Eigenvectors of a Lie group invariant covariant matrix Suppose you have a $n\times n$ covariance matrix $C$ that is commuting with all group elements, $g$, of a non abelian Lie group $G$, i.e. $[C,g]=0$ for all $g \in G$. Can we derive explicitly the form of the eigenvectors of $C$? If not can we derive any constrain on the form of the eigenvectors of $C$ from the invariance property of $C$?
Thanks!
Fabio 
 A: I assume you have $G$ acting on an $n$-dimensional complex vector space $V$ and $A$ is some linear transformation from ${\rm End}(V)$. Over an algebraically closed field, two linear maps commute iff one preserves the eigendecomposition of the other. (Can you prove this fact? Try!)
Hence $A\in{\rm End}_G(V)$ is $G$-equivariant (i.e. $[A,\rho(g)]=0$ for all $g\in G$) if and only if its eigenspaces are all invariant subspaces. This provides a fairly constructive recipe for determining all such intertwiners $A$ (assuming we have the decomposition of $V$ into irreducible representations): take the decomposition into irreducibles and partition it, then add the irreducibles in each cell together to form a collection of invariant subspaces, then attach a distinct scalar to each invariant subspace, then decompose each subspace into Jordan blocks associated to its given scalar as desired.
Since every vector $v\in V$ is contained in an invariant subspace, every vector is an eigenvector of some intertwiner. The real question is which sets of vectors can be eigenvectors simultaneously, which are those contained in some family of invariant subspaces of some decomposition of $V$; I am not sure how to answer that directly at the moment. Perhaps it might be useful to note that the isotypical components of $V$ are an invariant, but the decomposition of each isotypical component into irreducible summands is not unique. I am not sure how to leverage this.
