Where is commutativity of $b$ needed?

Question

I have a question about the following proof:

If $e^{ia}-e^{i\lambda}=(a-\lambda)be^{i\lambda}$ and $(a-\lambda)$ is not invertible then $(a-\lambda)x$ is not invertible for all $x$. Why "since $b$ commutes with $a$"?

Answer 1

The canonical example where this happens is given by the unilateral shift in $\ell^2(\mathbb N)$. If $$ S(a_1,a_2,\ldots)=(0,a_1,a_2,\ldots), $$ then $S^*S$ is the identity, so invertible. But $$ SS^*(a_1,a_2,\ldots)=(0,a_2,a_3,\ldots) $$ so not invertible (it has kernel).

There are C$^*$-algebras where it is true that $xy$ invertible implies $yx$ invertible. This happens for example in the presence of a faithful tracial functional, i.e. a positive functional $f$ such that $f(ab)=f(ba)$ for all $a,b$ (tracial) and that $f(a^*a)=0$ implies $a=0$ (faithful).

Examples of C$^*$-algebras with this property are all finite-dimensional ones, and more. Concretely, any residually finite dimensional C$^*$-algebra has a faithful tracial state (proven by Choi in the 80s). Also, all II$_1$-factors have a faithful trace.