This is a Problem 4.12 of "Tensors and Manifolds w Applications to Mechanics and Relativity" by Robert Wasserman. I'm going through it on my own, independent of any courses (ie it's not homework).
"If A is a tensor of type (r,s), and A has the same components in every basis, then either A=0 or r=s."
I've got so far: the dual space of tensors of type (r,s) is isomorphic to tensors of type (s,r). If we let V(r,s) be tensors of type r,s over a finite dimensional space V, ie V(r,s) = V(x)...(x)V(x)V*(x)...(x)V* (where there are r V's and s V*'s); then V(r,s) is a vector space and V(s,r) is isomorphic to it's dual so <,>: V(r,s)xV(s,r) -> R is nondegenerate. So I was thinking of using the nondegeneracy property:
Let A be in V(r,s).
Let B be in V(s,r).
Then if <A,B> = 0 for all B, A = 0 by nondegeneracy of <,>. So I would need to show that if A != 0 there exists some B such that <A,B> != 0, but this implies r=s (SOMEHOW...).
For working with <A,B> though, wouldn't I need to wrap B in an isomorphism f, like <A, f(B)> ?