The result of Example 16 suggests that perhaps the matrix
\[ A = \left[ \begin{array}{c c c c} 1 & \frac {1}{2} & \dots & \frac {1}{n} \\ \frac {1}{2} & \frac {1}{3} & \dots & \frac {1}{n + 1} \\ \vdots & \vdots & & \vdots \\ \frac {1}{n} & \frac {1}{n + 1} & \dots & \frac {1}{2 n - 1} \end{array} \right] \]
is invertible and \(A^{-1}\) has integer entries. Can you prove that?
2. Vector Spaces
Solution
Write \(H_n=(h_{ij})_{1\le i,j\le n}\) with \[ h_{ij}=\frac1{i+j-1}. \] We first prove the inverse formula for a general Cauchy matrix and then specialize it.
1. Let \(x_1,\ldots,x_n\) and \(y_1,\ldots,y_n\) be scalars such that the \(x_i\) are distinct, the \(y_i\) are distinct, and \(x_i+y_j\ne0\) for all \(i,j\). For\[ C_{ij}=\frac1{x_i+y_j}, \]one has\[ (C^{-1})_{ij} = \frac{ \displaystyle\prod_{k=1}^n(x_j+y_k) \displaystyle\prod_{k=1}^n(x_k+y_i) }{ (x_j+y_i) \displaystyle\prod_{k\ne j}(x_j-x_k) \displaystyle\prod_{k\ne i}(y_i-y_k) }. \]
Fix \(j\). Define \[ R_j(z)= \frac{\displaystyle\prod_{\ell\ne j}(z-x_\ell)} {\displaystyle\prod_{\ell=1}^n(z+y_\ell)} \cdot \frac{\displaystyle\prod_{\ell=1}^n(x_j+y_\ell)} {\displaystyle\prod_{\ell\ne j}(x_j-x_\ell)}. \] Then \(R_j(x_m)=\delta_{mj}\). Since \(R_j\) is a proper rational function with simple poles at \(-y_i\), it has a partial-fraction expansion \[ R_j(z)=\sum_{i=1}^n\frac{b_{ij}}{z+y_i}. \] The coefficient at the pole \(-y_i\) is \[ b_{ij} = \lim_{z\to-y_i}(z+y_i)R_j(z). \] Substituting the definition of \(R_j\) gives \[ b_{ij} = \frac{ \displaystyle\prod_{k=1}^n(x_j+y_k) \displaystyle\prod_{k=1}^n(x_k+y_i) }{ (x_j+y_i) \displaystyle\prod_{k\ne j}(x_j-x_k) \displaystyle\prod_{k\ne i}(y_i-y_k) }. \] Evaluating the partial-fraction expansion at \(z=x_m\) yields \[ \sum_{i=1}^n C_{mi}b_{ij}=R_j(x_m)=\delta_{mj}. \] Thus the matrix \(B=(b_{ij})\) satisfies \(CB=I\), hence \(B=C^{-1}\).
2. For the Hilbert matrix take\[ x_j=j-1,\qquad y_i=i. \]Then\[ (H_n^{-1})_{ij} =(-1)^{i+j} \frac{(n+i-1)!(n+j-1)!} {(i+j-1)(i-1)!^2(j-1)!^2(n-i)!(n-j)!}. \]
In the formula of 1, \[ \prod_{k=1}^n(x_j+y_k)=\frac{(n+j-1)!}{(j-1)!}, \qquad \prod_{k=1}^n(x_k+y_i)=\frac{(n+i-1)!}{(i-1)!}. \] Also \[ \prod_{k\ne j}(x_j-x_k) =(-1)^{n-j}(j-1)!(n-j)!, \] and \[ \prod_{k\ne i}(y_i-y_k) =(-1)^{n-i}(i-1)!(n-i)!. \] Since \((-1)^{2n-i-j}=(-1)^{i+j}\), substitution gives the displayed expression.
3. Equivalently,\[ (H_n^{-1})_{ij}=(-1)^{i+j}(i+j-1) \binom{n+i-1}{n-j} \binom{n+j-1}{n-i} \binom{i+j-2}{i-1}^{\!2}. \]
Expanding the three binomial coefficients into factorials and cancelling gives exactly the expression in 2.
4. Hence \(H_n\) is invertible and every entry of \(H_n^{-1}\) is an integer.
The matrix in 3 is an inverse by <1>1–<1>3. Every factor in the formula of 3 is an integer, so every inverse entry is an integer.