Let $A$ a nonderogatory matrix and $B$ that commutes with $A$. How can I show that exists a polinomyal $p(x)$ such that the set of the eigenvalues of $B$ can be expressed as $\{p(\lambda_i): \lambda_i\in\sigma (A)\}$, where $\sigma(A) $ is the spectrum of $A$.\
I have, with the same hipoteses, exists a polynomial $p(x)$ such that $B=p(A)$ but I don't kmow how to use it.
Fact 1: $A$ is non-derogatory (the minimum pol. of $A$ is the characteristic pol. of $A$) IFF $AB=BA\Leftrightarrow B$ is a polynomial in $A$.
FACT 2: Let $A$ be any square matrix and $p$ be any polynomial. If $spectrum(A)=(\lambda_i)_i$, then $spectrum(p(A))=(p(\lambda_i)_i$. Proof: triangularize the matrix $A$.