Let $A: \mathbb C^4 \to \mathbb C^4$ be a linear operator and let $f(x)$ be a polynomial with complex coefficents. If $c$ is an eigenvalue for $f(A)$, does there exists a eigenvalue $a$ of $A$ such that $f(a) = c$?
Please, explain why this is true or false.
The statement is true. One proof is as follows: let $f(z) - c = (z-z_1)\cdots(z-z_d)$ be a factorization into linear factors. Each $z_i$ satisfies $f(z_i) = c$. If $f(A)$ has $c$ as an eigenvector, then $f(A) - cI$ is not invertible. Applying our above factorization, this means that the matrix product $$ (A - z_1 I) \cdots (A - z_d I) $$ fails to be invertible. Thus, $(A - z_i I)$ is non-invertible for some $i$. That is, $z_i$ must be an eigenvector of $A$.