# 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? $$\begin{pmatrix} \frac{1}{\beta + 1} & \frac{1}{2} & \frac{1}{3} & \dots & \frac{1}{p+1}\\ \frac{1}{\beta + 2} & \frac{1}{3} & \frac{1}{4} & \dots & \frac{1}{p+2}\\ \frac{1}{\beta + 3} & \frac{1}{4} & \frac{1}{5} & \dots & \frac{1}{p+3}\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ \frac{1}{\beta + p + 1} & \frac{1}{p+2} & \frac{1}{p+3} & \dots & \frac{1}{2p+1} \end{pmatrix}.$$

• This may be a generalized Hilbert matrix. Maybe math.univ-lille1.fr/~otrt/otrt/Aleman.pdf will help. If not, anyway, I've given you a search term. Apr 22, 2020 at 1:08
• Maybe take a look at Christian Krattenthaler's "Advanced Determinant Calculus" (Séminaire Lotharingien Combin. 42 (1999), Article B42q, 67 pp. Apr 22, 2020 at 3:16
• I suspect you could also prove this using the fact that if the determinant is zero, then the left column must be a linear combination of the remaining columns. I don't have the time to math it out right now, though.
– JDL
Apr 23, 2020 at 7:52

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: $$\det ~\left(\frac{1}{x_i+y_j}\right)_{i,j} = \frac{\prod_{1\leq i Thus the determinant of $$\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.

• Alternative proof of the identity? Multiply both sides by $\prod_{1\leq i\neq j\leq n}(x_i+y_j)$, it's clear that the left-hand side is a polynomial in the ring $\mathbb{C}[x_i,y_i]$, which is a unique factorization domain. It clearly vanishes if any $x_i=x_j$, and ditto for $y_i=y_j$. so it is divisible by $\prod_{1\le i<j\le n}(x_i-x_j)(y_i-y_j)$. Both sides are homogeneous polynomials,and its easy to check that they have the same total degree, hence are constant multiples of one another. Now just look at one monomial on both sides to check that the multiple is $1$. Apr 22, 2020 at 17:50
• @Joe Silverman: Is it obvious how to check the coefficient of a monomial (which one?) in the determinant multiplied by $\prod(x_i+y_j)$?
– abx
Apr 24, 2020 at 16:22
• @abx Not obvious, no, but I haven't given it a lot of thought. Alternatively, if you know the determinant of the matrix $(1/(i+j))$, you can set $x_i=it$ and $y_j=jt$ and cancel the powers of $t$. I think that that matrix is fairly standard. Apr 24, 2020 at 19:23

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 $$p$$.

• +1. Nice proof!
– Hans
Apr 25, 2020 at 19:10
• @Hans: Thanks! I had just checked that a generalized Vandermonde with positive $x_i$ and real exponents is $\ne 0$, cannot be evaluated except for integer exponents, so that was the only way there. Apr 29, 2020 at 4:20