I'm trying to prove that if a linear operator $f$ is diagonalisable then its minimal polynomial is the product of distinct linear factors. This is what I have so far:
Let $f$ be diagonalisable. So there exists a basis relative to which $f$ has a diagonal matrix, say $D$. So the characteristic polynomial of $f$ is given by $p_f(x)=(x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_s),$ where $\lambda_i$ are the diagonal entries of $D$.
I know that the minimal polynomial must divide the characteristic polynomial and have the same linear factors. Without loss of generality let the first $i$ linear factors be distinct. So I claim that the minimal polynomial $m_f(x)=\pm (x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_i).$
However, how can I now verify that $m_f(D)=0$ ?