Is $N_T \subseteq Im(T^l)$?

Question

If $T\in \mathcal L(V)$ is a nilpotent linear map (ie. $T^k=0_V$ for some $k \in \mathbb{N}_{\ne 0}$), let $N_T$ be the nullspace of $T$, $Im(T^l)$ be the image of $V$ under $T^l$.

Question: Is $N_T \subseteq Im(T^l)$ for all $l \in \mathbb{N}$ such that $\dim(Im(T^l)) \geq \dim(N_T)$?

Answer 1

Let: $$ T=\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$

Clearly $N_T = \{e_1, e_2\}$. Also, $im(T)=\{e_2, e_3\}$. Both are $2$-dimensional, but $N_t \subsetneq im(T)$.

One way to come up with this counterexample is to think about Jordan normal form of nilpotent matrices, assuming we are over the complex numbers (the $T$ in this example has $2$ Jordan blocks, one of dimenision $1$ and one of dimension $3$).