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how to prove A is invertible or $\ detA\neq 0$ $$A=\begin{pmatrix} \frac11 & \frac12 & \frac13 & \cdots & \frac1n \\ \frac12 & \frac13 & \frac14 & \cdots & \frac{1}{n+1} \\ \vdots & \vdots& \vdots & \ddots & \vdots \\ \frac1n & \frac{1}{n+1} & \frac{1}{n+2} & \cdots & \frac{1}{2n-1} \end{pmatrix}$$

Thanks in advance

M.H
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2 Answers2

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It is a positive definite symmetric matrix.

As such it can be diagonalized, and its eigenvalues are all real and positive. So, its diagonalized form is nonsingular, and since it is similar to a nonsingular matrix... it is nonsingular!

rschwieb
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  • Does this (simply) follow from the calculation of $x^TAx$ or something else? – Lord Soth Apr 18 '13 at 20:54
  • @LordSoth Yeah: $xAx^T>0$ for a nonzero eigenvector $x$ implies $x\cdot x\lambda=0$, whence $\lambda|x|^2>0$ implies $\lambda>0$. – rschwieb Apr 18 '13 at 20:57
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This is the Hilbert matrix, which is a special case of Cauchy matrix. The determinant is given by $$\dfrac{\left(\displaystyle\prod_{k=1}^{n-1} k!\right)^4}{\displaystyle\prod_{k=1}^{2n-1} k!}$$