Let \(n\) be a positive integer and \(A\) the \(n \times n\) matrix \[ A = \begin{bmatrix} 1 & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n} \\ \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \cdots & \frac{1}{n+1} \\ \vdots & \vdots & \vdots & & \vdots \\ \frac{1}{n} & \frac{1}{n+1} & \frac{1}{n+2} & \cdots & \frac{1}{2n-1} \end{bmatrix}. \] Prove that \(A\) is positive.
Solution
1. \(A_{ij} = \frac{1}{i+j-1} = \int_0^1 t^{i+j-2}\, dt\).
\(\int_0^1 t^{i+j-2}\, dt = \frac{1}{i+j-1}\).
2. For any nonzero vector \(x = (x_1, \ldots, x_n)\), \(x^T A x = \int_0^1 \left( \sum_{i=1}^n x_i t^{i-1} \right)^2 dt\).
\(x^T A x = \sum_{i,j} x_i x_j A_{ij} = \sum_{i,j} x_i x_j \int_0^1 t^{i+j-2}\, dt = \int_0^1 \left(\sum_i x_i t^{i-1}\right)^2 dt\).
3. \(\int_0^1 \left( \sum_{i=1}^n x_i t^{i-1} \right)^2 dt > 0\).
the polynomial \(p(t) = \sum_{i=1}^n x_i t^{i-1}\) is not identically zero (since \(x \neq 0\)), so \(p(t)^2 \ge 0\) and is positive on a set of positive measure; hence its integral is strictly positive.
4. Hence \(x^T A x > 0\) for all \(x \neq 0\), so \(A\) is positive definite.
2 and 3.
5. Q.E.D.
4.