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

$$\DeclareMathOperator\rank{rank}\DeclareMathOperator\im{im}$$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?

• For 1. implies 2., observe that the image is a subspace of the kernel. Jun 27, 2018 at 15:09
• @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. Jun 27, 2018 at 15:15

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?
• 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. Jun 27, 2018 at 15:29
• I don't know if we can pass from $rank(f) = \dim ker(f)$ to $Im(f) = Ker(f)$. Jun 27, 2018 at 15:32
• 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. Jun 27, 2018 at 15:38
• 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 $U=V$. Jun 27, 2018 at 15:40