Timeline for Bounding the second derivative of the log-determinant
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Mar 26, 2013 at 19:15 | vote | accept | Paul Christiano | ||
| Mar 26, 2013 at 1:49 | comment | added | Suvrit | Nice question! +1 | |
| Mar 26, 2013 at 0:43 | answer | added | Suvrit | timeline score: 5 | |
| Mar 26, 2013 at 0:38 | comment | added | duetosymmetry | Does it help at all that—from the cyclicity of trace and the fact that A is positive definite, the LHS can be written as $\Tr[(A^{-1/2}BA^{-1/2})^2]$? (where $A^{-1/2}$ is the unique positive definite matrix square root of $A^{-1}$) Clearly $M\equiv A^{-1/2}BA^{-1/2}$ is symmetric. Then the LHS is $Tr(M^2)$ which is just the square of the Frobenius norm of M. Then the question is: what are the singular values of $A^{-1/2}BA^{-1/2}$? | |
| Mar 25, 2013 at 20:56 | comment | added | Peter Michor | $A$ is positive definite (since invertible). Viewing $A$ as an inner product on $\mathbb R^n$, your formula describes the natural induced inner product on the space of symmetric bilinear forms. This is the background for my description of the natural invariance group above. | |
| Mar 25, 2013 at 20:28 | history | edited | Paul Christiano | CC BY-SA 3.0 |
added 89 characters in body; added 29 characters in body
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| Mar 25, 2013 at 18:57 | comment | added | Paul Christiano | Sorry for the ambiguity, A is positive semidefinite (and symmetric). The large entry of B is not necessarily on the diagonal. Also, note that the claim is easy by rote computation if B has only one large entry. This is the observation that inspired me to try to prove this bound (that, and hope). Peter: I don't see how to use the large entry of B after making such a transformation. Does this translate into some nice property of $n^T B n$ or $g^T B g$? | |
| Mar 25, 2013 at 18:51 | history | edited | Paul Christiano | CC BY-SA 3.0 |
added 13 characters in body
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| Mar 25, 2013 at 9:23 | comment | added | Denis Serre | About the entry of $B$ larger than $1$, is it diagonal ? | |
| Mar 25, 2013 at 8:32 | comment | added | Peter Michor | A is positive, thus symmetric. The problem is invariant under the $GL(n)$-action $X\mapsto g^T.X g$ for $g\in GL(n)$. Any $g$ is of the form $g=n.a.k$ where $n$ is lower triangular with 1's on the diagonal, $a$ is diagonal with positive entries, and $k$ is orthogonal (Gram-Schmidt or Iwasawa). Choose $a=I$ so to not mess up too much your assumptions. There is an $n$ such that $n^T.A.n$ is diagonal with positive entries. Compute this entries. Check $n^t.B.n$. I hope this helps. | |
| Mar 25, 2013 at 7:43 | comment | added | Pietro Majer | A is not assumed to be symmetric, is it? | |
| Mar 25, 2013 at 6:41 | comment | added | Robert Israel | By "positive" you mean "positive definite"? | |
| Mar 25, 2013 at 5:31 | history | asked | Paul Christiano | CC BY-SA 3.0 |