Do isometries of regular convex polytopes generate everything? Let $D$ be a regular convex $d$-dimensional polytope that is symmetric $(D=-D$) in $\mathbb R^d$ centred at the origin. Let $G$ be the group of all isometries of $D$, that is, linear maps/matrices $T\in M_d$ such that $T[D]=D$. Does the linear span of $G$ in $M_d$ generate everything, that is, ${\rm span}\, G = M_d$, the space of all $d\times d$-matrices?
It is the case for $d$-dimensional cubes and octahedrons but is it true in general?
 A: Yes, this indeed the case. The proof has several ingredients.


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*Burnside's theorem: Suppose that $G< GL(n, {\mathbb C})$ is an irreducible subgroup, i.e. a subgroup which does not preserve any proper linear subspace in ${\mathbb C}^n$. Then the linear span of $G$ is the entire group $M_n({\mathbb C})$ of complex $n\times n$ matrices. (Take a look for instance here for a simple proof.) 


A consequence of this is that if $G< GL(n, {\mathbb R})$ is absolutely irreducible, i.e. it is irreducible as a subgroup of $GL(n, {\mathbb C})$, then the linear span of $G$ is the entire group $M_n({\mathbb R})$ of real $n\times n$ matrices. 


*Suppose that $G< GL(n, {\mathbb R})$ is an irreducible finite reflection group. Then $G$ is absolutely irreducible. See for instance Lemma 2.10 in 


W. G. Dwyer , C. W. Wilkerson, Centers and Coxeter elements. 


*The symmetry group $G$ of any regular convex polytope is a finite reflection group, see references given here. 
One can further show that $G$ is irreducible. (I will chase a reference when I have more time.)  


When you put all this together, you obtain the proof that the symmetry group of every complex polytope in ${\mathbb R}^n$ spans $M_n({\mathbb R})$.  
