Prove that $\text{rank } T = \operatorname{rank} T^2 \iff \operatorname{Im}T \cap \ker T = \{ \vec 0\}$

$\newcommand{\r}{ \operatorname{rank} } $

Let $T: V\to V$ be a linear transformation with $\dim V< \infty$. Prove that: $$ \r T = \r T^2 \iff \operatorname{Im} T \cap \ker T = \{ \vec 0 \}.$$

$"\Rightarrow"$ Let $\r T = \r T^2$. Then, by rank - nullity theorem we have that $$\dim \ker T =\dim \ker T^2 \tag 1.$$ But it is always true that: $\ker T \subseteq \ker T^2 .\tag 2$

By $(1),(2)$ we have that $\ker T = \ker T^2.$ So, instead of $\r T = \r T^2$ we can say that $\ker T = \ker T^2$ and we need to prove that $\operatorname{Im} T \cap \ker T = \{ \vec 0\}$.

Proof:

Suppose that there is a $z \in \operatorname{Im}T \cap \ker T$ with $z \neq 0$. Since $z \in \ker T \implies T(z) = 0$. Also, since $z \in \operatorname{Im}T \implies \exists y\in V$ such that $T(y) = z \implies T^2(y) = T(z) = 0.$ But this implies that $y \in \ker T^2 $ and by our hypothesis we have that $y \in \ker T \implies T(y) = 0 = z, $ which is absurd, because we assumed that $z \neq 0$.

$"\Leftarrow"$

We need to prove that $\ker T = \ker T^2$ or $\ker T^2 \subseteq \ker T.$

Proof:

Let $x \in \ker T^2$, which implies $T^2(x) = T\left(T(x)\right) = 0$. It is implied $T(x) \in \ker T,$ but also $T(x) \in \operatorname{Im}T.$ Thus, $T(x) \in \operatorname{Im}T \cap \ker T = \{0\}$. Thus, $T(x) = 0 \implies x \in \ker T.$

I would like to know if my reasoning is correct and if all the points are clear. Also, I would like to know if there is any shorter proof.

Solution 1:

Let $ W = \textrm{Im}\, T $, then the given conditions imply that $ T $ restricts to a surjective linear map $ T_W : W \to W $. Surjective linear maps from a vector space onto itself are invertible, which means that $ T_W $ has trivial kernel. The result follows as $ \ker T_W = W \cap \ker T $.