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bio website math.ucla.edu/~tao
location Los Angeles
age 39
visits member for 5 years, 8 months
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Professor of Mathematics at UCLA

7h
comment Expected determinant of a random NxN matrix.
Update: the central limit theorem for the log-determinant was worked out carefully by Nguyen and Vu, arxiv.org/abs/1112.0752
1d
comment Can phase significantly concentrate a function's spectrum?
The calculations here look similar to those used to disprove the (related) Hardy-Littlewood majorant conjecture, which was first done by Bachelis: ams.org/mathscinet-getitem?mr=320636
Jun
29
comment Probability that random nonnegative integer matrix is singular
Corollary 1.2 of Bourgain-Vu-Wood arxiv.org/pdf/0905.0461.pdf gives an upper bound of $(1 / \sqrt{k+1}+o(1))^n$ in the regime where $k$ is fixed and $n$ goes to infinity. This can be compared with the trivial lower bound of $(1/(k+1))^n$, coming from the event that the first two rows (say) agree.
Jun
28
awarded  Nice Answer
Jun
28
comment A curious determinantal inequality
I found this argument after playing around with the k=1 and k=2 cases for a while. Roughly speaking, $\lambda_k I + D$ represents the "largest" or "worst" that $A+B$ can be if one only constrains the top $k$ eigenvalues of $A+B$, which is what one is doing when trying to prove majorisation. (When $k=1$, $D$ is not present, and when $k=2$, $D$ is a rank one operator; after seeing these two cases I was able to extrapolate to the general case.)
Jun
28
answered A curious determinantal inequality
Jun
27
comment A curious determinantal inequality
Apply Fischer's inequality to $\begin{pmatrix} A^{1/2} & B^{1/2} \\ -B^{1/2} & A^{1/2} \end{pmatrix} \begin{pmatrix} A + B & 0 \\ 0 & A+B \end{pmatrix} \begin{pmatrix} A^{1/2} & -B^{1/2} \\ B^{1/2} & A^{1/2} \end{pmatrix}$. By Schur complement, the first and last matrix have determinant $\det( A + A^{1/2} B^{1/2} A^{-1/2} B^{1/2} )$, giving $\det( A^{1/2} (A+B) A^{1/2} + B^{1/2} (A+B) B^{1/2} ) \geq \det(A+B) \det( A + A^{1/2} B^{1/2} A^{-1/2} B^{1/2} )$ (I had some typos in the previous inequality as I had changed notation in my computations by squaring $A,B$).
Jun
27
comment A curious determinantal inequality
You are right, of course; I had mistakenly identified a vector space with its dual when thinking about the problem, which translated into the sign error here when converted back into matrices. I can establish the weaker inequality $\det(A^{1/2} (A^2+B^2) A^{1/2} + B^{1/2} (A^2+B^2) B^{1/2}) \geq \det(A^2+B^2) \det(A^2 + A B A^{-1} B)$ with this approach, but it does not appear strong enough to recover the full inequality.
Jun
27
revised A curious determinantal inequality
added 183 characters in body
Jun
27
awarded  Nice Answer
Jun
27
revised A curious determinantal inequality
added 113 characters in body
Jun
27
answered A curious determinantal inequality
Jun
25
comment Number of $\mathbb F_p$ points constant mod $p$?
Reminds me a bit of Higman's PORC conjecture, groupprops.subwiki.org/wiki/Higman's_PORC_conjecture, though this conjecture is now believed to be false in general.
Jun
19
comment A weakening of the Littlewood conjecture
yes; it also keeps the $p_j$ of magnitude comparable to $q_j$, allowing the pigeonholing to work.
Jun
18
answered A weakening of the Littlewood conjecture
Jun
17
comment Do relaxed Liouville functions violate Chowla's conjecture?
As far as I know the Chowla conjecture may well be true only assuming multiplicativity at small primes, and this could well be a viable route towards proving this conjecture (Kaisa, Maks and I have some work in progress that advances a little bit in this direction). Note that a lot of recent progress on the related Sarnak conjecture only uses small prime multiplicativity. This is in contrast with the superficially similar Hardy-Littlewood prime tuples conjecture, which is easy to disrupt by modifying the primes on a set of positive relative density, and so appears to be strictly harder.
Jun
15
comment Do relaxed Liouville functions violate Chowla's conjecture?
I think the current version of your notion of "relaxed Liouville function" may be too strong. If one has $f(nm) = \lambda(n) f(m)$ for most $n,m$, then one also has $f(nm) = \lambda(m) f(n)$ for most $n,m$, and so $f(n)/\lambda(n) = f(m)/\lambda(m)$ for most $n,m$, and so $f(n)$ is a constant multiple of $\lambda(n)$ for most $n$, and then the Chowla-type conjecture for $f$ will follow from that of $\lambda$ (with an error proportional to $\varepsilon$).
Jun
15
answered Do relaxed Liouville functions violate Chowla's conjecture?
Jun
11
awarded  Notable Question
Jun
8
awarded  Enlightened