^2(\mathbf{v}) \textrm{ are linearly independent. } $$
Let $$N_{1} = \text{span}\{\mathbf{u},\mathbf{v}\} \quad \textrm{ and } \quad N_{2} = \text{span}\{T(\mathbf{u}), T(\mathbf{v})\} \quad \textrm{ and } \quad N_{3} = \text{span}\{T^2(\mathbf{u}),T^2(\mathbf{v})\}$$
We know and can prove that $Y = N_{1} ⊕ N_{2} ⊕ N_{3}$.
My question is knowing that $\quad Y=N1⊕N2⊕N3 \quad$ how could we explain why $\quad im(T^2) \cap ker(T^2) \quad$ is not trivial but $\quad im(T^k) \cap ker(T^k) \quad$ is trivial for all $k \geq 3$?
$Y = N_{1} ⊕ N_{2} ⊕ N_{3}$ tells you a lot of information.
for $k\geq 3$ you have $T^k$ annihilates $N_i$ for arbitrary $i$ so $T^kY =\mathbf 0\implies \ker T^k = Y \text{ and } T^k =\mathbf 0$. The intersection is zero dimensional.
If $\text{im}T^2 \cap \ker T^2=\{0\}$ then $\text{rank}\big(T^2\big)=\text{rank}\big(T^3\big)=0$ so
$T^2(\mathbf{u}) \textrm{ and } T^2(\mathbf{v})$ are both zero and hence not linearly independent, a contradiction. Thus $\text{im }T^2 \cap \ker T^2\neq \{0\}$