Eigenvalue of a polynomial evaluated in a operator Suppose $T:V\to V$, $p\in \mathcal{P}(\mathbb{C})$ (polynomials with complex coefficients), and $a\in \mathbb{C}$. Prove that $a$ is an eigenvalue of $p(T)$ if and only if $a=p(\lambda)$ for some eigenvalue $\lambda$ of $T$.
I can prove: if $a=p(\lambda)$ then $a$ is a eigenvalue of $p(T)$ because:
$$Tv=\lambda v$$
$$T^kv=\lambda^k v$$
$$p(T)v=p(\lambda)v=av$$
But, how can I justify the other direction?
Thanks for your help.
 A: 
Claim 1: $a$ is an eigenvalue of $p(T)$ $\Leftarrow$ $a=p(\lambda)$ for some eigenvalue $\lambda$ of $T$.

Proof: For any eigenvalue $\lambda$ of $T$, we have $Tv = \lambda v$. It is easy to show by linearity and induction that $p(T)v = p(\lambda)v$. Hence if $a = p(\lambda)$ then $a$ is an eigenvalue of $p(T)$. $\square$

Claim 2: $a$ is an eigenvalue of $p(T)$ $\Rightarrow$ $a=p(\lambda)$ for some eigenvalue $\lambda$ of $T$.

Proof: Over $\Bbb{C}$, we can factor (where $u \in \Bbb{C} - \{0\}$): $$p(x) - a = u \prod_{i=1}^{n} (x-\lambda_i) \\ \text{so } a = p(\lambda_i) \text{ for some } i.\tag{1}$$ If $a$ is an eigenvalue of $p(T)$ then $p(T) - aI$ is singular. But from $(1)$ we have $$p(T) -aI = u \prod_{i=1}^{n} (T - \lambda_iI)$$ So$^\dagger$ for some $i$ we have $T - \lambda_iI$ is singular, hence $\lambda_i$ is an eigenvalue for $T$. Recall from $(1)$ that $a = p(\lambda_i)$. $\square$

$^\dagger$ It's easy to show that if $AB$ is singular, then at least one of $A$ and $B$ is singular. Switch between a linear transformation and its matrix in some bases, then use the fact that determinant is multiplicative. 
