A set of $d\times d$ real or complex matrices is called “commonly irreducible” if those matrices do not jointly preserve a linear subspace with dimension strictly between $0$ and $d$.
I wanted to know whether the Kronecker product preserves this property. In other words, given two sets of commonly irreducible matrices $S$ and $Q$, is the set $ S\otimes Q = \{P_1 \otimes P_2 \mid P_1 \in S, P_2\in Q\} $ commonly irreducible as well?
The question can be rephrased as follows:
This question is answered in Tensor products of simple modules over algebras. The assertion is true if $$ is algebraically closed (a consequence of Burnside’s theorem on matrix algebras), but false if $$ is not algebraically closed.
So yes, over $ℂ$ the set $S ⊗ Q$ will be commonly irreducible again.
Over $ℝ$ we can use the linked question to find an explicit counterexample: the rotation matrix $$ J ≔ \begin{pmatrix} 0 & -1 \\ 1 & \phantom{-}0 \end{pmatrix} $$ preserves no one-dimensional linear subspace of $ℝ^2$, but the matrix $$ J ⊗ J = \begin{pmatrix} 0 & -J \\ J & \phantom{-}0 \end{pmatrix} = \begin{pmatrix} 0 & \phantom{-}0 & \phantom{-}0 & 1 \\ 0 & \phantom{-}0 & -1 & 0 \\ 0 & -1 & \phantom{-}0 & 0 \\ 1 & \phantom{-}0 & \phantom{-}0 & 0 \end{pmatrix} $$ will have to preserve some proper, non-zero linear subspaces of $ℝ^4$. We can actually read off a decomposition of $ℝ^4$ into one-dimensional linear subspaces that are preserved by $J ⊗ J$, namely $$ ℝ^4 = \langle e_1 + e_4 \rangle ⊕ \langle e_1 - e_4 \rangle ⊕ \langle e_2 + e_3 \rangle ⊕ \langle e_2 - e_3 \rangle \,. $$