$\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 $\operatorname{Im} T \cap \ker T = \{ \vec 0\}$.


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 $z \neq 0$.


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


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.

  • 1
    $\begingroup$ Your proof is very good. I can't see a faster way, at least. $\endgroup$ – Ivo Terek May 27 '16 at 0:10
  • $\begingroup$ 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. $\qquad$ $\endgroup$ – 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 $.

  • $\begingroup$ Yes, didn't see that! $\endgroup$ – thanasissdr May 27 '16 at 0:46

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