Sets of matrices which are irreducible but not strongly irreducible A set of $d \times d$ real or complex matrices is commonly called irreducible if those matrices do not jointly preserve a linear subspace with dimension strictly between zero and $d$. A stronger hypothesis which is useful in multiplicative ergodic theory - for example, in Furstenberg's theorem on random matrix products - is that the matrices do not jointly preserve a finite union of nontrivial proper subspaces. This condition is referred to as strong irreducibility. Both of these properties also pass to the semigroup generated by the set. I would like to ask:

Is there a precise characterisation of sets (or semigroups) of matrices which are irreducible but not strongly irreducible?

An irreducible set of matrices which generates a finite semigroup will not be strongly irreducible. (Clearly if we can achieve this situation by multiplying each matrix by a nonzero scalar then the matrix set is also not strongly irreducible.)  Another example of irreducibility without strong irreducibility is that in which $\mathbb{R}^d$ or $\mathbb{C}^d$ splits into a direct sum of $k$-dimensional subspaces which are permuted by the different matrices. Do these two mechanisms exhaust the range of possible examples?
Thanks in advance!
 A: The answer to your second question is no.
Let $E_1$, $E_2$, $E_3$ be pairwise transverse $2$-dimensional subspaces of $\mathbb{R}^4$.
Consider the following semigroup:
$$
\Sigma := \{M \in \mathrm{Mat}(4,4) ; \; M(E_i)=E_i, i=1,2,3\}.
$$
[Edit] The following remark will be useful: Any linear transformation $T_1 : E_1 \to E_1$ can be uniquely extended to an element of $\Sigma$. Indeed, with respect to the splitting $E_1\oplus E_2$, the space $E_3$ is the graph of some isomorphism $L \colon E_1 \to E_2$. Then $\Sigma$ is exactly the set of linear transformations $T\colon \mathbb{R}^4 \to \mathbb{R}^4$ whose block form with respect to the splitting $E_1\oplus E_2$ is:
$$
T = 
\begin{pmatrix} 
T_1 & 0 \\ 
0 & L T_1 L^{-1} 
\end{pmatrix}.
$$
Now, for definiteness, we choose the following three subspaces, given in coordinates $(x_1,x_2,x_3,x_4)$ by:
\begin{align*}
E_1 &:= \{ x_1 = 0 , \ x_2 = x_4 \}, \\
E_2 &:= \{ x_2 = 0 , \ x_3 = x_4 \}, \\
E_3 &:= \{ x_3 = 0 , \ x_1 = x_4 \},
\end{align*}
Consider the following orthogonal matrix which permutes the $E_i$'s:
$$
R := 
\begin{pmatrix}
0 & 0 & 1 & 0 \\ 
1 & 0 & 0 & 0 \\ 
0 & 1 & 0 & 0 \\ 
0 & 0 & 0 & 1 
\end{pmatrix}
$$
Note that $R$ preserves the orthogonal splitting $F \oplus G$, where 
\begin{align*}
F &:=  \{x_1+x_2+x_3=0, \ x_4=0\},\\
G &:=  \{x_1=x_2=x_3\}, 
\end{align*}
being a rotation of $120$ degrees on $F$, and the identity on $G$.
Let $\Gamma := \Sigma \cup \{R\}$.
Then:


*

*$\Gamma$ is not strongly irreducible, since $E_1 \cup E_2 \cup E_3$ is preserved.

*$\Gamma$ is irreducible. Indeed:


(a) $\Sigma$ preserves no $1$- nor $3$-dimensional subspace. This is easy to check using the "useful remark" above.
(b) the only $2$-dimensional subspaces preserved by $R$ are $F$ and $G$, but these are not preserved by $\Sigma$.


*The semigroup generated by $\Gamma$ is infinite (even after identifying rescaled matrices).

*$\Gamma$ preserves no union of subspaces forming a splitting of $\mathbb{R}^4$.  


Therefore $\Gamma$ is a counterexample to your proposed classification.
A: In case your set consists of non-singular (=invertible) matrices, the condition of strong irreducibility is equivalent to the condition that the connected component of identity of the Zariski closure of the group generated by your set acts irreducibly on ${\mathbb C}^d$. In both your cases, this is satisfied. 
Suppose $G$ is the Zariski closure and $G^0$ is the connected component of identity; it is a fact that $G/G^0$ is finite. Suppose that $G^0$ does not act irreducibly, and therefore, suppose that $W$ is an irreducible invariant subspace. Then the set of $G$ translates of $W$ is finite (since $G/G^0$ is finite) and hence all of $G$ preserves the union $\cup _{g\in G} g(W) $. In particular, all your matrices preserve this finite union.
Conversely, suppose the group $\Gamma $generated by the matrices preserves a finite union of proper subspaces. Such a union is a Zariski closed subspace of the projective space associated to the vector space. Hence the Zariski closure of $\Gamma$ preserves this finite union, hence so does $G^0$. 
If a connected group preserves a finite union of proper subspaces, it is (tedious but) easy to show that it preserves a proper subspace.  
