Simple $C^*$ algebras with invariant subspace property Edit: According to the valuable  comment of  Yemon Choi I  revise the  question by  replacing "faithful" with "irreducible". 
We  say  that  a  $C^*$ algebra  $A$  satisfies the invariant  subspace property  if  there exist  an  irreducible   representation $\phi: A \to B(H)$, for  some  Hilbert space $H$, such that   $\forall  a \in A,\; \phi(a)$ has  a  nontrivial  closed invariant subspace of $H$.

What  is  an example  of  a simple $C^*$  algebra  with this property not isomorphic  to the matrix algebra or the  algebra of  compact operators?

In particular, does $C^*_{red}(F_2)$  satisfy this property?
 A: Any unital C*-algebra has this property. This follows from these two facts: 
Fix $a\in A$.
(1) If $a-\lambda\cdot 1$ is not left invertible then there is an irreducible representation $\pi$ such that $\lambda$ is an eigenvalue of $\pi(a)$.
(2) The set of $\lambda$ such that $a-\lambda \cdot 1$ is not left invertible contains  the boundary of the spectrum of $a$. In particular, this set is nonempty.
Proof of (1): Assume first that $a$ is not left invertible, so that $\overline{Aa}$ is a proper left ideal in $A$. Choose a maximal left ideal $L$ containing $\overline{Aa}$. Then there exists a pure state $\phi$ such that 
$$
L=\{x\in A:\phi(x^*x)=0\}.
$$
See Theorem 2.9.5 in Dixmier's "C*-algebras". Now do the GNS construction with this state. We get that $\pi_{\phi}(a)\xi_\phi=0$ (because $a\cdot 1\in L$). So $\pi_\phi(a)$ has kernel. Now, in general,  if $a-\lambda\cdot 1$ is not left invertible then we apply the same argument to $a-\lambda \cdot 1$.
Notice that, conversely, if $\lambda$ in the spectrum of $a$ can be realized as an eigenvalue in some representation, then $a-\lambda \cdot 1$ cannot be left invertible.
Proof of (2): (This is well known.) Suppose that $a$ is left invertible and in the closure of the set of invertible elements. Say $ba=1$ and $c_n\to a$, with $c_n$ invertible. Then $bc_n\to 1$. So $bc_n$ is invertible for large $n$. Since $c_n$ is invertible, we get that $b$ is invertible, whence $a$ is invertible.
Somewhat digressing into the question of realizing the full spectrum as eigenvalues:
(1) If $A$ is finite (i.e., the unit is a finite projection) then left invertible implies invertible so the full spectrum can be realized as eigenvalues. (Pointed out by Yemon in the comments.)
(2) If $a$ is normal, then again left invertible implies invertible, so the same works. This case is a theorem in Pedersen's book "C*-algebras and their automorhism groups". 
