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Is there any difference between tensor product and Kronecker Product?

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The two notions represent operations on different objects: Kronecker product on matrices; tensor product on linear maps between vector spaces.

But there is a connection: Given two matrices, we can think of them as representing linear maps between vector spaces equipped with a chosen basis.

The Kronecker product of the two matrices then represents the tensor product of the two linear maps.

(This claim makes sense because the tensor product of two vector spaces with distinguished bases comes with a distinguish basis.)

All this and more is explained on wikipedia.

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    $\begingroup$ While true, it does not help to answer the question. I also stumbled here for help, but my guess is that you mean to say, "if you limit A and B to be linear maps, then the Kronecker product is the tensor product; but the tensor product is more general and is not equal to the Kronecker product when they aren't linear maps." Two things would help with your answer: (1) answer the actual question, and (2) please provide an example. Like the OP, I would like to understand the answer better. Thanks! $\endgroup$ – Mike Williamson Aug 1 '17 at 22:06
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    $\begingroup$ I expanded my answer. Note that I meant to say what I said, despite your speculation to the contrary. $\endgroup$ – Rasmus Aug 1 '17 at 22:26
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    $\begingroup$ @MikeWilliamson: how would you even define the tensor proudct of nonlinear maps? $\endgroup$ – darij grinberg Aug 1 '17 at 22:27
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    $\begingroup$ Every matrix represents a linear map. $\endgroup$ – Rasmus Aug 2 '17 at 17:11
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    $\begingroup$ @MikeWilliamson Linear maps are in some sense more general than matrices. They are also different (types of) objects, even though matrices can be used to represent some linear maps. The matrix representation of a particular linear map depends on the choice of basis of the domain and target space, for example, whereas the linear map itself is invariant under such choices. On the other hand a matrix can also represent a bilinear form, not just a linear map. Consider reading Axler's Linear Algebra Done Right for a good explanation of some of such subtleties. $\endgroup$ – Chill2Macht Oct 18 '17 at 17:20

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