Prove an $n\times n$ matrix is negative definite

Question

I wonder is there any way to prove the $n\times n$ matrix with elements below is negative definite:

$$ \sigma_{ij} = \frac{a_ia_j}{\sum_k s_ka_k} \space; i \neq j \text{ (off diagonal terms)}$$ $$\sigma_{ii} = \frac{a_ia_i}{\sum_k s_ka_k} - \frac{a_i}{s_i} \text{ (diagonal terms)}$$

Here $1\leq i,j,k\leq n$, $0 < a_i < 1$ and $s_k$ is the weight of $a_k$ such that: $0 < s_k < 1$ and $\sum_k s_k = 1$.

Any suggestions or references are highly appreciated.

Why the separate definitions for $i\neq j$ and $i=j$? They seem the same... — Thomas Andrews, Sep 14 '14 at 02:44
i \ne j is for off-diagonal terms, and i = j is for diagonal terms. — user175861, Sep 14 '14 at 02:54

score 1 · Answer 1 · edited Jul 26 '15 at 20:22

1

The idea is to use the Cauchy-Schwarz inequality in the form $$\left(\sum_i a_i x_i\right)^2 \leq \left(\sum_i \frac{a_i}{s_i} x_i^2\right) \left(\sum_i a_i s_i\right). $$

edited Jul 26 '15 at 20:22

Daniel Fischer

206,697

answered Sep 14 '14 at 05:16

Yuval Filmus

57,157

Prove an $n\times n$ matrix is negative definite

1 Answers1