Show that any tensor, say the third-rank tensor $\textbf{T}$, can be expanded in terms of tensor products of the basis vectors: $$\textbf{T} = T_{ijk} e_i \otimes e_j \otimes e_k$$ and that the components $T_{ijk}$ of $\textbf{T}$ can be computed by the generalization $$T_{ijk} = \textbf{T} (e_i, e_j, e_k).$$
My attempt: I plugged in some vectors $A,B,C$ to get $$\textbf{T} (A,B,C) = A_i B_j C_k \textbf{T} (e_i, e_j, e_k),$$ which follows from the linearity of the tensor. But then I don't know how to connect this to either of the requested things to be proven. I would greatly appreciate any help! Thanks, and all the best.
Identifying covariant and contravariant things using the standard dot product in $\Bbb R^n$ only ofuscates what is going on and violates Einstein's convention (whose true power lies in its built-in error detector given by the distinction between upper and lower indices). Recall that if $(e_i)_{i=1}^n$ is a basis for your space with $(e^i)_{i=1}^n$ a basis for the dual space, the space of covariant rank $3$ tensors has a basis given by $(e^i\otimes e^j\otimes e^k)_{i,j,k=1}^n$. This means that if $T$ is such a tensor, we may write $$T = T_{ijk}e^i\otimes e^j\otimes e^k.$$How to find $T_{ijk}$? Evaluate both sides at the triple $(e_r,e_s,e_t)$ to get $$T(e_r,e_s,e_t) = T_{ijk}e^i(e_r)e^j(e_s)e^k(e_t) = T_{ijk}\delta^i_r\delta^j_s\delta^k_l = T_{rst}.$$Since the indices were arbitrary, we conclude that $T_{ijk} = T(e_i,e_j,e_k)$ for all $i,j$ and $k$.