bio  website  math.ucla.edu/~tao 

location  Los Angeles  
age  39  
visits  member for  5 years, 8 months 
seen  1 hour ago  
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Professor of Mathematics at UCLA
7h

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Expected determinant of a random NxN matrix.
Update: the central limit theorem for the logdeterminant was worked out carefully by Nguyen and Vu, arxiv.org/abs/1112.0752 
1d

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Can phase significantly concentrate a function's spectrum?
The calculations here look similar to those used to disprove the (related) HardyLittlewood majorant conjecture, which was first done by Bachelis: ams.org/mathscinetgetitem?mr=320636 
Jun 29 
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Probability that random nonnegative integer matrix is singular
Corollary 1.2 of BourgainVuWood 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 
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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 
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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 
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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 
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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 
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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 HardyLittlewood 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 
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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 Chowlatype 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 