Hey MAX09.

Consider that C^n is an operator with det(C^n) != 0.

Now consider an eigen-decomposition of the matrix C^n.

Each matrix will have different eigenvalues assuming they are not all 1 or 0.

But this implies linear independence given the properties of the eigen-values.

If you can show how many distinct eigen-values there are you get the number of independent operators and thus the dimension of the entire space spanned by the set of operators.

I haven't given a complete proof but hopefully it is an alternative viewpoint to ponder for more ideas.