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.

  • $\begingroup$ I TeXified your question the best I could. Please check and edit and/or ask for help. $\endgroup$ – Jyrki Lahtonen Jun 24 '14 at 19:01
  • $\begingroup$ May I also inquire where the finite-field tag came from? Is $F$ finite? (Not that it affects anything here). $\endgroup$ – 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 $\ker\alpha^2\subset \ker \alpha.$


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