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

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

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. $\im f = \ker f$

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?

  • $\begingroup$ For 1. implies 2., observe that the image is a subspace of the kernel. $\endgroup$ Jun 27, 2018 at 15:09
  • $\begingroup$ @LordSharktheUnknown From $f^2 = 0$ we have $Im f \subset Kerf$. To get the equality, than I need to prove $Kerf \subset Imf$. I don't see how to proceed. $\endgroup$ Jun 27, 2018 at 15:15

1 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?

  • $\begingroup$ To get $Im(f) = Ker(f)$, we need to prove the other inclusion $Ker(f) \subset Im(f)$. Which I could not see how to get from your inequality. $\endgroup$ Jun 27, 2018 at 15:29
  • $\begingroup$ Does my edit help at all? $\endgroup$
    – Aweygan
    Jun 27, 2018 at 15:30
  • $\begingroup$ I don't know if we can pass from $rank(f) = \dim ker(f)$ to $Im(f) = Ker(f)$. $\endgroup$ Jun 27, 2018 at 15:32
  • 1
    $\begingroup$ Well if $x\in\ker(f)\setminus\text{Im}(f)$ and $\{e_1,\ldots,e_m\}$ is a basis of $\text{Im}(f)$, then $\{x,e_1,\ldots,e_m\}$ is a linearly independent subset of $\ker(f)$. But then $\dim\ker(f)\geq m+1>m=\text{rank}(f)=\dim\ker(f)$, a contradiction. $\endgroup$
    – Aweygan
    Jun 27, 2018 at 15:38
  • 1
    $\begingroup$ This is part of a more general phenomenon: If $U$ is a subspace of a vector space $V$ and $\dim U=\dim V<\infty$, then $U=V$. $\endgroup$
    – Aweygan
    Jun 27, 2018 at 15:40

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