Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

While experimenting with positive-definite functions, I was led to the following:

Let $n$ be a positive integer, and let $x_1,\ldots,x_n$ be sampled from a zero-mean, unit variance gaussian. Consider the (positive-definite) matrix $$M_{ij}=\frac{1}{1+|x_i-x_j|}.$$ Now I wish to know:

How do I obtain an estimate for the smallest eigenvalue $\lambda_n$ of $M$?

Preliminary experiments (see plot; x-axis: $n$, y-axis: $\lambda_n$) suggest that $\lambda_n \approx 1/n^2$, but how do I prove that or perhaps a more accurate result?

you should see plot here

share|improve this question
    
Do you mean $M_{ij}=\frac{1}{1+|x_i-x_j|}$? Because $M$ as you've defined doesn't seem to be a matrix. –  Gilead Nov 13 '10 at 23:00
1  
Here's a link: mathoverflow.net/questions/24287/… –  Gilead Nov 13 '10 at 23:02
    
@Gilead: Thanks for catching the typo; also, i am not looking for a numerical method, but rather a "closed-form" bound or approximation. –  Suvrit Nov 14 '10 at 9:09
add comment

1 Answer

I think I can get an upper bound of $O(1/n^2)$ by exhibiting a vector $v$ of magnitude comparable to $1$ which gets mapped to a vector of magnitude $O(1/n^2)$. The basic idea is to exploit the birthday paradox to find (with high probability) two indices $i \neq i'$ such that $x_i-x_{i'} = O(1/n^2)$. It should also be possible to then find another additional index $i''$ such that $x_{i''} = x_i + O(1/n)$.

Now look at the $i^{th}$ and $(i')^{th}$ rows, which have components $1/(1+|x_i-x_j|)$ and $1/(1+|x_{i'}-x_j|)$. These rows differ by $O(n^{-2})$ in each coefficient. This already gives an upper bound of $O(n^{-3/2})$ for the smallest eigenvalue, but one can do better by using Taylor expansion to note that the difference between the two components is $(x_i-x_{i'}) \hbox{sgn}( x_i-x_j ) / (1 + |x_i-x_j|)^2 + O(n^{-4})$ except when $x_j$ is very close to $x_i$, at which point we only have the crude bound of $O(n^{-2})$. Similarly, the difference between the $i''$ and $i$ rows is something like $(x_i-x_{i''}) \hbox{sgn}( x_i-x_j ) / (1 + |x_i-x_j|)^2 + O(n^{-4})$ except when $x_j$ is too close to $x_i$. So we can use a multiple of the second difference to mostly cancel off the first difference, and end up with a linear combination of three rows in which most entries have size $O(n^{-4})$ and only about $O(1)$ entries have size $O(n^{-2})$. This seems to give an upper bound of $O(n^{-2})$ for the least eigenvalue (or least singular value), though I have not fully checked the details.

To get a matching lower bound is trickier. One may have to move to a Fourier representation of the matrix as this would more readily capture the positive definiteness of the matrix (as suggested by Bochner's theorem).

share|improve this answer
10  
It's even easier than that: once you've found a pair of indices giving distance order $1/n^2$, they define a $2 \times 2$ matrix with an eigenvalue of order $1/n^2$, and then by interlacing of eigenvalues for Hermitian (or real symmetric) matrices the entire matrix has an eigenvalue no larger than that. –  Tracy Hall Nov 15 '10 at 4:32
    
@Terry: thanks for the nice argument for an upper bound (and @Tracy, good point of invoking Cauchy's interlacing theorem to argue about the upper bound on the eigenvalues). As for the Fourier connection, indeed this matrix is generated by the posdef function $f(x)=1/(1+|x|)$. In fact, perhaps even more can be proved, because actually the said matrix is infinitely divisible. –  Suvrit Nov 16 '10 at 8:53
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.