Powers of an operator Let $T$ be a liinear operator on a vector space $V \to V$. Let $K_r$ and $W_r$ denote the kernel and image , respectively, of $T^r$.
(a) Show that $K_1 \subset K_2\subset\cdots$ and that $W_1 \supset W_2 \supset\cdots$. (From the context here $\subset$ should mean subset instead of proper subset, and $\supset$ means superset but need not be proper superset.)
(b) The following conditions might or might not hold for a particular value of $r$:
(1) $K_r=K_{r+1}$, (2)$W_r=W_{r+1}$, (3) $W_r\cap K_1=\{0\}$, (4) $W_1+K_r=V.$
Find all implications among the conditions (1)-(4) when $V$ is finite dimensional.
(c) Do the same thing when $V$ is infinite dimensional.
$\textbf{My answer:}$
(a) is trivial.
(b)(1) If $K_r=K_{r+1}$ then it implies $T$ restricted to $W_r$ is an automorphism, which also means $W_{r+1}=W_r$.
(b)(2) The same as (b)(1), i..e, $W_r=W_{r+1}$, implying $T$ restricted to $W_r$ is an automorphism.
(b)(3) It implies $T(v\in W_r)=0$ only if $v=0$. Hence $W_r=W_{r+1}$
$\textbf{My problem:}$
I am stuck at (b) 4 and (c).
Any help? Thanks in advance!
 A: I will first address condition (4) in part (b). To begin, let me observe the following:
Since $T(K_{r+1}) \subseteq K_r$, the map $T: V \to V$ induces a map $\overline T: V/K_{r+1} \to V/K_r$. It is always injective because $$\overline T(\overline v) = \overline 0 \iff T(v) \in K_r \iff v \in K_{r+1} \iff \overline v = 0$$ $\overline T$ is surjective iff for every $v \in V$, $v = T(v') + v''$ for some $v' \in V$ and some $v'' \in K_r$, which is the case iff $V = W_1 + K_r$. Moreover, because all of these vector spaces are finite dimensional and $\overline T$ is always injective, it follows that it is surjective iff $$\dim V/K_{r+1} = \dim V/K_{r} \iff \dim K_{r+1} = \dim K_r \iff K_{r+1} = K_r$$ Hence condition (4) is equivalent to the others.
For part (c), the same observation as in the finite dimensional case tells you that condition (1) is equivalent to $T:W_r \to W_{r+1}$ being injective, which is of course equivalent to condition (3). There are no logical implications between conditions (1) and (2) as the following examples show:
Let $V$ be a vector space with basis $v_0, v_1, \ldots$ and let $T:V \to V$ be defined by $T(v_i) = v_{i+1}$ for all $i$. Then $K_r = 0$ for all $r$, hence condition (1) holds for all $r$, but $W_r = \text{span}\left\{v_r, v_{r+1}, \ldots\right\}$, so $W_r \neq W_{r+1}$ for all $r$, so condition (2) doesn't hold for any $r$.
Let $V$ be the same vector space, but let $T: V\to V$ now be defined by $T(v_i) = v_{i-1}$ for $i > 0$, and $T(v_0) = 0$. Then $W_r = V$ for all $r$, hence condition (2) holds for all $r$, and yet $K_r = \text{span}\left\{v_0, v_1, \ldots, v_{r-1}\right\}$, so condition (1) doesn't hold for any $r$.
Finally, we can see that (2) and (4) are equivalent as follows. The same reasoning as in the finite dimensional case shows that (2) is equivalent to the condition that $\overline{T}: V/K_{r+1} \to V/K_r$ is surjective. Since $K_r$ is the kernel of the map $T^r : V\to V$ and $W_r$ is its image, the third isomorphism theorem says that $T^r$ induces an isomorphism $\overline{T^r}: V/K_r \to W_r$. Likewise $\overline{T^{r+1}} :V/K_{r+1} \to W_{r+1}$ is an isomorphism too. Thus we have the sequence of linear maps $$W_{r+1} \xrightarrow{(\overline{T^{r+1}})^{-1}} V/K_{r+1} \xrightarrow{\overline{T}} V/K_{r} \xrightarrow{\overline{T^{r}}} W_{r}$$ As you will need to check for yourself, the composition is just the inclusion map $W_{r+1} \hookrightarrow W_r$. It follows that $\overline{T}:V/K_{r+1} \to V/K_r$ is surjective iff $W_{r+1} = W_r$. In fact, this proof works to show that (2) $\iff$ (4) in the finite dimensional case too.
