Prove that if $R(T)\cap N(T)=\{0\}$ then $\operatorname{rank}(T)=\operatorname{rank}(T^2)$

Question

In the Linear Transformations chapter in the book "Linear Algebra" by Michael O'Nan, there's this question :

$R(T)\cap N(T)=\{0\}$ iff $\operatorname{rank}(T)=\operatorname{rank}(T^2)$

I'm ok with the part trying to prove

$r(T) = r(T^2) \Rightarrow R(T)\cap N(T)=\{0\}$

The part where I don't really understand at all, is trying to prove the opposite.

Can you please explain the proof for me? thanks

Answer 1

Let $y\in R(T)$. Then $y=T(x)$ for some $x\in V$. Now observe that $R(T)\bigcap N(T)=\{0\}$ impies $V=R(T)\oplus N(T)$. So $x=w+x_0$ for some $x_0\in N(T), w\in R(T)$. Thus $y=T(x_0+w)=T(w)=T(T(x'))$ as $w\in R(T)$.

Aternatively, we can show that $N(T)=N(T^2)$. Clearly $N(T)\subseteq N(T^2)$. Let $x\in N(T^2)$. Then $T(T(x))=0\Rightarrow T(x)\in N(T)\bigcap R(T)\Rightarrow T(x)=0\Rightarrow x\in N(T)$. By rank-nullity theorem $rank(T)=rank(T^2)$. In fact $R(T^2)\subseteq R(T)$ implies $R(T)=R(T^2)$.