# Prove positive definiteness

I want to prove that the matrix $$\begin{pmatrix} 1 &\cfrac{1}{2} &\cfrac{1}{3} &\cdots &\cfrac{1}{n} \\ \cfrac{1}{2} &\cfrac{1}{3} &\cfrac{1}{4} &\cdots &\cfrac{1}{n+1} \\ &\vdots &&\ddots &\vdots \\ \cfrac{1}{n} &\cfrac{1}{n+1} &\cfrac{1}{n+2} &\cdots &\cfrac{1}{2n-1} \end{pmatrix}$$ is positive definite. Using mathematical induction, I only need to show that its determinant is positive. But I can't find the way out.

• This is a Hilbert Matrix, and in the link the determinant is computed in the closed form; but I think this is not needed in its fullest strength, in the mere order to show that the determinant is positive. – awllower May 6 '14 at 6:35
• oeis.org/A005249 tabulates (the inverse of) the determinant of this matrix and gives a formula, which maybe you can prove by induction. It also gives some references that might be helpful. – Gerry Myerson May 6 '14 at 10:49
• – Martin Sleziak Sep 11 '18 at 17:40

Hint. Call your matrix $H$. Then $H=\left(\int_0^1 x^{i+j}dx\right)_{i,j=1,2,\ldots,n}=\int_0^1 v(x)v(x)^Td x$, where $v(x)=(1,x,x^2,\ldots,x^{n-1})^T$. Can you see that $H$ is positive semi-definite? Now, if $u^THu=0$ for some vector $u$, then $\int_0^1 u^Tv(x)v(x)^Tu dx=0$. Can you infer that $u=0$ (think about Vandermonde matrix)?
Remark. From a wider perspective, $H$ is an instance of matrices of the form $A=\left(\frac1{1+x_i+x_j}\right)_{i,j=1,2,\ldots,n}$ where $x_i\ge0$ for all $i$ (in your case, $x_i=i-1$), which is known (q313249) to be positive semidefinite. The same trick in the above hint can be used to show that $A$ is positive definite iff all $x_i$s are distinct.
Edit. Alternatively, $H$ is a Hilbert matrix. Every leading principal submatrix of $H$ is again a Hilbert matrix. So, by Sylvester's criterion, it suffices to prove that every Hilbert matrix has a positive determinant. As Gerry Myerson has pointed in his comment, mathematical induction may be useful in this case.