# A couple of proofs on a spectrum

Let $T$ be a normal bounded operator. Let ${\lambda}$ be in $({\sigma}(T))$. Without invoking general algebra theories, show that:

a) $p({\lambda},{\lambda}^*)$ is in $({\sigma}(T))$ for all polynomials $p$

My thoughts : I know both ${\lambda},{\lambda}^*$ are in the spectrum but why should it be closed under polynomialsI suppose I can reduce any such polynomial down to a scalar multiple of ${\lambda}$ or ${\lambda}^*$ but not sure that helps.

b) show that $max(|p({\lambda},{\lambda}^*)|$ is equal to $max(|{\lambda}|)$ where the max in both cases is take over all ${\lambda}$ in the spectrum

My thoughts: This may be obvious when it is shown why a) is true.

Thanks

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a) isn't right. It should probably be $p(\lambda,\lambda^\ast) \in \sigma(p(T,T^\ast))$. –  Daniel Fischer Feb 14 '14 at 14:36
can you help with b)? –  user108605 Feb 14 '14 at 15:16
/cleans glasses/ That's wrong too, just take $p(z,w) = 3z$ to have a counterexample. Something's seriously amiss here, better re-read the exercise carefully, and if you didn't forget some constraint here, check with the professor/teaching assistant. –  Daniel Fischer Feb 14 '14 at 15:28
how can i contact you when i've double checked? –  user108605 Feb 14 '14 at 15:36
You can ping me here. You know how @-pings work, I presume? –  Daniel Fischer Feb 14 '14 at 15:42