Timeline for $(AB)^+\approx B^+A^+$ for $B$ "fat" enough?
Current License: CC BY-SA 4.0
7 events
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| Jan 15, 2019 at 16:44 | comment | added | Ludwig | I see, thank you for clarifying! So the missing step is to show that $\|(AB_n)^+-B_n^+A^+\|$ tends to zero (in probability?) as $n$ goes to infinity, and this should follow from the fact that $B_n$ is almost orthogonal as $n$ tends to infinity. Am I correct? | |
| Jan 11, 2019 at 12:30 | comment | added | user114668 | Note that this limit is just a heuristic and not used; the important step is to find a suitable $n$ later on, which you would have to make rigorous. Nevertheless, if you want to know the limit, I would expect (at least) convergence in probability relatively straightforwardly from Markov's inequality for $\Pr[\frac{\sigma^2}{\sqrt{n}} \Vert R \Vert > \epsilon]$ plus some variance-type inequality for $\mathrm{E}[\Vert R \Vert^k]$ in the assumed norm, since $R$ is of fixed size $m \times m$ and does not grow with $n$. | |
| Jan 11, 2019 at 10:15 | comment | added | Ludwig | Oh I see, sorry for the silly question. One last comment: which type of convergence are you considering when you write $\lim_{n\to \infty} \frac{1}{n} B_n B_n^*=\sigma^2 I$? | |
| Jan 11, 2019 at 10:07 | comment | added | user114668 | You get the expectation and covariance from the fact the entries of $B_n$ are independent and normal. I have added the steps to the answer, below the line. | |
| Jan 11, 2019 at 10:06 | history | edited | user114668 | CC BY-SA 4.0 |
added 1215 characters in body
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| Jan 11, 2019 at 9:26 | comment | added | Ludwig | Thanks for your answer. Could you please elaborate a little more (or provide a reference) on the derivation of the first two displayed equations? | |
| Jan 4, 2019 at 1:35 | history | answered | user114668 | CC BY-SA 4.0 |