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?
 A: 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. 
