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$\def\im{\operatorname{im}}$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 $\ker T^2 \le \ker T$?

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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 $\leq 1$.

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  • $\begingroup$ Great solution. $\endgroup$
    – corner3
    Mar 25 '20 at 15:33

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