Timeline for Tensor matricizations and their decompositions
Current License: CC BY-SA 3.0
15 events
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| May 10, 2022 at 8:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Jan 10, 2022 at 8:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 12, 2021 at 7:05 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Aug 13, 2021 at 6:36 | answer | added | Sebastian K. | timeline score: 1 | |
| Apr 5, 2016 at 17:50 | comment | added | Nathaniel Johnston | @qbit- For question 1, I'm not quite sure... you can get some very extreme behavior, but I'm not sure about the intermediate behavior. For example, if rank(M) = 1, it could be the case that rank(N) = 1 or rank(N) = n^2 (the largest possible value: n is the dimension of each of the 4 tensor indices). For one extreme, let N be the n^2-by-n^2 identity matrix: then M has rank 1. However, it's not clear to me (for example) whether or not it's possible that if rank(M) = 3 then rank(N) could be arbitrary. If you're interested in absolute separability/ever want to discuss it, feel free to e-mail me. | |
| Apr 1, 2016 at 19:02 | comment | added | qbit- | @Nathaniel Johnston Oh, I've found your paper here arxiv.org/pdf/1405.5853 :) | |
| Apr 1, 2016 at 18:56 | history | edited | qbit- | CC BY-SA 3.0 |
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| Apr 1, 2016 at 18:54 | comment | added | qbit- | @Nathaniel Johnston, thanks for commenting. Indeed, I'm interested in this problem because it's relation to quantum entanglement. I've corrected the question to clarify it | |
| Apr 1, 2016 at 18:50 | history | edited | qbit- | CC BY-SA 3.0 |
added 218 characters in body
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| Mar 31, 2016 at 17:20 | comment | added | Nathaniel Johnston | Unfortunately I'm not sure how to answer this question right now because it is so general. For example, yes, the SVD of $N$ can be calculated using the SVD of $M$: use the SVD of $M$ to reconstruct $M$ itself, which can be used to construct $N$, which can be used to get the SVD of $N$. Of course, this certainly isn't what you actually want, but I'm not really sure what you do want either. It seems like you're perhaps interested in how the realignment map from quantum entanglement affects the singular values of a matrix. Is this correct? | |
| Mar 24, 2016 at 17:28 | comment | added | qbit- | Just following on the idea of using multilinear SVD (HOSVD). In HOSVD the tensor is decomposed to a "core" tensor and orthogonal factor matrices along each dimension. $t_{ijkl} = \sum_{abcd} \sigma_{abcd} U_{ai} V_{bj} W_{ck} Y_{dl}$. Permuting the original tensor would mean permuting the core tensor and the order of factor matrices. However, I don't see how the core tensor from HOSVD is connected to the singular value matrix $S$ of $M$. In fact, if we set the size of the dimensions to be $n$, $\sigma$ can have up to $n^4$ elements, whereas $S$ will have only $n^2$ at most | |
| Mar 24, 2016 at 9:27 | comment | added | qbit- | @Surb I've read it some time ago, but thank you for bringing this point | |
| Mar 24, 2016 at 9:00 | comment | added | Surb | Do you know the paper A multilinear singular value decomposition? It gives some nice insight about SVD of tensors unfoldings. | |
| Mar 24, 2016 at 7:06 | review | First posts | |||
| Mar 24, 2016 at 8:30 | |||||
| Mar 24, 2016 at 7:04 | history | asked | qbit- | CC BY-SA 3.0 |