# Show $\ker(\alpha)=\ker(\alpha)^2 \ \iff \ \ker(\alpha)\cap \mathrm{Im}(\alpha)=\{0\}$

Let $$V$$ be a vector space over a field $$F$$ and let $$\alpha$$ be an element of $$\operatorname{End}(V)$$. Show $$\ker(\alpha)=\ker(\alpha^2)$$ iff $$\ker(\alpha)$$ and $$\operatorname{im}(\alpha)$$ are linearly disjoint.

so, i know that $$\ker(\alpha)\subseteq \ker(\alpha^2)$$ is always true

If $$v$$ is an element of $$\ker(\alpha)\implies \alpha(r)=0_V$$ $$\implies\alpha(\alpha(r))=\alpha(0)=0_V$$

$$\implies \alpha^2(v)=0_V \implies v \in \ker(\alpha^2)$$

$$\ker(\alpha)=ker(\alpha^2)\Leftrightarrow \ker(\alpha^2) \subseteq \ker(\alpha)$$

Now, I know that I'm suppose to take some $$u$$ element of $$\ker(\alpha)\cap \operatorname{im}(\alpha)$$ and show that =$$0_V$$

Then I'm suppose to take $$u$$ element of $$\ker(\alpha^2)$$ and show $$u$$ element of $$\ker(\alpha)$$ where $$\alpha^2(u)=0$$, and $$\alpha(\alpha(u))=0$$

Im just not sure how to proceed from here.

• I TeXified your question the best I could. Please check and edit and/or ask for help. – Jyrki Lahtonen Jun 24 '14 at 19:01
• May I also inquire where the finite-field tag came from? Is $F$ finite? (Not that it affects anything here). – Jyrki Lahtonen Jun 24 '14 at 19:02

If $v$ is a nonzero vector in both $im(\alpha)$ and $ker(\alpha)$ then let $u$ be such that $\alpha(u)=v$. Then $\alpha(u)=v\not=0$ so $u\notin ker(\alpha)$, but $\alpha^{2}(u)=\alpha(v)=0$ so $u\in ker(\alpha^{2})$.
If $u\in ker(\alpha^{2})\backslash ker(\alpha)$ then $\alpha^{2}(u)=0$ so $\alpha(u)\in ker(\alpha)$ and $\alpha(u)\not=0$ so $\alpha(u)\in im(\alpha)\bigcap ker(\alpha)$ and is nonzero.
Consider $v\in\ker \alpha^2.$ So $\alpha(\alpha(v))=0,$ that is, $\alpha(v)\in \ker\alpha.$ Moreorever $\alpha(v)\in\alpha(V).$ Since $\ker \alpha\cap \alpha(V)=\{0\}$ it must be $\alpha (v)=0,$ that is, $v\in \ker \alpha.$ This shows that $\ker\alpha^2\subset \ker \alpha.$