Show that $v$ is a eigenvector of $q(A)$ with respect to $p(\lambda)$.

Question

Let $K$ a field, $p=\sum_{i=0}^k \alpha_i t^i \in K[t]$ and $A\in K^{n,n}, \quad n\in \mathbb N\setminus\{0\}$. Define $q:K^{n,n}\to K^{n,n}, q(A)=\sum_{i=0}^n \alpha_i A^i$. Let $\lambda \in K$ a eigenvalue of A with eigenvector $v$. Show that $v$ is a eigenvector of $q(A)$ with respect to $p(\lambda)$.

I tried to show $q(v)=p(\lambda)v$, but I am not sure whether this is what is to show? Does this exercise has to do with $\cal Cayley -Hamilton$?

Answer 1

$$Av = \lambda v \implies A^2v = \lambda^2v \implies \cdots \implies A^iv = \lambda^iv, \forall i \in \mathbb{N}$$

Also recall that $(\alpha A + \beta B)x = \alpha Ax + \beta Bx$.

Now it should be clear that:

$$\left(\sum_{i=0}^n \alpha_iA^i\right)v = \sum_{i=0}^n \alpha_iA^iv = \sum_{i=0}^n\alpha_i\lambda^i v = \left(\sum_{i=0}^n\alpha_i\lambda^i\right) v$$

Answer 2

Hint: If $Av = \lambda v$, then $A^i v = \lambda^i v$.