Theorem - Let $\mathcal{A}$ and $\mathcal{B}$ be $C^*$ -algebras and $\Phi : \mathcal{A} \longrightarrow \mathcal{B}$ an injective *-homomorphism. Then $\|\Phi(x)\|=\|x\|$ and $\sigma(\Phi(x)) = \sigma(x)$ for every $x \in \mathcal{A}$ , where $\sigma(y)$ denotes the spectrum of the element $y$ .

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Proof: It suffices to prove the result for unital $C^*$ -algebras, since the general case follows directly by considering the minimal unitizations of $\mathcal{A}$ and $\mathcal{B}$ . So we assume that $\mathcal{A}$ and $\mathcal{B}$ are unital and we will denote their identity elements by $e$ , being clear from context which one is being used.

Let us first prove the second part of the theorem for normal elements $x \in \mathcal{A}$ . It is clear that $\sigma(\Phi(x)) \subseteq \sigma(x)$ since if $x - \lambda e$ invertible for some $\lambda \in \mathcal{C}$ , then so is $\Phi(x) - \lambda e = \Phi(x - \lambda e)$ . Suppose the inclusion is strict, then there is a non-zero function $f \in C(\sigma(x))$ whose restriction to $\sigma(\Phi(x))$ is zero (here $C(\sigma(x))$ denotes the $C^*$ -algebra of continuous functions $\sigma(x) \longrightarrow \mathbb{C}$ ). Thus we have, by the continuous functional calculus, that $f(x) \neq 0$ and also that

Let $R_{\sigma}(z)$ denote the spectral radius of the element $z$ . From the norm and spectral radius relation in $C^*$ -algebras we know that, for an arbitrary element $x \in \mathcal{A}$ , we have that

Since $\Phi$ is isometric, $\Phi(\mathcal{A})$ is closed *-subalgebra of $\mathcal{B}$ , i.e. $\Phi(\mathcal{A})$ is a $C^*$ -subalgebra of $\mathcal{B}$ , and it is isomorphic to $\mathcal{A}$ . Using the spectral invariance theorem we conclude that $\sigma(x)=\sigma(\Phi(x))$ for every $x \in \mathcal{A}$ . $\square$