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

Question

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 $\textbf{spectrum}$ of an element $a\in A$ is the set \begin{equation} \sigma(a) = \{\lambda\in \mathbb C : a-\lambda \text{ is not invertible in $A$}\}. \end{equation} 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 $a\in A$.)

Answer 1

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.

Answer 2

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