One sided ideal in finite dimensional $\mathrm{C}^*$-algebras

Question

Let $\mathcal{A}$ be a finite dimensional $\mathrm{C}^*$-algebra and $\mathcal{A}p$ a one-sided, non-self-adjoint left ideal. That is there exists $f_1,f_2\in\mathcal{A}$ such that $$f_1pf_2\not\in\mathcal{A}p.$$

Can the "not a right ideal"ness of $\mathcal{A}p$ be exhibited by a projection $q=f_1=f_2$?

Answer 1

Lemma. If $p$ and $f$ are projections in a C*-algebra such that $fpf$ is idempotent, then $p$ and $f$ commute.

Proof. We have $$ (pf-fp)^3= (pfpf - fpf - pfp + fpfp)(pf-fp) = $$$$ = pfpfpf - fpfpf - pfpf + fpfpf - pfpfp + fpfp + pfpfp - fpfpfp = $$$$ = pfpf - fpf - pfpf + fpf - pfpfp + fpfp + pfpfp - fpfp = 0. $$ Since $pf-fp$ is skew-adjoint, it follows that $pf-fp=0$, so $p$ and $f$ commute. $\qquad\square$

Proposition. Let $A$ be a C*-algebra generated by projections and let $p$ be a fixed projection in $A$. If $$ fpf\in Ap, $$ for every projection $f$, then $p$ is central (and hence $Ap$ is a two sided ideal).

Proof. If $fpf \in Ap$, then $fpf = fpfp$. Multiplying on the right by $f$ gives $fpf = fpfpf$, which means that $fpf$ is idempotent, and hence $f$ and $p$ commute by the Lemma. Since $f$ is an arbitrary projection in $A$, and $A$ is generated by projections, then $p$ is central. $\qquad\square$