$\text{null}\,T^k\subsetneq\text{null}\,T^{k+1}$ and $\text{range}\,T^k\supsetneq\text{range}\,T^{k+1}$ for all $k\in\mathbb{N}$

Question

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?

Answer 1

We can zip your examples for strictly increasing null spaces and for strictly decreasing ranges together to obtain an example of an endomorphism having both.

If $A \colon X \to X$ has $\operatorname{null} A^k \subsetneq \operatorname{null} A^{k+1}$ for all $k \in \mathbb{N}$, and $B \colon Y \to Y$ has $\operatorname{range} B^k \supsetneq \operatorname{range} B^{k+1}$ for all $k \in \mathbb{N}$, then the operator $C \colon X \times Y \to X \times Y$ given by $C(x,y) = (Ax,By)$ has strictly increasing null spaces - $x \in \operatorname{null} A^{k+1} \setminus \operatorname{null} A^k \iff (x,0) \in \operatorname{null} C^{k+1}\setminus \operatorname{null} C^k$ - and strictly decreasing ranges - $y \in \operatorname{range} B^k \setminus \operatorname{range} B^{k+1} \iff (0,y) \in \operatorname{range} C^k \setminus \operatorname{range} C^{k+1}$.

In your examples you use $X = Y = W$, and then we can weave the two examples together using an isomorphism $W\times W \to W$, for example the operator

$$T \colon (a_1, a_2, a_3,\dotsc) \mapsto (a_3, 0, a_5, a_2, a_7, a_4, \dotsc )$$

shifting odd-indexed entries two positions left and even indexed entries two positions right has the desired property.

We can also use an "infinite matrix" with nilpotent Jordan blocks of unbounded size on the diagonal to achieve the goal, for example

$$M = \begin{pmatrix} 0 & 1 & 0 & 0 & 0 & 0 &\cdots\\ 0 & 0 & 0 & 0 & 0 & 0 &\cdots\\ 0 & 0 & 0 & 1 & 0 & 0 &\cdots\\ 0 & 0 & 0 & 0 & 1 & 0 &\cdots\\ 0 & 0 & 0 & 0 & 0 & 0 &\cdots\\ \vdots & & & & & &\ddots \end{pmatrix}$$

having a Jordan block of size $2$ followed by a block of size $3$, then a block of size $4$ and so on. Since each row contains at most one nonzero entry, the product $Mx$ is well-defined for $x \in W$, and $T\colon x \mapsto Mx$ is a well-defined linear operator on $W$ with the desired property.

This differs from the previous example, since there we had $\dim (\operatorname{null} T^{k+1} / \operatorname{null} T^k) = 1$ and $\dim (\operatorname{range} T^k/\operatorname{range} T^{k+1}) = 1$ for all $k$ while in the later example the dimensions are all infinite.