The spectral permanence theorem (parent entry) relates the spectrums $\sigma_{\mathcal{B}}(x)$ and $\sigma_{\mathcal{A}}(x)$ of an element $x \in \mathcal{B} \subseteq \mathcal{A}$ relatively to the Banach algebras $\mathcal{B}$ and $\mathcal{A}$ .

For $C^*$ -algebras the situation is quite simple.

Spectral invariance theorem - Suppose $\mathcal{A}$ is a unital $C^*$ -algebra and $\mathcal{B} \subseteq \mathcal{A}$ a $C^*$ -subalgebra that contains the identity of $\mathcal{A}$ . Then for every $x \in \mathcal{B}$ one has

The spectral invariance theorem is a straightforward corollary of the next more general theorem about invertible elements in $C^*$ -subalgebras.

Theorem - Let $x \in \mathcal{B} \subset \mathcal{A}$ be as above. Then $x$ is invertible in $\mathcal{B}$ if and only if $x$ invertible in $\mathcal{A}$ .

Proof :

• $(\Longrightarrow)$

  If $x$ is invertible in $\mathcal{B}$ then it is clearly invertible in $\mathcal{A}$ .

• $(\Longleftarrow)$

  If $x$ is invertible in $\mathcal{A}$ , then so is $y=x^*x$ . Thus, $0 \notin \sigma_{\mathcal{A}}(y)$ .

  Since $y$ is self-adjoint, $\sigma_{\mathcal{A}}(y) \subseteq \mathbb{R}$ (see this entry), and so $\mathbb{C} - \sigma_{\mathcal{A}}(y)$ has no bounded connected components.

  By the spectral permanence theorem we must have $\sigma_{\mathcal{B}}(y)=\sigma_{\mathcal{A}}(y)$ . Hence, $0 \notin \sigma_{\mathcal{B}}(y)$ , i.e. $y$ is invertible in $\mathcal{B}$ .

  It follows that $x^{-1}=(x^*x)^{-1}x^*=y^{-1}x^* \in \mathcal{B}$ , i.e. $x$ is invertible in $\mathcal{B}$ . $\square$