Timeline for A rank inequality
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 12, 2018 at 15:58 | comment | added | Tobias Fritz | @NathanielJohnston: good point, thanks! I confess that I hadn't actually written it out. | |
| Mar 12, 2018 at 12:38 | comment | added | Nathaniel Johnston | @Tobias - That's not true -- the partial transpose of the standard rank-1 maximally-entangled state in $M_2 \otimes M_2$ has rank 4. The partial transpose swaps the $b_i$ and $y_j$ terms in your sum, so you can no longer necessarily split up the sum over $i$ and the sum over $j$. | |
| Mar 11, 2018 at 20:10 | comment | added | Tobias Fritz | The operation that you're performing there is also called 'partial transposition'. It's not hard to show that partial transposition preserves rank, by simply computing that the partial transpose of a rank one matrix again has rank one. Just write the rank one matrix explicitly as $(\sum_i a_i\otimes b_i)(\sum_j x_j \otimes y_j)^T$ and compute the partial transpose. | |
| S Mar 11, 2018 at 16:33 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor edits because $\prime$ did not denote transposition
|
| Mar 11, 2018 at 11:22 | review | Suggested edits | |||
| S Mar 11, 2018 at 16:33 | |||||
| S Mar 11, 2018 at 11:19 | history | suggested | Rodrigo de Azevedo | CC BY-SA 3.0 |
Minor improvements
|
| Mar 11, 2018 at 10:23 | review | Suggested edits | |||
| S Mar 11, 2018 at 11:19 | |||||
| Mar 11, 2018 at 7:59 | comment | added | Federico Poloni | For the benefits of other readers, the same matrices written with Kronecker product notation: $M = \sum_{i=1}^r A_i \otimes B_i, \, M' = \sum_{i=1}^r A_i \otimes B_i^T$. | |
| Mar 11, 2018 at 5:35 | history | edited | SMD | CC BY-SA 3.0 |
added 13 characters in body
|
| Mar 11, 2018 at 4:58 | history | asked | SMD | CC BY-SA 3.0 |