TITLE: Are the invertible elements dense in a finite dimensional Banach Algebra?
QUESTION [5 upvotes]: Let $A$ be a Complex unital Banach algebra which is finite dimensional. Let $G(A)$ be the set of all invertible elements in $A$. I want to know if the $G(A)$ is dense in $A$ in the norm topology.

REPLY [3 votes]: Let $a \in A$.
Case 1: $a \in G(A)$. Then $a \in \overline{G(A)} $.
Case 2: $a \notin G(A)$. Then $0 \in \sigma(a)$ ( = spectrum of $a$). Since $A$ is finite dimensional, $\sigma(a)$ is a finite set. Hence $0$ is an isolated point of $\sigma(a)$. 
Therefore we get a sequence $( \mu_n)$ in $ \rho(a) $ (= resolvent set of $a$) such that $ \mu_n \to 0$.
This gives $a- \mu_n e \in G(A)$ and $a- \mu_n e \to a$. Hence again $a \in \overline{G(A)} $.
These two cases show $A = \overline{G(A)} $.