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Let $V$, $W$ be vector spaces over a field $K$ of finite dimension with bases $e_1, \ldots, e_n \in V$ and $f_1, \ldots, f_m \in W$. Then the tensor product $V \otimes W$ is defined as

$$ V \otimes W := \langle e_i \otimes f_j : 1 \leq i \leq n, 1 \leq j > \leq m \rangle_K.$$

The elements $e_i \otimes f_j$ are a base.

Thats the definition I don't understand. The definition is not complete, is it? How is $e_i \otimes f_j$ for elements $e_i \in V$ and $f_j \in W$ defined?

ATW
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  • $e_i\otimes f_j$ is just a new symbol for the base vectors, it does not have any further definition. – anankElpis Jul 03 '21 at 12:23
  • @StefanAlbrecht But what is $V \otimes W$ then? $e_i \otimes f_j$ are base vectors of what? For me, it looks like the set $V \otimes W$ is constructed by some mysterious new, undefined base vectors and I think thats my problem of understanding. – ATW Jul 03 '21 at 12:27
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    I suggest to consult a different text. This is not a definition, and even if we make it precise (see other comments: $V \otimes W$ is the free vector space generated by $B \times C$, where $B$ is a basis of $V$ and $C$ is a basis of $W$), it depends on a basis and is therefore useless. I always recommend https://kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod.pdf – Martin Brandenburg Jul 03 '21 at 12:28
  • You can see $e_i\otimes f_j$ as the matrix $e_if_j^T$, i.e. $(e_i\otimes f_j)_{k\ell}=\delta _{ik}\delta _{j\ell}.$ – joshua Jul 03 '21 at 12:28
  • Does this answer help? – Hans Lundmark Jul 03 '21 at 12:40
  • @MartinBrandenburg : I also recommend the chapter 14 in the "Advanced Linear Algebra" book of Prof. Roman. The representation of "Tensor product" as an universal pair in this chapter is easy quite easy to read, and it doesn't require reading other chapters in his book to understand. – Paresseux Nguyen Jul 03 '21 at 12:43
  • So $V \otimes W = V \times W$ provided with a structure of a vector space? – ATW Jul 03 '21 at 12:45
  • @ATW: Nevertheless, if you find these texts too general for you, you can just accept that $V \otimes W$ is a new vector space of dimension $mn$ which has a basis $ ( e_{i,j} )$ indexed by the pair the pair $(i,j)$ with $1\le i \le n , 1\le j \le m$( instead of $e_1,...,e_{mn}$) in usual). Moreover, this space has some interesting properties and that we can define a product $\otimes$ on this linear space so that this product has the associativity, distributivity, etc. – Paresseux Nguyen Jul 03 '21 at 12:49
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    @ATW: in answer to your last comment: No, the underlying set of $V \otimes W$ is not $V \times W$. The link supplied by Martin Brandenburg gives details. – Michael Cohen Jul 03 '21 at 12:55
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    One of troubles for you if you choose to understand this way is that you cannot make sense of most of demonstration concerning tensors but accept it as some special properties of an object called "tensor products" (who knows, perhaps it is not a downside but a good sign). P.s: Then if you face something like $U \otimes V \otimes W$ , may be you should see the tensor product $\otimes$ as some product that make you have a ring out of vector spaces. – Paresseux Nguyen Jul 03 '21 at 12:55
  • The article Martin Brandenburg supplied is helpful - so let me try to formulate it again: The tensor produkt $V \otimes W$ of vector spaces $V, W$ with bases $B_V \subseteq V$ and $B_W \subseteq W$ is a vector space with a basis $B_V \times B_W$ where the operations are given by bilinear continuation of the operators on $V$ and $W$? – ATW Jul 03 '21 at 13:36

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