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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