Element $a$ in a unital C$^{*}$-algebra $A$ with $\overline{\langle a\rangle}=A$, but $a$ not left-invertible

Question

Given $a$ in a C$^{*}$-algebra $A$, let $\langle a\rangle$ denote the (not necessarily closed) left ideal generated by $a$. My question is

Is there a unital C$^{*}$-algebra $A$ containing a non-left-invertible element $a$ such that $\overline{\langle a\rangle}=A$?

Clearly if $A=\langle a\rangle$, then $a$ must be left-invertible since $\{ca:c\in A\}=\langle a\rangle=A\ni 1_{A}$. But I can't seem to rule out the case when $A\not=\langle a\rangle$.

Answer 1

Your conditions guarantee that there exists a sequence $\{c_n\}$ with $c_na\to 1_A$. So, for any $n$ big enough, $\|c_na-1_A\|<1$ and so $c_na$ is invertible. Taking any such fixed $n$, we have obtained $c\in A$ such that $ca$ is invertible. So there exists $b\in A$ with $bca=1_A$. In particular, $a$ is left-invertible.

You cannot expect more than that. For instance, if $A=B(H)$ and $a$ is the unilateral shift, then $\langle a\rangle=A$, but $a$ is not right-invertible.