Given a Banach algebras: $$1_\mathcal{A}=1_\mathcal{B}\in\mathcal{A}\subseteq\mathcal{B}$$
For C*-algebras it holds: $$A\in\mathcal{A}:\quad \sigma_\mathcal{A}(A)=\sigma_\mathcal{B}(A)$$
But what about in general:
Are there elements for which this fails?
Of course $\sigma_B(T) \subset \sigma_A(T)$ for any $T \in A$. But in general, they need not be equal. For example, take $B = C(\mathbb{T})$, with $\mathbb{T}$ the unit circle in the complex plane. If we consider the identity function $z$, then $\sigma_B(z) = \mathbb{T}$, but consider the restriction to the subalgebra $A$ of $B$ which consists of functions with continuous extension to the disk and analytic on the interior (i.e. the closure of the subalgebra generated by 1 and $z$). Then we have $\sigma_A(z) = \bar{\mathbb{D}}$, the closed disk. In general, larger algebras can have 'holes' in spectra relative to the smaller subalgebra. See theorem 1.2.8 from Murphy's C*-algebra book for the precise statement about holes and more on this example.