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?

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Maybe I am missing something, don't we need to prove this for an arbitrary $p$ rather than just $p(x)=x^k$? And for that matter why not use $p(x)=x$ which would work just as well... – axblount Aug 27 '12 at 17:46
– user2468 Aug 27 '12 at 22:53

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.

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Now that I read it all over, I could've written it much more concisely in 3-4 lines! – user2468 Aug 27 '12 at 22:42
Remark: the real trick about the other direction is (1) factoring $p(T) -aI$ into products of $T - \lambda_i I$ (2) showing that one of the factors is singular. (3) Hence $\lambda_i$ is an eigenvalue of $T$ (4) using the factoring to show that $a = p(\lambda_i)$. – user2468 Aug 27 '12 at 22:44