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.
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2I think you want to say $V_i^* \otimes W_i$. Anyway, doesn't the $(p,q)$ tensor thing only make sense if you have one fixed vector space? – Hoot Jul 08 '16 at 22:56
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@Hoot You are right I think, feel free to edit it. what I wrote is with a big abuse of language(notation). – Kavim Jul 08 '16 at 23:13
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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.
Ben Grossmann
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@N.Li Why would $T_1,T_2$ be a (0,2) tensors? They're linear transformations, so they should be (1,1) tensors. Anyway, as the comment on the question states, the (p,q) tensor thing only makes sense if you have one fixed vector space, but we're considering a linear transformations between tensor products of spaces, i.e. from the tensor product $V_1 \otimes V_2$ to the tensor product $W_1 \otimes W_2$. It turns out such a linear transformation can be canonically identified with a (2,2) tensor over $V_1\times V_2\times W_1^\times W_2^$, but that is not we're doing here. – Ben Grossmann Dec 17 '24 at 16:49
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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 $f_i \otimes f'_j$.
Nate
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