Timeline for Eigenvalue perturbation of a symmetric matrix by a random orthogonal projection
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Mar 10, 2016 at 19:37 | comment | added | Richard Zhang | Thanks Suvrit. Looking at the rank-1 case immediately proves that $\mathbf{Pr}(\mu=0) = 0$, because the event can only occur if $u$ is exactly orthogonal to the rank-1 eigenspace associated with the smallest eigenvalue. Still working on statements 2 and 3.... | |
| Mar 9, 2016 at 15:17 | comment | added | Suvrit | I removed the previous comments because they don't help. Perhaps this proves helpful, though am not sure about it. Considering $D+uu^T$ may still be interesting, in particular, because we know that the eigenvalues of this perturbed matrix are the roots of $f(\lambda) := 1+\sum_i \frac{u_i^2}{d_i-\lambda}$. Alternatively, since the eigenvalues of the perturbed matrix are of the form $d_i+\nu_i$ for $0\le \nu_i \le 1$, we can consider solving for $\nu$. | |
| Mar 8, 2016 at 16:15 | history | edited | Richard Zhang | CC BY-SA 3.0 |
Shortened the title
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| Mar 8, 2016 at 15:58 | comment | added | Richard Zhang | Assuming that $\lambda_n \equiv \lambda_\min$, of course, $\lambda_\min(QQ^T)=0$ for all $m<n$, so that bound simply gives the trivial result that $\mu\ge0$. In the case of $m=n$, we have $QQ^T=UU^T=I_n$, but that yields the other trivial result that $\mu \le 1$. (I had defined $Q$ to be the first $m$ columns of $U$. This is just a more rigorous way of saying "$m$ random orthonormal columns", using language taken from Jiang's paper) | |
| Mar 8, 2016 at 13:55 | history | edited | Richard Zhang | CC BY-SA 3.0 |
mistake in description of numerical experiment
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| Mar 8, 2016 at 6:24 | history | edited | Richard Zhang | CC BY-SA 3.0 |
Added numerical example
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| Mar 8, 2016 at 6:12 | history | asked | Richard Zhang | CC BY-SA 3.0 |