Theorem 29.4.6 (Gelfand-Naimark).label Let $A$ be a unital Banach algebra. If every non-zero element of $A$ is invertible, then $A$ is isometrically isomorphic to $\complex$.
Proof. Let $x \in A$. By Proposition 29.4.5, there exists $\lambda \in \sigma_{A}(x)$. Since $\lambda \cdot 1 - x$ is not invertible and every non-zero element of $A$ is invertible, $x = \lambda \cdot 1$. Therefore the mapping $\complex \to A$ defined by $\lambda \mapsto \lambda \cdot 1$ is an isometric isomorphism.$\square$
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