# Im $𝑇=$Im $𝑇^2⟹ \ker𝑇=\ker𝑇^2$

$$\def\im{\operatorname{im}}$$Assume $$V$$ is finite dimensional and let $$T:V\to V$$ be a linear operator. If $$\im T=\im T^2$$, then $$\ker T=\ker T^2$$

I am so confused about this proof. I am only able to know that $$\ker T \leq \ker T^2$$ and $$\im T^2 \leq \im T$$. How can I get $$\ker T^2 \le \ker T$$?

We have $$\dim(V) = \dim ker(T) + \dim im(T) = \dim ker(T^2) + \dim im(T^2)$$. Since $$im(T) = im(T^2)$$, we have $$\dim im(T) = \dim im(T^2)$$ and hence $$\dim ker(T) = \dim ker(T^2)$$. Since, as you said, $$ker(T) \subset ker(T^2)$$, we obtain $$ker(T) = ker(T^2)$$ from the equality of their dimensions. Observe that we use in a crucial way that $$V$$ is finite-dimensional!
Edit: To show that the statement does not hold true if $$V$$ is infinite-dimensional, it suffices to give a surjective linear map $$T \colon V \to V$$ such that $$ker(T)$$ is a proper subset of $$ker(T^2)$$. Let for instance $$V$$ be the real vector space of polynomial functions $$\mathbb R \to \mathbb R$$ and $$T(p) := p'$$ the map which maps a $$p \in V$$ to its derivative. This map is easily seen to be surjective. Moreover, $$ker(T)$$ consists of the constant functions, but $$ker(T^2)$$ consists of the polynomial functions of degree $$\leq 1$$.