# Does $\text{null } T^m = \text{null }T^{m+1}$ if and only if $\text{range } T^m = \text{range }T^{m+1}$ hold in infinite dimensional vector spaces?

Let $T \in \mathcal{L}(V)$ and $m$ be a nonnegative integer. Prove that $\text{null } T^m = \text{null }T^{m+1}$ if and only if $\text{range } T^m = \text{range }T^{m+1}$

This is Exercise 8A.19 in Sheldon Axler's Linear Algebra Done Right.

In this case, if I assume that $V$ is a finite dimensional vector space, the result can easily be proved using rank-nullity.

Is this result true in general? Intuitively, I feel like it should be, but I don't know how to prove it.

No. Consider the right shift on $\ell_2$, $$S: (x_1, x_2, \dots) \mapsto (0, x_1, x_2, \dots).$$ $S^n$ is always injective, but the range of $S^{n+1}$ is properly contained in the range of $S^n$.
You get the "opposite" phenomenon with the left shift $S^*$. The powers of $S^*$ are always surjecive but the kernel of $(S^*)^{n+1}$ properly contains the kernel of $(S^*)^{n}$.