# Showing that $(A_{ij})=\left(\frac1{1-x_ix_j}\right)$ is positive semidefinite

Consider the matrix $A$ whose elements are $A_{ij} = \frac{1}{1-x_i x_j}$ where we have $-1 < x_i < 1$ and $-1 < x_j < 1$ for $i,j=1,2,...n$. For example, when $n=3$ the matrix becomes

\begin{pmatrix} \frac{1}{1-x_1^2} & \frac{1}{1-x_1x_2} & \frac{1}{1-x_1x_3} \\ \frac{1}{1-x_1x_2} & \frac{1}{1-x_2^2} & \frac{1}{1-x_2x_3} \\ \frac{1}{1-x_1x_3} & \frac{1}{1-x_2x_3} & \frac{1}{1-x_3^2} \end{pmatrix}

For $n=2,3$ I considered the principal minors and concluded that this matrix is positive semidefinite. However, I could not generalize it for $n \geq 3$. How can I prove that this matrix is positive semidefinite for arbitrary $n$ ?

• What is the source of the problem? – Jonas Meyer Jan 24 '13 at 2:27
• If you multiply the $i$-th row by $x_i$ for each $i$, what you have is a Cauchy matrix and you can write down the determinant of any principal sub matrix. (The relevant information is on the wikipedia page for Cauchy matrices.) – Chris Godsil Jan 24 '13 at 2:39
• @JonasMeyer : In my harmonic analysis notes, it was written that this matrix is obviously positive semidefinite. That's why I asked – neticin Jan 24 '13 at 9:52

In the real inner product space $\ell^2(\mathbb Z_{\geq 0},\mathbb R)=\{(a_0,a_1,a_2,\ldots):\sum_n a_n^2<\infty\}$ with inner product $\langle(a_n)_n,(b_n)_n\rangle=\sum_n a_nb_n$, note that $\dfrac{1}{1-x_ix_j}=\langle(1,x_i,x_i^2,\ldots),(1,x_j,x_j^2,\ldots)\rangle$.