Let $f:L \rightarrow L$ be a diagonalizable operator. This is a question from the book "Linear Algebra and Geometry", of Kostrikin. I would like an idea how to get started.

Let $f:L \rightarrow L$ be a diagonalizable operator with distinct eigenvalues. Prove that any operator $g: L\rightarrow  L$ such that $gf = fg$ can be represented in the form of a polynomial of $f$.


Suppose that $f$ is the identity. Then the equality $g\circ f=f\circ g$ holds for every $g\colon L\longrightarrow L$. However, it is not true that every operator is a polynomial on $f$. So, the statement is false.
