**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?



Simple $C^*$ algebras with invariant subspace property
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". 


