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Let $ V$ be an inner product space and let $\alpha\in \operatorname{End}(V )$ be selfadjoint. Show that $\ker(\alpha) = \ker(α^h)$ for all $h \geq 1$.

Let $v\in \ker(\alpha)$, then $\alpha(v)=0$, so for any integer $h\geq1$, we have $\alpha^h(v)=\alpha^{h-1}(\alpha(v))=0$, which implies $v\in \ker(\alpha^h)$.

For the other part, I guess, it would be better to prove by indution on $h$.

The assumption is true for $h=2$.

Let $v\in \ker(\alpha^ 2)$, hence $\langle \alpha^2(v),v\rangle=\langle \alpha(v),\alpha(v)\rangle=0$ and this implies that $\alpha(v)=0$. Thus $v\in \ker(\alpha)$.

For the $h>2$ I don't know how to solve it and I also would like to have a prove with using the idea of endomorphism rather than matrices.

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    The oldest trick in the book. With $\alpha$ being self-adjoint, so are all powers of $\alpha$. Now, if $\alpha^h(v) = 0$ for some $h\geq 2$, then $\alpha^{2\left(h-1\right)}(v) = 0$ (why?) and thus $\alpha^{h-1}(v) = 0$ (why?). Keep going down until the exponent becomes $1$. – 2017-01-03
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    @darijgrinberg I tried to write what I got from your hint. I guess it must fine. – 2017-01-03

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We assume the assumption is true for all integers less than $h$.

Let $v\in ker (\alpha^h)$.

$$\alpha^{2(h-1)}(v)=\alpha^{h-2}(\alpha^h(v))=0\implies\langle\alpha^{2(h-1)}(v),v\rangle=\langle\alpha^{h-1}(v),\alpha^{h-1}(v)\rangle.$$

Thus, $\alpha^{h-1}(v)=0$ which implies $v\in ker(\alpha^{h-1})$. By the induction we have $v\in ker(\alpha).$

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    Good job -- just don't forget to require $h \geq 2$. – 2017-01-03