Are the invertible elements dense in a finite dimensional Banach Algebra? 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.
 A: Yes, $G(A)$ is dense in $A$.  Here's one way to prove it.  Given $a\in A$, there is a unique $\mathbb{C}$-algebra homomorphism $\mathbb{C}[x]\to A$ sending $x$ to $a$.  Since $A$ is finite-dimensional, this homomorphism is not injective, and its kernel is generated by some nonzero polynomial $p(x)$.  The subalgebra of $A$ generated by $a$ is then isomorphic to $\mathbb{C}[x]/(p(x))$, with $a$ corresponding to $x$.
Now note that if $\lambda\in\mathbb{C}$ is not a root of $p(x)$, then $x-\lambda$ is invertible in the ring $\mathbb{C}[x]/(p(x))$ (proof: by polynomial long division, there is some polynomial $q(x)$ such that $p(x)=q(x)(x-\lambda)+p(\lambda)$; then in $\mathbb{C}[x]/(p(x))$, $-p(\lambda)^{-1}q(x)$ is an inverse for $x-\lambda$).  Thus if $\lambda$ is not a root of $p(x)$ then $a-\lambda$ is invertible in $A$.  Since $p(x)$ only has finitely many roots, there exists $\lambda$ arbitrarily close to $0$ such that $a-\lambda$ is invertible.  Thus there are elements of $G(A)$ arbitrarily close to $a$.  Since $a\in A$ was arbitrary, this means $G(A)$ is dense in $A$.
A: 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)} $.
