# Spectrum of an element in sub-algebra: $\sigma_A(b)\setminus \{0\}\subseteq \sigma_B(b) \setminus \{0\}$

Please help me to prove this:(or give me some references for this.) Thanks very much!

Let $A$ be a (unital) algebra and $B\subset A$ a (unital) sub-algebra. Then for all $b\in B$: $\sigma_A(b)\setminus \{0\}\subseteq \sigma_B(b) \setminus \{0\}.$

Recall that if $A$ is any unital algebra, the {\bf spectrum} of an element $a\in A$ is the set $$\sigma(a) = \{\lambda\in \mathbb C : a-\lambda \text{ is not invertible in A}\}.$$ If $A$ is not unital, then $\sigma(a)$ is defined to be the spectrum of $a$ in the unitization $\widetilde A$. (It follows that if $A$ is non-unital, then $0\in\sigma(a)$ for every $a\in A$.)

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How are you defining the spectrum of an operator in a non-unital algebra? – Kevin Carlson Aug 9 '12 at 6:26
see above remark. – Spring Xiao Aug 26 '12 at 9:15

## 2 Answers

The unital case doesn't take much proving. If $b-\lambda I$ is invertible in $B$, it has the same inverse in $A$; so by contraposition we get that if it's singular in $A$, it's singular in $B$, and we don't even have to worry about leaving 0 out of the spectra.

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The units maybe different – Spring Xiao Aug 9 '12 at 10:03
Sure-I just meant the case in which the inclusion itself is unital. – Kevin Carlson Aug 9 '12 at 22:11

For the unital case, see Theorem 11.29 of Rudin, "Functional Analysis".

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