Timeline for $(AB)^+\approx B^+A^+$ for $B$ "fat" enough?
Current License: CC BY-SA 4.0
11 events
when toggle format | what | by | license | comment | |
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Jan 4, 2019 at 1:35 | answer | added | user114668 | timeline score: 2 | |
Jan 3, 2019 at 18:55 | comment | added | Federico Poloni | @Ludwig Yes, sorry, I was imprecise. The result holds if $B$ has orthonormal rows, as is stated also in the wiki page, but it's easy to see that it also holds if a multiple of $B$ has orthonormal rows, i.e., if the rows all have the same norm. But this property should hold again approximately thanks to LLN/CLT-type results. One would need to check if the normalization factor (which should be something like $\sqrt{n}$ or $n$) is tame or if it makes the whole expression diverge/converge. | |
Jan 3, 2019 at 16:53 | comment | added | Ludwig | @FedericoPoloni: Thanks for your comment! I was wondering whether the rows of $B$ need to be orthonormal for $(AB)^+=B^+ A^+$ to hold true. | |
Jan 3, 2019 at 16:07 | comment | added | user114668 | @Ludwig Ah, there is my mistake. $A$ has full row rank. Apologies! | |
Jan 3, 2019 at 15:46 | comment | added | Ludwig | @N.T.: Yes, I would say that $B_n$ is of full (row or column, depending on $m$) rank with probability 1 for every $n$. Why? | |
Jan 3, 2019 at 15:25 | comment | added | user114668 | I do not know much about random matrix theory, but is $B_n$ not of full rank with probability 1 independent of $n$? | |
Jan 3, 2019 at 9:19 | history | edited | Ludwig | CC BY-SA 4.0 |
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Jan 3, 2019 at 8:46 | history | edited | Ludwig | CC BY-SA 4.0 |
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Jan 2, 2019 at 22:12 | history | edited | Ludwig | CC BY-SA 4.0 |
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Jan 2, 2019 at 13:10 | comment | added | Federico Poloni | I guess it's because $B$ has 'almost orthogonal' rows (by LLN/CLT-type properties), and $(AB)^+=B^+A^+$ is a property that holds true when $B$ has orthogonal rows. Probably someone more versed than me with random matrices knows how to make this more formal. | |
Jan 2, 2019 at 12:58 | history | asked | Ludwig | CC BY-SA 4.0 |