Find an operator such that $kerT^k$ is a strict subset of $kerT^{(k+1)}$ ... Find a vector space U and operator $T\in \scr L$(U) such that $\ker T^k$ is a strict subset of $\ker T^{(k+1)}$ and $\text{ran}\,T^{k+1}$ is a strict subset of $\text{ran}\,T^k$ for every k $\in$ \mathbb N. 
I've tried so many examples and nothing seems to work! I think it is easiest to work with infinite sequences, so starting with $(x_1, x_2,...)$ then possibly removing something and then maybe adding somethin? Not sure, any help needed!!!!  
 A: Take $\mathbb \ell^\infty(\mathbb N)$,  and $R$ the reverse unilateral shift, that is
$$
R(x_1,x_2,\ldots)=(x_2,x_3,\ldots).
$$
Then 
$$
R^k(x_1,x_2,\ldots)=(x_{k+1},x_{k+2},\ldots).
$$
So
$$
\ker R^k=\{(x_1,\ldots,x_k,0,0,\ldots):\ x_1,\ldots,x_k\in\mathbb R\},
$$
But $R$ and its powers are onto, so the condition on the ranges will fail. 
Let $S$ be the unilateral shift,
$$
S(x_1,x_2,\ldots)=(0,x_1,x_2,\ldots).
$$
Here both $S$ and its powers are injective, so the kernels will all the same. But the ranges of $S^k$ are decreasing:
$$
\text{ran}\,S^k=\{(\overbrace{0,\ldots,0}^k,x_1,x_2,\ldots)\}
$$
Now we can combine the two: let $U=\ell^\infty(\mathbb N)\oplus \ell^\infty(\mathbb N)$, and $T=R\oplus S$. Then
$$
\ker T^k=\{(x_1,\ldots,x_k,0,0,\ldots)\oplus 0:\ x_1,\ldots,x_k\in\mathbb R\},
$$
$$
\text{ran}\,T^k=\ell^\infty(\mathbb N)\oplus \{(\overbrace{0,\ldots,0}^k,x_1,x_2,\ldots):\ x_1,\ldots,x_k\in\mathbb R\}.
$$
A: I think this works . . .

Let $U = F^{\infty}$, where $F$ is a field, and let $\{e_1,e_2,e_3,...\}$ be a basis for $U$.

For each integer $n > 1$, let


*

*$p(n)$ be the least prime factor of $n$.

*$d(n)$ be the largest divisor of $n$ which is relatively prime to $p(n)$.


Let $T$ be the unique linear transformation from $U$ to $U$ such that
$$T(e_n) =
\begin{cases}
0 &\text{if}\;n=1\\[4pt]
{\large{e_{d(n)}}}&\text{if}\;n>1\\[4pt]
\end{cases}
$$
Then for each positive integer $k$, the kernel of $T^k$ is the subspace of $U$ generated by 
$$\{e_n \mid n\;\text{has at most}\;k-1\;\text{distinct prime factors}\}$$
and the image of $T^k$ is the subspace of $U$ generated by
$$\{e_n \mid n\;\text{has no prime factors among the first $k$ primes}\}$$
It follows that for all positive integers $k$, we have the strict inclusions
\begin{align*}
\text{ker}(T^k)\;&\subset\;\text{ker}(T^{k+1})\\[4pt]
\text{im}(T^{k+1})\;&\subset\;\text{im}(T^k)\\[4pt]
\end{align*}
as required.
