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Looking at this paper here (see page 5, just above equation (7)) https://arxiv.org/pdf/1606.09225v1.pdf...

It shows an equation:

$$ X_{3of4} = I \otimes I \otimes X \otimes I $$ Where $I$ is a 2x2 identity matrix, and X is the matrix: $$ X = \left[ \begin{matrix} 0 & 1\\1 & 0\end{matrix} \right] $$

How would you work this out, as once you did $I \otimes I$, you would have a 4x4 matrix, so you couldn't then do it to the $X$ matrix...

Any help would be appreciatied

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    $\begingroup$ I don't understand what your confusion here is. Certainly, you can compute $$ ((I\otimes I)\otimes X)\otimes I $$ Or, if there's an issue there, what is it? Nothing about the Kronecker product necessitates that the two matrices have the same shape or size. $\endgroup$ – Omnomnomnom Aug 20 '16 at 23:42
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just for fun, \begin{align} I \otimes I \otimes X \otimes I = \begin{pmatrix} 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 \end{pmatrix} . \end{align}

As @Omnomnomnom points out, having computed one of the Kronecker products, you can continue with the others.

You can do this e.g. in Matlab/Octave with

I = eye(2); X = [0 1; 1 0]; kron(kron(kron(I, I), X), I).

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  • $\begingroup$ Thanks for that ;) did you just do them sequentially? The comment above said you don't actually need them to be the same size, that true? $\endgroup$ – Adam Kelly Aug 21 '16 at 0:48
  • $\begingroup$ Ah yes! Thanks for that! $\endgroup$ – Adam Kelly Aug 21 '16 at 0:49
  • $\begingroup$ @ user66081 , your equality $I\otimes I\otimes X\otimes I=\cdots$ is not quite correct. More precisely, $(E\otimes F)\otimes G$ and $E\otimes(F\otimes G)$ are not the same space but are isomorphic under a canonical isomorphism. Then, if you want explicitly write such a multiple tensor product, then you must choose some bracketing. Clearly, the explicit result depends on such a choice. $\endgroup$ – loup blanc Aug 29 '16 at 11:15
  • $\begingroup$ @loupblanc: I understand; but we are talking about the Kronecker product of matrices here $\endgroup$ – user66081 Aug 29 '16 at 21:49
  • $\begingroup$ @ user66081 , yes, you are right. $\endgroup$ – loup blanc Aug 29 '16 at 22:06

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