[Math] If $A$ has $n$ distinct eigenvalues, then every matrix that commutes with $A$ commutes with each other.

Question

$A$ is a $n\times n$ matrix with entries from $\mathbb{R}$. If the characteristic polynomial for $A$ has $n$ distinct roots, then every $n\times n$ matrix (with real entries) that commutes with $A$ commutes with each other.

I know that $n$ distinct roots implies diagonalizable. But I am not sure where to go from there. I have played around with the identities $AB=BA, AC=CA, A=P^{-1}DP$ to try and get to $BC=CB$ but I am not seeing it. I would love any help. This is self study and not assigned.

Answer 1

Hint. Let $B=P^{-1}B_1P$. Then $AB=BA$ if and only if $DB_1=B_1D$. Using the fact that $D$ is a diagonal matrix with distinct diagonal entries, you may try to prove that $B_1$ is a diagonal matrix.