Let $V$ be finite dimension vector space and let $T:V \rightarrow V$. Why is it true that $\operatorname{Image}(T) = \operatorname{Image}(T^2)$ $⟹$ $\ker (T)\cap\operatorname{Image}(T)=0$.
my attempt:
Assume $v\in\ker (T)\cap\operatorname{Image}(T)$ Then $v\in \operatorname{Image}(T^2)$. Therefore, there exist $u,w \in V$ s.t $$T(u)=v$$ $$T^2(w)=v$$ $$T(v)=0$$
From here I have no idea how to show that $v=0$
Best Answer
$\newcommand{\im}{\operatorname{Im}}\newcommand{\span}{\operatorname{span}}$In finite dimensional space, a linear transformation is invertible if and only if it is surjective - this follows from the rank nullity theorem. Then $\im(T)=\im(T^2)$ implies that the restriction of $T$ to $\im T$, the map $T:\im T\to\im T^2$, is a surjection and thus a bijection; the kernel of this restriction is trivial.
You then have $\ker T\cap\im T=\{0\}$.
If you want a more "direct" way of seeing this, note that if $u\in\ker T\cap\im T\setminus\{0\}$, then as both $\ker T,\im T$ are linear subspaces we have that $\lambda u\in\ker T\cap\im T$ for all $\lambda\in\Bbb R$ (or whatever field you're working in). Then by the exchange/replacement theorems, I can form a basis of $\im T$ from $u,v_1,v_2,\cdots,v_{n-1}$, where $\dim\im T=n$, and then: $$\im T^2=T(\im T)=\span\{T(u),T(v_1),\cdots\}=\span\{0,T(v_1),\cdots\}=\span\{T(v_1),\cdots\}$$And you get that $\dim\im T^2=\dim\span\{T(v_1),\cdots,T(v_{n-1})\}\le n-1$ but $\dim\im T=n$, so $\im T^2\neq\im T$, a contradiction.