TITLE: Eigenvector of polynomial
QUESTION [7 upvotes]: Suppose that $T: V \rightarrow V$ is an endomorphism of the linear space V (about $\mathbb{K}$) and that $p(X)$ is a polynomial with coefficients in $\mathbb{K}$. Show that if $x$ is an eigenvector of $T$ than it is also an eigenvector of $p(T)$.

My attempt:
So if $x$ is an eigenvector of $T$ that means that $T(x) = \lambda x$ ($\lambda$ being the eigenvalue associated to $x$).
Ok so my next step is the one I feel is not correct 
$p (T(x)) = p (\lambda x)$ so $\lambda$ is an eigenvalue of the polynomial.
I don't feel this is a correct assumption, that you can't immediately conclude this, can  we?

REPLY [3 votes]: Adding a bit of generality to @Surb 's excellent accepted answer.
This exercise is just a glimpse of a very important technique in linear algebra. When $P$ is a polynomial with coefficients in $\mathbb{K}$ and $T$ is an endomorphism then $P(T)$ is also an endomorphism whose properties are closely related to those of $T$. When $\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$ then you can often make sense of $P(T)$ when $P$ is a convergent power series. That allows you to define useful endomorphisms like 
$$
e^T \quad \text{and} \quad \frac{1}{1-T}\ .
$$