Let $f:L \rightarrow L$ be a diagonalizable operator.

Question

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 $f$.

Answer 1

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.

Answer 2

If the eigenvalues are distinct then you should edit the question; in that case it's simple: Say $b_1,\dots,b_n$ is a basis for $L$ and $fb_j=\lambda_jb_j$. Now $$f(gb_j)=g(fb_j)=\lambda_jgb_j.$$Since the $\lambda_j$-eigenspace has dimension $1$ this shows that $gb_j=\alpha_jb_j$ for some scalar $\alpha_j$.

There is a polynomial $p$ such that $p(\lambda_j)=\alpha_j$ for all $j$; hence $gb_j=p(f)b_j$, so $g=p(f)$.