how the Kronecker product is a tensor product?

Question

I have seen this question but still don't understand how the Kronecker product is a tensor product. If $T_1 : V_1 \to W_1$ and $T_2 :V_2 \to W_2$ are linear transformations their matrix representation with respect to chosen bases are $[T_1]_{m \times n} $ and $[T_2]_{m' \times n'} $ and their K-product is a $(mm') \times (nn')$ matrix. On the other hand there is a natural isomorphism between $L (V_i, W_i)$ and $V_i^* \otimes W_i$ for $i=1,2$. Now let $\widetilde{T}_1$ and $\widetilde{T}_2$ be the corresponding tensors, $\widetilde{T}_1 \otimes \widetilde{T}_2$ is like a $(2,2)$ tensor which is a rank two tensor so is not a matrix.

Answer 1

Think about what you are doing when you write down a matrix for a linear map $\ell: V \to W$. You are just writing the linear map as a linear combination of maps of the form $e_i^* \otimes f_j$, where $e_1,e_2, \dots$ is a basis for $V$ with dual basis $e_1^*, e_2^* \dots$ and $f_1,f_2, \dots$ is a basis for $W$. The entries of the matrix are just the coefficients of the linear combination.

Now given two linear maps $\ell: V \to W$, $\ell': V' \to W'$ we get a linear map $\ell \otimes \ell': V \otimes V' \to W \otimes W'$ which sends $v\otimes v'$ to $\ell(v)\otimes \ell'(v')$. The Kronecker product of matrices is just the matrix for this linear map expressed in the bases $e_i \otimes e'_j$ and $f_i \otimes f'_j$.

Answer 2

Let $A_i$ denote the matrix of the transformation $T_i$ with respect to the canonical basis of $V_i$ and $W_i$ for each $i$. Let $(e_{i,j})$ denote the canonical basis of $V_i$, and $f_{i,j}$ the basis of $W_i$.

The matrix of $T_1\otimes T_2$ with respect to the bases $\{e_{1,j}\otimes e_{2,k}\}$ and $\{f_{1,j} \otimes f_{2,k}\}$ (each taken in lexicographical order) is the Kronecker product.