Prove that if $f^2 = 0_{E}$ and $n = 2\operatorname{rank} f$ then $\operatorname{im} f = \ker f$.

Question

$\DeclareMathOperator\rank{rank}\DeclareMathOperator\im{im}$Let $f$ be a linear map from $E$ unto $E$.

$E$ is a vector space that has dimension $n$.

Prove that the following two statements are equivalent:

1. $f^2 = 0_{E}$ ($0_{E}$ is the zero map) and $n = 2\rank(f)$
2. $\im f = \ker f$

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I proved $(2) \implies (1)$

From rank-nullity theorem we have: $ \dim \im f + \dim \ker f = n $

$\implies \dim \im f + \dim \im f = n $ (because $\im f = \ker f$).

$\implies n = 2\rank(f) $

Also We have: $x \in \im f \implies x \in \ker f \implies f(x) = 0 \implies f^2(x) = f(0) = 0$

I don't know how to prove $ (1) \implies (2) $.

Is my first argument correct? How can I prove the other way?

Answer 1

Your argument for $(2) \implies (1)$ is fine.

For the other implication, note that $f^2=0$ implies $\text{Im}(f)\subset\ker(f)$. Thus $\dim\ker(f)\geq\text{rank}(f)$, so we have $$n=\text{rank}(f)+\dim\ker(f)\geq2\cdot\text{rank}(f)=n.$$ so equality holds, and thus $\text{rank}(f)=\dim\ker(f)$. What can you infer from this?