Let $f : \mathbb{C}^{n}\rightarrow \mathbb{C}^{n}$ be a linear operator with a simple spectrum, furthermore, let $g : \mathbb{C}^{n}\rightarrow \mathbb{C}^n $ be a linear operator such that $f$ and $g$ commute.
Show that there is $P$ a polynomial such that $g=P(f)$.
Remark: The spectrum is simple when the characteristic polynomial has not multiple roots.
If the characteristic polynomial of $f$ has no multiple roots, then there exist distinct complex numbers $\lambda_1, \dots, \lambda_n$ and a basis $v_1, \dots, v_n$ of $\mathbb{C}^n$ such that $$ fv_k = \lambda_k v_k $$ for $k = 1, \dots, n$. Now \begin{align*} f(gv_k) &= g(fv_k) \\ &= \lambda_k (gv_k), \end{align*} which means that $gv_k$ is in the eigenspace of $f$ corresponding to the eigenvalue $\lambda_k$. This implies that there exists $\alpha_k \in \mathbb{C}$ such that $$gv_k = \alpha_k v_k. $$ Now let $P$ be a polynomial such that $$ P(\lambda_k) = \alpha_k $$ for $k = 1, \dots, n$. Then \begin{align*} \bigl( P(f) \bigr) v_k &= P(\lambda_k) v_k \\ &= \alpha_k v_k \\ &= gv_k \end{align*} for $k = 1, \dots, n$. Because $P(f)$ and $g$ agree on a basis, we conclude that $P(f) = g$, as desired.