From Herstein's "Topics in Algebra" ($2^{\text{nd}}$ edition):
We have just seen that $N^{*}$ commmutes with all the linear transformations that commute with $N$, when $N$ is normal; this is enough to force $N^{*}$ to be a polynomial expression in $N$. However, this can be shown directly as a consequence of Theorem $6.10.4$ (see Problem $14$).
Here $N^{*}$ refers to the adjoint of $N$. $N^{*} = p(N)$ can be shown using the fact that when $N$ is normal, $N$ is diagonal in some orthonormal basis consisting of the eigenvectors of $N$ (this is precisely the content of Theorem $6.10.4$). Such a proof can be found, for example, here: Normal operator $f \in L(V,V)$ adjoint as a polynomial in $f$, $f^{*} = p(f)$. The quoted passage from Herstein's book seems to imply that there is a proof of this which doesn't use Theorem $6.10.4$, namely one that uses the fact that $N^{*}$ commutes with all linear transformations which commute with $N$ (when $N$ is normal). How can this be done?
