Let $V$ be a vector space over a field $\mathbb{F}$. Suppose that $T: V \rightarrow V$ is a linear operator with an eigenvalue $\lambda$, and $v$ is an eigenvector of $T$ corresponding to $\lambda$.
Why is it true that, for every $p(x) \in \mathbb{F}[x]$, the scalar $p(\lambda)$ is an eigenvalue of the operator $p(T)$ and $v$ is also an eigenvector of $p(T)$ corresponding to the eigenvalue $p(\lambda)$?
Best Answer
If $p = a_0 + a_1 X + ... + a_n X^n$, and $Tv = \lambda v$ ($v \neq 0$), then
$$ p(T) v = a_0 Iv + a_1 Tv + ... + a_n T^n v = a_0v + a_1 \lambda v + ... + a_n \lambda^n v = p(\lambda) v. $$
This uses the definition of $p(T),$ and $T^n v = \lambda T^{n-1}v = ... = \lambda^n v.$