Reading the book Introduction to group rings by Polcino and Sehgal, in this page I stuck with this statement about matrix representation:
If $a$ is an algebraic element of $A$, that is, if there exists a non-zeo polynomial $f(X) \in K[X]$ such that $f(a)=0$, then the eigenvalues of the matrix $\rho(a)$ also satisfy $f(X)$.
I'm facing difficulties in proving that. If I choose 2 different basis the polynomial $f$ will change? Can you explain me how to prove these statement?
If $\lambda$ is an eigenvalue of $\rho(a)$, there is a $b\in A\setminus\{0\}$ such that $ab=\lambda b$. On the other and, if $\mu$ is a scalar and $n\in\mathbb{Z}^+$, then$$(\mu a^n)b=\mu\lambda^nb$$and therefore, for each polynomial $p(x)$, $p(a)b=p(\lambda)b$. In particular, $0=f(a)b=f(\lambda)b$, and so $f(\lambda)=0$.