I have a question regarding tensors and basis changes, however upon searching the web I've found an infinity of definitions for tensors so I'll have to give the one I know first:
Given a vector space $E$ over a (commutative) field $F$, we call, for $p\ge 1$ a $p$-covariant tensor a map of the type $$ f:E\times\overset{\text{(p)}}{\dotsb}\times E\to F $$ where $f$ is $p$-linear (linear in each of the $p$ components). We call the set of $p$-covariant tensors (over $E$) $ T_p(E)$, which is the set of $p$-linear maps from $E$ to $F$, in other words, $T_p(E) = \mathcal L_p(E,F)$.
Next we define for $f \in T_p(E),\ g \in T_q(E)$ the tensor product of $f$ and $g$, as $f \otimes g \in T_{p+q}(E)$, where: $${f\otimes g : E\times\overset{\text{(p)}}{\dotsb}\times E\times\overset{\text{(q)}}{\dotsb}\times E \to F} \atop {(u_1,\dots,u_p,v_1\dots,v_q) \mapsto f(u_1,\dots,u_p)\cdot g(v_1\dots,v_q)} $$
For example, for $p=1, \ T_1(E) = E^*$. For arbitrary $p$, it can be shown that $T_p(E)$ is a vector space over $F$, and if $e_1,\dots,e_n$ is a basis for $E$, with dual basis $e_1^*,\dots,e_n^*$ ($e_i^*(e_j) = \delta_{ij}$) then a basis for $T_p(E)$ is $\{e_{i_1}^*\otimes\dots\otimes e_{i_p}^*\}_{1\le i_1,\dots, i_p\le n}$
My question regards 2-covariant tensors. For simplicity, say we have $\varphi\in T_2(E)$, and $E$ is two-dimensional with basis $e_1,e_2$. Then $T_2(E)$ has the ordered basis $B_e = \{e_1^*\otimes e_1^*, \ e_1^*\otimes e_2^*, \ e_2^*\otimes e_1^*, \ e_2^*\otimes e_2^*\}$ and $\varphi = (a,b,c,d)$ in this basis.
What I want to know is, if $B_u$ is a similar basis of $T_2(E)$ relative to basis $u_1, u_2$ of $E$, how can I express $\varphi$ in basis $B_u$, using the fact that I know the basis change matrix that changes from $e_1,e_2$ to $u_1,u_2$ in $E$? I know it has to do with a Kronecker product of matrices, but I don't quite understand how it works.