How to prove the determinant of a Hilbert-like matrix with parameter is non-zero Consider some positive non-integer $\beta$ and a non-negative integer $p$. Does anyone have any idea how to show that the determinant of the following matrix is non-zero?
$$
\begin{pmatrix}
\frac{1}{\beta + 1} & \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{p+1}\\
\frac{1}{\beta + 2} & \frac{1}{3} & \frac{1}{4} & \dots & \frac{1}{p+2}\\
\frac{1}{\beta + 3} & \frac{1}{4} & \frac{1}{5} & \dots & \frac{1}{p+3}\\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{1}{\beta + p + 1} & \frac{1}{p+2} & \frac{1}{p+3} & \dots & \frac{1}{2p+1}
\end{pmatrix}.
$$
 A: I think the reference "Advanced Determinant Calculus" has a pointer to the answer. But I'll still elaborate for it is ingenious. 
Suppose $x_i$'s and $y_j$'s, $1\leq i,j \leq N$, are numbers such that $x_i+y_j\neq 0$ for any $i,j$ combination, then the following identity (called Cauchy Alternant Identity) holds good:
$$
\det ~\left(\frac{1}{x_i+y_j}\right)_{i,j} = \frac{\prod_{1\leq i<j\leq n}(x_i-x_j)(y_i-y_j)}{\prod_{1\leq i\neq j\leq n}(x_i+y_j)}.
$$
Thus the determinant of 
$$
\begin{pmatrix}
\frac{1}{\beta + 1} & \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{p+1}\\
\frac{1}{\beta + 2} & \frac{1}{3} & \frac{1}{4} & \dots & \frac{1}{p+2}\\
\frac{1}{\beta + 3} & \frac{1}{4} & \frac{1}{5} & \dots & \frac{1}{p+3}\\
\vdots & \vdots & \vdots & \dots & \vdots \\
\frac{1}{\beta + p + 1} & \frac{1}{p+2} & \frac{1}{p+3} & \dots & \frac{1}{2p+1}
\end{pmatrix}
$$
can be obtained by choosing $[x_1,\cdots, x_{p+1}] = [1, \cdots, (p+1)]$ and $[y_1,\cdots, y_{p+1}] = [\beta, 1, \cdots, p]$. This is certainly not zero as $\beta$ is not an integer. 
The proof of the identity is ingenious. Perform the basic column operation where, $C_j = C_j-C_n$, and remove common factors from the rows and columns. Then perform the row operations, $R_j = R_j-R_n$. This renders the matrix block diagonal of 2 blocks with size n-1 and 1. The first block is the the principal submatrix of the orignal matrix, and the second block is the element 1. This then induces a recursion for the determinant, which yields the desired result.  
Thanks for the good question and the reference.
A: Rows linearly dependent means for some $c_1$, $\ldots$, $c_{p+1}$ the non-zero rational function 
$\sum_{k=1}^{p+1}  \frac{c_k}{x+k}$ has $p+1$ roots $\beta$, $1$, $2$, $\ldots$, $p$, not possible, since its numerator has degree at most $p$.
