Are there any nice ways of encoding tensor products of matrices using index notation? I.e. for standard matrix multiplication we have
$$(AB)_{ij} = \sum_k A_{ik} B_{kj} $$
Is there anything similar for tensor products?
$$(A\otimes B)_{ij} = ???$$
[Math] Tensor Product Between Matrices Index Notation
tensor-products
Related Solutions
There is no canonical way of having $4\times4$ matrices act on $2\times2$ matrices, but notice the space of $2\times2$ matrices is four-dimensional, and linear transformations of this space can be encoded as - you guessed it - $4\times4$ matrices. Write down the relations $(e_{ab}\otimes e_{cd})(e_u\otimes e_v)=\delta_{bu}\delta_{dv}(e_a\otimes e_c)$, where $e_{ij}$ are the elementary matrices and $e_k$ the obvious coordinate vectors, then convert everything on both sides to the relevant matrices via Kronecker products. This tells you how to apply elementary $4\times4$ matrices to elementary $2\times2$ matrices; everything else follows from using the distributive property (write matrices as sums of elementary matrices).
Tensors of vectors can be thought of in different ways. One way is as multidimensional arrays of numbers (usual vectors are one-dimensional arrays, matrices are two-dimensional, etc.), possibly with the "axes" of this array partitioned into upper or lower indices, or covariant versus contra-variant dimensions. This would be highly relevant in physics.
To a pure mathematician, $V\otimes W$ is first and foremost defined implicitly via universal properties, which comes with a canonical explicit construction in terms of the "symbols" $v\otimes w$ which satisfy bilinearity relations. This is my preferred method, as it captures the fact we're trying to left-multiply elements of $V$ against elements of $W$, make scalars in the base field commute with everything, and impose a distributive property.
Finally, as vector spaces there is an isomorphism $K^n\otimes_KK^m\cong M_{n\times m}(K)$ which takes pure tensors of coordinate vectors to their Kronecker product. I suppose this might be useful computationally in some settings, but I am not used to this method.
Fix a basis $\{e_1, \ldots, e_n\}$ of $V$, and consider the dual basis $\{f_1, \ldots, f_n \}$ of $V^\ast$. Then we have a basis $$\{e_1\otimes f_1,\ldots, e_i \otimes f_j, \ldots, e_n \otimes f_n\}$$ for $V \otimes V^\ast$, and the matrix $$A = (a_{ij})$$ is just a way of representing the element $$\sum_{i=1}^n \sum_{j=1}^n a_{ij} \; e_i \otimes f_j \in V \otimes V^\ast.$$
Of course an element of $V \otimes V^\ast$ gives a linear map $V \to V$ by
$$(w \otimes f)(v) := f(v) w$$
and extending by linearity. Given two such elements, we can compose the corresponding functions:
$$(w' \otimes f')(w \otimes f)(v) = (w' \otimes f')(f(v) w) = f(v) f'(w) w' = f'(w) \; (w' \otimes f)(v)$$
so composition of linear maps is given by
$$(w' \otimes f') \circ (w \otimes f) = f'(w) \; (w' \otimes f)$$
extended by linearity. If you write your elements in the $e_i \otimes f_j$ basis and apply this operation to them, you'll see that the usual definition of matrix multiplication pops right out.
Of course all the calculations with explicit tensors above can be rephrased in terms of the universal property of the tensor product if you like.
This is all assuming you want the matrix to represent an element of $V \otimes V^\ast$ rather than an element of $V \otimes V$ or $V^\ast \otimes V^\ast$. But you can work out what should happen in cases like that the same way.
Best Answer
If by "tensor product" you mean the Kronecker product, then we can write in block-matrix notation $$ (A \otimes B) = [a_{ij}B]_{i,j = 1}^n $$ Another thing you can say is that if $B$ is $m \times n$, then $$ (A \otimes B)_{(i-1)m + p,(j-1)n + q} = a_{ij}b_{pq} $$