Talk:Tensor product

Ack! Too hard!!! My brain hurts! Can someone rewrite this so that a mere physicist like myself can understand it? -- Tim Starling 04:15 Mar 4, 2003 (UTC)

hmm... there seems to be a section under matrix multiplication on Kronecker product/direct product, which is the same thing as tensor product as far as i know, with a slightly different definition.(the rank is neglected) I know that my definition is correct. (the rank is relevant). The question is, how do we remedy this duplicity? Kevin Baas 2003.03.14

How's that for a start? See people, like this. Why are mathematicians bad prose writers? Sigh... I don't mean to invade on this. It just needs to be written clearly. Kevin Baas 2003.03.14

Thank you, yes it's certainly a good start. I like the split format ("pretentious part starts here") - it's ugly enough to encourage contributors to properly integrate the article.

As for duplicity, personally I like repetition in Wikipedia. After all, there's essentially no space limit and a huge pool of contributors, so you may as well give the reader the most specific, tailor made information possible. I think it would be enough to explain the connection in both articles, with links of course. Since matrix multiplication is a more general subject, the Kronecker product section should be kept brief, perhaps indicating that more information is to be found here. See meta:Consolidating v/s breaking up. -- Tim Starling 06:57 Mar 15, 2003 (UTC)

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The statement about the rank can't be possibly true:

${\displaystyle {\begin{bmatrix}a\end{bmatrix}}\otimes {\begin{bmatrix}b\end{bmatrix}}={\begin{bmatrix}ab\end{bmatrix}}}$

or still worse:

${\displaystyle {\begin{bmatrix}0\end{bmatrix}}\otimes {\begin{bmatrix}b\end{bmatrix}}={\begin{bmatrix}0\end{bmatrix}}}$

Also, I think the example should be given as a tensor product of matrices from which the vector case can be derived easily. -- looxix 14:12 Mar 15, 2003 (UTC)

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Maybe a tensor with 1 dimension has undefined rank? this is no disproof. i have very authoritave sources that verify the fact that the ranks are summed by a tensor product. can you tell me whether [0] has rank 1 or 2, or even 7? it's a 1x1x1x1x1x1.... tensor. But isn't 1 also ${\displaystyle e^{i*2\pi }}$?

The 'rank' of a tensor is actually not defined by the columns, rows, ect., but by an equation such as:

${\displaystyle {\bar {T}}^{ij}=T^{rs}{\frac {\partial {\bar {x}}^{i}}{\partial x^{r}}}{\frac {\partial {\bar {x}}^{j}}{\partial x^{s}}}}$

where the rank of such T is 2 because it has order 2, regardless of how many ${\displaystyle x^{i}}$'s or ${\displaystyle x^{j}}$'s there are.

kevin -2003.03.15

Nothing is wrong if the rank is the rank_(tensor), but it is linked to Rank of a matrix, which is (loosely speaking) the degree of linear dependency of a matrix and was what I talked about in my remark. -- looxix 17:25 Mar 15, 2003 (UTC)

That's my fault -- Kevin linked to rank and I disambiguated. I haven't studied tensors before, so I wouldn't know one from the other. -- Tim Starling 23:02 Mar 15, 2003 (UTC)

i fixed that. looxix, i don't understand what you mean by "giving the example as a tensor prodcut of matrices from which the vector result can be derived easily" - how would i express a tensor with rank>2 without using embedded matrices, which would be potentially confusing? -- Kevin Baas -2003.03.15

What I meant was an example like:

${\displaystyle {\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\end{bmatrix}}\otimes B={\begin{bmatrix}a_{11}B&a_{12}B&a_{13}B\\a_{21}B&a_{22}B&a_{23}B\end{bmatrix}}=\dots }$

But this is in fact what is called matrix direct product, sometimes also called matrix tensor product.

What should be better is a general formula such:

<math>A_{ij\dots}^{k\dots} \otimes B_{m\dots}^{npq\dots} = C_{ijm\dots}^{knpq\dots} \Rightarrow c_{ijm\dots}^{knpq\dots} = a_{ij\dots}^{k\dots} \times b_{m\dots}^{npq}