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It is easy to see that for any finite $n$, the invertible elements in the algebras algebras $\mathbb{C}^n$ (edowed with the maximum norm, say) and $M_n$ (all square matrices) are dense. Thus, by the Artin-Wedderburn all finite-dimensional C*-algebras have dense invertible groups.

Is there an example of a unital, finite-dimensional Banach algebra where the invertible elements are not dense?

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No. The spectrum of a non-invertible element is finite. So you can find sequence of numbers in the resolvent set converging to zero. From this, you can cook up a sequence of invertible elements converging to your non-invertible element.

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  • $\begingroup$ What if the algebra is real? $\endgroup$ – Tomasz Kania Apr 17 '17 at 20:51
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    $\begingroup$ The spectrum of an element of a real algebra equals the spectrum in the complexification. You can choose the sequence to be made of real numbers. $\endgroup$ – vap Apr 17 '17 at 21:16
  • $\begingroup$ If $a \in A$ has finite spectrum (wit $A$ real or complex), then there is a very small $\epsilon > 0$ with $0 \notin \sigma(a + \epsilon)$. $\endgroup$ – user42761 Apr 18 '17 at 7:33

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