Let $E$ be a Euclidean Space and $u\in\mathcal L(E)$. We let $u^{\star}\in\mathcal L(E)$ such that $\forall x,y\in E, \langle u(x)|y\rangle = \langle x|u^{\star}(y)\rangle$.
We want to show that if $u\circ u^{\star} = u^{\star}\circ u$ then $\ker(u^{\star})=\ker(u)$.
I know that: $\ker(u^{\star}) = Im(u)^{\bot}$ and $\ker(u) = Im(u^{\star})^{\bot}$
Do you have some ideas?
Let $x\in Ker(u)$. By definition $u(x)=0$ and also $(u^{\ast}\circ u)(x)=u^{\ast}(u(x))=0$. Using this, we get $$0=\langle 0 \vert x \rangle= \langle (u^{\ast}\circ u)(x) \vert x \rangle= \langle (u\circ u^{\ast})(x) \vert x \rangle = \langle u^{\ast}(x) \vert u^{\ast}(x) \rangle=||u^{\ast}(x)||$$ and hence $x \in Ker(u^{\ast})$. The other direction follows with the same argument.