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!



Sets of matrices which are irreducible but not strongly irreducible
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:


1. $\Gamma$ is not strongly irreducible, since $E\_1 \cup E\_2 \cup E\_3$ is preserved.
2. $\Gamma$ 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 $\Sigma$.


3. The semigroup generated by $\Gamma$ is infinite (even after identifying rescaled matrices).
4. $\Gamma$ preserves no union of subspaces forming a splitting of $\mathbb{R}^4$.


Therefore $\Gamma$ is a counterexample to your proposed classification.


