Let $A$ be invertible, and let $f$ be a function of the form $f(z)=\sum_{j=-m}^{m}c_jz^j$. Show that if $\lambda$ is an eigenvalue of $A$

Question

Let $A$ be invertible, and let $f$ be a function of the form $f(z)=\sum_{j=-m}^{m}c_jz^j$. Show that if $\lambda$ is an eigenvalue of $A$ then $f(\lambda)$ is an eigenvalue of $f(A)$.

I have to prove that $(\sum_{j=-m}^{m}c_jA^j)x=(\sum_{j=-m}^{m}c_j(\lambda)^j)x$ knowing that $Ax=\lambda x$, but I do not know how to do this, could someone help me please? It would be enough to show that $\sum_{j=-m}^{m}c_jA^j=\sum_{j=-m}^{m}c_j(\lambda)^j$, how do I do this? Thank you very much.

Answer 1

If $\lambda$ is an eigenvalue of $A$ and $n\in\mathbb Z$, then $\lambda^n$ is an eigenvalue of $A^n$, with the same eigenvector (you should show this). If $c\in\mathbb C$, then $c\lambda$ is an eigenvalue of $cA$, with the same eigenvector (again, you should show this).

If now $Ax=\lambda x$, then $$f(A)x=\left(\sum_{j=-m}^mc_jA^j\right)x=\sum_{j=-m}^m\left(c_jA^jx\right)=\sum_{j=-m}^m\left(c_j\lambda^jx\right)=\left(\sum_{j=-m}^m c_j\lambda^j\right)x=f(\lambda)x.$$