I'm trying to prove the following statement:
Let $N$ be a normal transformation and let $S$ be a linear transformation. Prove that if $SN=NS$ then $S^*N=NS^*$
We know that $N^*N=NN^*$, because N is a normal transformation. Also, from the properties of the adjoint operator, we get: $$SN=NS\quad/()^*$$ $$(SN)^*=(NS)^*$$ $$N^*S^*=S^*N^*$$
I'm not sure what to do from here. I've tried multiplying the last line by $N^*N$, but I get stuck later on. I'd appreciate any help, thanks!
As $N$ is normal the space decomposes into the orthogonal sum $E_1\oplus E_2\oplus \ldots \oplus E_k$ of eigenspaces of $N$ corresponding to the mutually different eigenvalues $\lambda_1,\lambda_2,\ldots, \lambda_k.$ Moreover each $E_j$ is the eigenspace of $N^*$ corresponding to the eigenvalue $\overline \lambda_j.$ For any $v\in E_j$ we have $$NSv=SNv=\lambda_jSv.$$ Therefore $Sv\in E_j.$ Similarly $$N^*S^*v=S^*N^*v=\overline \lambda_j S^*v,$$ hence $S^*v\in E_j.$ Thus $$N^*S^*v= \overline \lambda_jS^* v,\qquad S^*N^*v=S^*(\overline \lambda_j v)=\overline \lambda_j S^*v.$$ Therefore $$N^*S^*v=S^*N^*v,\qquad v\in E_j.$$ As the subspaces $E_j$ span the whole space we get $N^*S^*=S^*N^*.$
