Timeline for Name for an operation on matrices?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| Oct 1, 2015 at 13:09 | comment | added | Grigory Yaroslavtsev | Treating j as an $n$-dimensional vector $j[t]$ is its $t$-th entry. | |
| Oct 1, 2015 at 7:19 | comment | added | Dima Pasechnik | what are these $B_{j[1]}$, $B_{j[2]}$, etc? | |
| Sep 30, 2015 at 18:10 | history | edited | darij grinberg | CC BY-SA 3.0 |
added 4 characters in body
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| Sep 30, 2015 at 11:55 | comment | added | Grigory Yaroslavtsev | Thanks. I should have fixed this earlier but can't edit the comment. The multiplicativity of rank only holds as an inequality $rank(A \dagger B) \ge rank(A) rank(B)$. | |
| Sep 30, 2015 at 6:32 | comment | added | მამუკა ჯიბლაძე | I've posted a question about the latter | |
| S Sep 30, 2015 at 0:04 | history | suggested | Tadashi |
Added relevant tag
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| Sep 29, 2015 at 23:46 | review | Suggested edits | |||
| S Sep 30, 2015 at 0:04 | |||||
| Sep 29, 2015 at 20:56 | comment | added | მამუკა ჯიბლაძე | As a starter, one needs an operation on vector spaces something like $V^{\otimes W}$ of dimension $\dim(V)^{\dim(W)}$; of course with chosen bases it is immediate, but is there an invariant version? | |
| Sep 29, 2015 at 20:37 | comment | added | Grigory Yaroslavtsev | Thanks, just for the rank argument tensor product is definitely enough, I was just wondering about the matrix itself. | |
| Sep 29, 2015 at 20:09 | comment | added | მამუკა ჯიბლაძე | If you just need rank multiplicativity, there is a more economic one - tensor product (of size $ab\times nm$, with $(A\otimes B)_{\langle i,k\rangle,\langle j,l\rangle}=A_{i,j}B_{k,l}$) | |
| Sep 29, 2015 at 18:42 | comment | added | Thomas Rot | I'm in favor of the yaroslavtsev super-slam. You can cite this MO question:) | |
| Sep 29, 2015 at 16:50 | comment | added | Grigory Yaroslavtsev | I just need the fact that $rank(A \dagger B) = rank(A)rank(B)$ which is easy to show directly. However, it would be helpful to know if this operation and its properties are already known so that I can just cite an appropriate source. I like your "super-slam" idea though :) | |
| Sep 29, 2015 at 16:33 | comment | added | Chris Ramsey | How is this operation arising? Or are you hoping to call it the "Yaroslavtsev super-slam"? | |
| Sep 29, 2015 at 16:15 | history | asked | Grigory Yaroslavtsev | CC BY-SA 3.0 |