# 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.

• Your proof is very good. I can't see a faster way, at least. – Ivo Terek May 27 '16 at 0:10
• You wrote \textrm{rank }, with a blank space after the "k". If you write \operatorname{rank}, with no blank space, then the spacing before and after it depends on the context, so that if you write A\operatorname{rank}B you see $A\operatorname{rank}B$ and if you write A\operatorname{rank}(B) you see $A\operatorname{rank}(B)$, with a much smaller space before the $B$. I.e. it works just like \sin and \max and \log and \det, etc. That's the right way to do that. $\qquad$ – Michael Hardy May 27 '16 at 0:11

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 $\ker T_W = W \cap \ker T$.