I am self-studying Bishop and Goldberg's nice book on Tensors and Manifolds, but got stuck on Exercise 2.13.3: If $A$ is a tensor of type $(r,s)$ such that the components of $A$ are the same with respect to every basis, show that either $A=0$ or $r=s$.
My thoughts so far:
If $A=0$, then there is nothing to show.
If $A$ is nonzero, then I can see this for example if $(r,s)=(1,0)$: Then $A^{f,i}=a^i_j A^{e,j}$, where $A^{e,i}$ denotes the components of the $A$ with respect to a basis ${e_i}$ and likewise $A^{f,i}$ for the basis ${f_i}$, and $P=(a^i_j)$ is the matrix taking the basis ${e_i}$ to ${f_i}$. Now we can take any invertible $P$, and this yields with the choice of $P=\textrm{diag}(2,\cdots, 2)$, that all the components of $A$ must be zero, a contradiction. A similar proof can also be done for the case when $(r,s)=(0,1)$. But I don't yet see clearly the general case. Any help is appreciated. Thanks!