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.


2 Answers 2


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 $A$.

  • $\begingroup$ That doesn't really matter--all you need to know for this argument is that $\lambda\mapsto a-\lambda$ is continuous from $\mathbb{C}\to A$, which is immediate from the continuity of addition and scalar multiplication. I never used any topology on anything until the very last step. $\endgroup$ Dec 13, 2016 at 8:23

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)} $.

  • $\begingroup$ Can you tell why $\sigma(a)$ will be finite. $\endgroup$
    – user346635
    Dec 13, 2016 at 10:28
  • $\begingroup$ Let $a \in A$ and define $T:A \to A$ by $Tx=ax$. Then $T$ is a bonded linear operator on $A$ and we have $\sigma(a)=\sigma(T)$. This is easy to check (for a reference see: Bonsall, Duncan: Complete Normed Algebras,Proposition 19 in §3). Thus, if $A$ is finite-dimensional, then $T$ is a finite-dimensional operator. Therefore $T$ has finite spectrum. This shows that $ \sigma(a)$ is finite.The following result, due to Hirschfeld and Johnson (1971) might be of interest: let $B$ be a complex Banach algebra for which every element has finite spectrum. Then $B/rad(B)$ is finite-dimensional. $\endgroup$
    – Fred
    Dec 14, 2016 at 7:18

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