Let $ P \in \mathbb{F}[X] $ a polynomial, $T:V \to V$ a linear operator..
Prove or disprove:
$ \lambda $ is an eigenvalue of $T$ iff $ P(\lambda)$ is an eigenvalue of $P(T)$.
Given that $ \lambda $ is an eigenvalue of $T$, its quite easy to prove the second half, however, given the second half gives me nothing to work with to prove the first half, so I'll assume its not correct.
What would be a way to construct a sufficient counterexample? Every polynomial I can find shows that this claim is true, but I can't manage to prove it. Could I somehow use the minimal polynomial as an example perhaps?
Best Answer
Let $T$ be the identity, and let $P(x)=x^2$.
Then $P(-1)=1$ is an eigenvalue of $T^2$, but $-1$ is not an eigenvalue of $T$.