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.
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 $v\in ker(\alpha).$