Spectral permanence for Hermitian Banach algebras (${\sigma }_B(b)={\sigma }_A(b)$)

Let $B$ be a closed star-subalgebra of a unital Hermitian Banach algebra $A$, containing the same identity. For Hermitian $h\in B$, the complement of ${\sigma }_A(h)\subseteq \mathbb{R}$ is connected. The resolvent belongs to $B$ at large modulus by the Neumann series; its membership in $B$ is both open and closed in that complement, so ${\sigma }_B(h)={\sigma }_A(h)$. If $b$ is invertible in $A$, both $b{b}^{\ast }$ and $b^{\ast }b$ are invertible there and Hermitian algebra elements, hence invertible in $B$. Their inverses supply respectively a right and a left inverse for $b$ in $B$, proving invertibility there. Applying this to $b-\lambda 1$ proves the spectrum equality.