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I understand how to do a tensor product of 2 vectors, but I am not quite sure how to incorporate a third one. Can someone explain to me how to compute tensor product if I have 3 vectors

$$u\otimes v\otimes w$$ $$u=(1,1)$$ $$v=(1,-2)$$ $$w=(-1,3)$$

I can't find any examples online, and I feel like I will understand it most if I see an example.

Thank you.

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  • $\begingroup$ Do the operation to u and v, then take the result of that, and do the operation to the result and w. $\endgroup$
    – Joppy
    May 2, 2018 at 9:00
  • $\begingroup$ But if tensor product of u⊗v is (1,-2,1,-2) then how can I do a ⊗ of that and w? $\endgroup$ May 2, 2018 at 9:29
  • $\begingroup$ Can you copy the method you used to do the last tensor product? What is the formula you are using? $\endgroup$
    – Joppy
    May 2, 2018 at 9:32
  • $\begingroup$ quantiki.org/wiki/tensor-product $\endgroup$ May 2, 2018 at 9:33

1 Answer 1

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Well, it's easier if you write them in index notation

$u^{i} = \begin{bmatrix}1\\1\end{bmatrix} $ $v^{i} = \begin{bmatrix}1\\-2\end{bmatrix} $ $w^{i} = \begin{bmatrix}-1\\3\end{bmatrix} $

then their tensor product $(u\otimes v\otimes w)$ simply has components: $$(u\otimes v\otimes w)^{ijk} = u^{i}v^{j}w^{k}$$

To be completely explicit:

$$\begin{array} (u\otimes v\otimes w)^{111} = u^{1}v^{1}w^{1} &= (1)(1)(-1) &= -1\\ (u\otimes v\otimes w)^{112} = u^{1}v^{1}w^{2} &= (1)(1)(3) &= 3\\ (u\otimes v\otimes w)^{121} = u^{1}v^{2}w^{1} &= (1)(-2)(-1) &= 2\\ (u\otimes v\otimes w)^{122} = u^{1}v^{2}w^{2} &= (1)(-2)(3) &= -6\\ (u\otimes v\otimes w)^{211} = u^{2}v^{1}w^{1} &= (1)(1)(-1) &= -1\\ (u\otimes v\otimes w)^{212} = u^{2}v^{1}w^{2} &= (1)(1)(3) &= 3\\ (u\otimes v\otimes w)^{221} = u^{2}v^{2}w^{1} &= (1)(-2)(-1) &= 2\\ (u\otimes v\otimes w)^{222} = u^{2}v^{2}w^{2} &= (1)(-2)(3) &= -6\\ \end{array}$$

So you see, in order to write the components of this tensor as a matrix you would need to arrange them in a cube. But the components are there.

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  • $\begingroup$ This is the most intuitive answer for what is the tensor product! $\endgroup$ Feb 16, 2019 at 6:00
  • $\begingroup$ @induction601 at least in coordinates, yes I think so. $\endgroup$ Feb 16, 2019 at 6:03
  • $\begingroup$ Is there a similar explanation for the tensor product of matrices? $\endgroup$ Feb 16, 2019 at 6:04
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    $\begingroup$ It's basically the same thing, but with more indices. If you do the tensor product of two 2x2 matrices $A_{ij}$ and $B_{ij}$ (with 4 components each) the result will be an object with 4 indices and 16 components. Three (2x2) matrices will give you 6 indices and 64 components, and so on. $\endgroup$ Feb 16, 2019 at 6:08
  • $\begingroup$ Now I see. For example, if we have $A$,$B$,$C$ of $2\times 2$ matrices, $$ (A\otimes B \otimes C)_{1:4,1:4} = \begin{bmatrix} a_{11}b_{11}C & a_{11}b_{12}C \\ a_{11}b_{21}C & a_{11}b_{22}C \end{bmatrix}$$ $\endgroup$ Feb 16, 2019 at 6:28

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