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Let $\mathcal A$ be a commutative, unital Banach algebra. Take $A \in \mathcal A$ such that $A$ is non-scalar, i.e. $A\neq \alpha \mathbb I $, where $\mathbb I$ is the unit element. Denote the spectrum of $A$ as $\sigma(A)$ and take $\lambda \in \sigma(A)$.

Why is is the following true?

$$\vert \vert (A-\lambda \mathbb I)B - \mathbb I \vert \vert \geq 1, \forall B \in \mathcal A$$

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    $\begingroup$ If $\|x\|<1$, then $1-x$ is invertible thanks to the Neumann series. $\endgroup$ – Julien Apr 28 '13 at 15:55
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Otherwise, using power series, $(A-\lambda I)B$ would be invertible for some $B\in\mathcal A$, hence by commutativity, so would be $A-\lambda I$.

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