Let $V$ be a vector space over $\mathbb{F}=\mathbb R$ or $\mathbb C$ and $T$ an operator on $V$.
It is well known that $$\forall k\in\mathbb{N},\,\text{null}\,T^k\subseteq\text{null}\,T^{k+1}\,\land\,\text{range}\,T^{k+1}\supseteq\text{range}\,T^k$$
Exercise $21$ page $251$ in Sheldon Axler's Linear Algebra Done Right is:
Find a vector space $W$ and $T\in\mathcal{L}(W)$ sich that $\text{null}\,T^k\subsetneq\text{null}\,T^{k+1}$ and $\text{range}\,T^k\supsetneq\text{range}\,T^{k+1}$ for every positive integer $k$.
It is well known that if $\dim V=n$ is finite then $$\text{null}\,T^n=\text{null}\,T^{n+1}=\text{null}\,T^{n+2}=\cdots$$ and that $$\text{range}\,T^n=\text{range}\,T^{n+1}=\text{range}\,T^{n+2}=\cdots$$ Therefore we must choose an infinite dimensional vector space. I chose $W=\mathbb{F}^\infty$ the set of all sequences $(a_1,a_2,\cdots)$ over $\mathbb F$. Consider $\mathcal{B_c}=\{e_1,e_2,\cdots\}$ its canonical basis.
When we define $T$ such as $T(a_1,a_2,\cdots)=(a_2,a_3,\cdots)$ we have $\forall k\in\mathbb{N}\backslash\{0\},\,\text{null}\,T^k=\text{span}\{e_1,\cdots,e_k\}\,\land\,\text{range}\,T^k=\mathbb F^\infty$, which satisfies only one condition. On the other hand, defining $T$ by $T(a_1,a_2,\cdots)=(0,a_1,a_2,\cdots)$ gives $\text{null}\,T^k=\{0\}\,\land\,\text{range}\,T^k=\text{span}\{e_{k+1},e_{k+2},\cdots\}$, $T$ satisfies the other. Projections are no good as $T=T^2$ and I tried some few other examples but I couldn't find a good one. The idea I kept in mind while looking for an example is that as we move on from $T^k$ to $T^{k+1}$, one vector is "transported" from $\text{range}\,T^k$ to $\text{null}\,T^k$. Unfortunately, I failed to find such an operator.
Could you please provide me with some examples?