In electromagnetism, which is an abelian gauge theory, we have the nice fact that all of the components of $$ F_{\mu\nu} = \partial_\mu A_\nu - \partial_\nu A_\mu $$ are gauge invariant quantities. We can equivalently talk about the field in terms of $\mathbf{A}_{\perp}$ and $\mathbf{E}_{||}$, the solenoidal part of the vector potential and the irrotational part of the electric field (respectively), as being a compete set of gauge invariant quantities that fully determines the state of the field everywhere without degeneracy.
When we move to non-abelian Yang-Mills theory the field strength components \begin{align} F_{\mu\nu} & = \frac{i}{g}\left[D_\mu,\, D_\nu\right] \\ & = \partial_\mu A_\nu - \partial_\nu A_\mu - ig \left[A_\mu,\, A_\nu\right] \\ & = T^a \left(\partial_\mu A^a_\nu - \partial_\nu A^a_\mu +gf^{abc} A^b_\mu A^c_\nu\right) \end{align} become matrices in the group space ($T^a$ are generators of transformations of the group). Thus the elements of the field strength indexed by space-time, $F_{\mu\nu}$, are no longer gauge invariant but gauge covariant — transforming as a rank two tensor in whatever representation of the gauge group the the generators are in.
When I want to know the group invariant parts of a matrix I think "eigenvalues", at least when the group is defined by preserving an inner product on some vector space (as $\operatorname{SU}(N)$ is). In other words, it seems to me like the physical information will be contained in the gauge invariant parts of $F_{\mu\nu}$ and finding the eigenvalues sounds like the way to go about it.
Has this been done? It seems unlikely that a single gauge transformation could simultaneously diagonalize all of the $F_{\mu\nu}$ at any given point in space-time, what are the consequences of this (if it's true)? Finally, would the eigenvalues of $F_{\mu\nu}$ also be independent of which representation the $T^a$ are in (i.e. would finding the eigenvalues in the defining representation give the same result as any other representation), assuming the representations all have the same normalization condition (e.g. $\operatorname{Tr}(T^a T^b)= \delta_{ab} / 2$)?