5
$\begingroup$

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.

  • 0
    [Google books search: Spectral Mapping Theorem](http://www.google.ca/search?q=Spectral+Mapping+Theorem&btnG=Search+Books&tbm=bks&tbo=1).2012-08-27

1 Answers 1

5

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.

  • 0
    **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 $a = p(\lambda_i)$.2012-08-27