TITLE: V is a linear space that has finite dimension $T : V \rightarrow V$ Proof if $Ker T = Ker T^2$ then $Im T = Im T^2$
QUESTION [2 upvotes]: I have the following question :
V is a linear space that has finite dimension $T : V \rightarrow V$ Proof if $Ker T = Ker T^2$ then $Im T = Im T^2$
I tried to proof it, but I got stuck, This is what I did :
So for all $v \in Ker T \Longleftrightarrow v \in Ker T^2$
We know that (from a theorem) $$dim(ImT)+dim(KerT)=dim(V)=n$$ and also: $$dim(ImT^2)+dim(KerT^2)=dim(V)=n$$
Therefore I can conclude that $dim(ImT^2)=dim(ImT)$
Let $w \in ImT$ so exist $u \in V$ so $T(u)=w \neq 0$

If $T^2(u)=T(T(u))=T(w) \neq 0$ Therefore we showed that for any $ w \in ImT \implies w \in ImT^2$ meaning that $ImT \subseteq ImT^2$ and we also know that $dim(ImT^2)=dim(ImT)$ Therefore $ImT=ImT^2$.
If $T^2(u)=T(T(u))=T(w) = 0$ Therefore $w \in KerT \Longleftrightarrow w \in KerT^2$ 

And here I'm stuck I don't think I realize what does it mean to be in $KerT^2$ if $w \in KerT$ then is clear that $T(w)=0$ and also its trivial that $T^2(w)=T(T(w))=T(0)=0$ therefore always if $w \in KerT \implies w \in KerT^2$ isn't that always true? Can someone explain?
Thank you.

REPLY [1 votes]: You need to prove first that 
$$
\operatorname{Ker}(T) = \operatorname{Ker}(T^{2})\implies \operatorname{Ker}(T) \cap \operatorname{Im}(T)=\{0\}
$$
Since $T$ maps each $x$ to $0$ or nonzero, $\dim(T^{2})\leqslant\dim(T^{})$. If $\dim(T^{2})=\dim(T^{})$, then $T^{}$ is invariant. So there is 
$$
\operatorname{Im}(T)\cap \operatorname{Ker}(T^{})=\{0\}
$$
for if not, let $x\in \operatorname{Im}(T^{})\cap \operatorname{Ker}(T^{})$, then $T(x)=0$ and $\dim(T^{2})<\dim(T^{})$.
Similarly
$$
\operatorname{Im}(T^2)\cap \operatorname{Ker}(T^{2})=\{0\}
$$
Since 
$$
\dim{\operatorname{Ker}(T)}+\dim{\operatorname{Im}(T)}=n\quad\text{and}\quad\dim{\operatorname{Ker}(T^2)}+\dim{\operatorname{Im}(T^2)}=n
$$
There is $\operatorname{Im}(T^2)=\operatorname{Im}(T)$.
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