Proving eigenvalue $\lambda$ is a root for an annihilating polynomial $p(t)$

Question

I have the next statement I need to prove:

Let $\lambda$ be an eigenvalue of a endomorphism $f$ and $p(t)$ an annihilating polynomial of $f$. Prove that $\lambda$ is a root for $p(t)$.

Thank you very much.

Answer 1

By assumption, there is a nonzero vector $v$ such that $f(v) = \lambda v$. Then $f^{2} (v) = f \circ f(v) = \lambda^{2} v$, etc.

Write $p(t) = t^n + a_{n-1} t^{n-1} + \dots + a_{1} t + a_{0}$.

Then $$ 0 = p(f) v = (f^n + a_{n-1} f^{n-1} + \dots + a_{1} f + a_{0}) v = (\lambda^n + a_{n-1} \lambda^{n-1} + \dots + a_{1} \lambda + a_{0}) v , $$ that is, $p(\lambda) = 0$.

Answer 2

Suppose that $v$ is the associated vector. To say $p$ annihilates $f$ is to say that $p(f) = 0$, i.e., that for every $w$ in the domain of $f$, $p(f)(w) = 0$, Apply this notion to compute $p(f)(v)$, where $v$ is the eigenvector for $\lambda$. What does $p(f)(v) = 0$ tell you?