Is an element in a homogeneous C$^{*}$-algebra whose image in each primitive quotient is invertible necessarily invertible?

Let $$N\in\mathbb{N}$$ and suppose $$A$$ is a unital C$$^{*}$$-algebra with the property that each irreducible representation of $$A$$ has dimension $$N$$. I.e., $$A$$ is $$N$$-homogeneous.

Suppose we have an element $$a\in A$$ such that $$\varphi(a)$$ is invertible in $$M_{N}(\mathbb{C})$$ for each irreducible representation $$\varphi$$ of $$A$$. Does it necessarily follow that $$a$$ will be invertible in $$A$$?

This is certainly true if $$A$$ is an algebra of the form $$C(X,M_{N}(\mathbb{C}))$$ (since the irreducible representations are equivalent to point evaluations), but what about more general homogeneous algebras?

If $$A$$ is separable, then any representation is (approximately unitarily equivalent to) a direct sum of irreps. So take a faithful representation $$\pi:A\to B(H)$$, then $$\pi\sim\bigoplus_j\pi_j$$. Then \begin{align} \sigma(a)&=\sigma(\pi(a))=\sigma(\bigoplus_j\pi_j(a))=\{\lambda:\ \bigoplus_j\pi_j(a)-\lambda I\ \text{ is not invertible }\}\\ \ \\ &=\{\lambda:\ \exists j, \pi_j(a-\lambda I)\ \text{ is not invertible }\}\\ \ \\ &=\bigcup_j\sigma(\pi_j(a)). \end{align} As $$\pi_j(a)$$ is invertible for all $$j$$, $$0\not\in\sigma(a)$$, and $$a$$ is invertible.
The fact that $$\sigma(\pi(a))=\sigma(\bigoplus_j\pi_j(a))$$ depends on the fact that approximate unitary equivalence preserves the spectrum. This in turn depends on the fact that approximate unitary equivalence preserves invertibility, since $$a=\lim_nu_nbu_n^*$$, then $$a-\lambda I=\lim_nu_n(b-\lambda I)u_n^*$$.
Now, let us show that if $$a=\lim_n u_nbu_n^*$$ and $$b$$ is invertible, then $$a$$ is invertible (which in turn, as mentioned, shows that approximate unitary equivalence preserves the spectrum). Indeed, note that $$a^*a=\lim_n u_nb^*b u_n$$ and that $$a$$ is invertible if and only if both $$a^*a$$ and $$aa^*$$ are invertible, so it is enough to check that approximate unitary equivalence preserves invertibility of positives. But this is easy: if $$b\geq0$$ and invertible, there exists $$c>0$$ with $$b-cI\geq0$$. But then $$a-cI\geq0$$, and so $$a$$ is invertible.
• Assuming $A$ is simple, is it true that we can only conclude that $\pi$ is approximately unitarily equivalent to a direct sum of irreducibles? If so wouldn't this only imply that $\pi(a)$ is in the closure of the invertibles in $\pi(A)$? Is it possible that since $A$ is $N$-homogeneous, we don't have to worry about the approximately unitarily equivalent part, and that, in fact, $\pi$ does decompose as a direct sum of irreducibles? – ervx Oct 27 '18 at 18:03