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Let's say, that we have a $6 \times 6$ matrix $M$. By conducting an SVD of $M$ we obtain $USV^\dagger$ matrices, where $U$ is of size $6 \times 6$ and is left-normalized, $S$ is diagonal with 6 singular values on the diagonal, and $V^\dagger$ is right-normalized, also of size $6 \times 6$.

Now, I want to reshape $U$ into a rank-3 tensor of size $6 \times 2 \times 3$, and similarly $V^\dagger$ into a $2 \times 3 \times 6$ tensor. My question is, is it possible to associate two diagonal matrices $S_0$ (of size $2 \times 2$) and $S_1$ (of size $3 \times 3$) to each of the newly obtained dimensions of $U$ and $V^\dagger$? By doing that, it would be best to somehow inference values of $S_0$ and $S_1$ from the initial $S$.

Using the tensor network notation I want to achieve the following: enter image description here

Or maybe there is some direct transition from the $M$ matrix to the final picture, which I just didn't see?

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If you still want $U$ and $V$ to be left/right normalized, then I don't think it is generally possible to split in your way. The singular value decomposition is unique up to phases if the singular values are not degenerate. Therefore, you are only allowed to change $S$ up to phases. According to your splitting, this means that $\mathrm{diag}(e^{i\theta}) S = S_0 \otimes S_1$. However, this is not possible if $S$ is entangled (that is, it has more than one nonzero singular value) regarded as a reshaped matrix of dimensions $4\times 9$, which is in general the case.

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  • $\begingroup$ You are right... Do you think that it would be possible to somehow apply in this picture disentanglers, to minimize the entanglement between bonds with size 2 and 3, to avoid the problem you mentioned? $\endgroup$ Jun 22 at 17:47
  • $\begingroup$ I'd say it is always possible to use disentanglers, but that will modify either U and V' and break their normalization conditions. For example, you can just let W = S*V' and then M = U * identity * W, and you are free to break the identity into tensor products. But that ruins the normalization of V so probably this is not useful. $\endgroup$
    – Yijian Zou
    Jun 23 at 10:45

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