# Prove a matrix has non-zero determinant

While working on a problem, I managed to reduce it to showing that the matrix $$A = [a_{ij}]$$, where $$2a_{ij} = \frac{1}{i+j-1} - \frac{(-1)^{i+j-1}}{i+j-1}$$, has non-zero determinant, for whatever size you choose. Here are a few cases: $$\det\begin{bmatrix} 1 & 0 & \frac{1}{3} \\ 0 & \frac{1}{3} & 0 \\ \frac{1}{3} & 0 & \frac{1}{5} \end{bmatrix} = \frac{4}{135};$$

$$\det\begin{bmatrix} 1 & 0 & \frac{1}{3} & 0 \\ 0 & \frac{1}{3} & 0 & \frac{1}{5}\\ \frac{1}{3} & 0 & \frac{1}{5} & 0 \\ 0 & \frac{1}{5} & 0 & \frac{1}{7} \end{bmatrix} = \frac{16}{23625}$$

I worked on a few cases using online calculators, but explicit determinant calculation in arbitrary sizes is very cumbersome. It seems to tend to $$0$$, but to always be positive. In fact, I suspect this is a positive-definite matrix, but, again, couldn't quite prove it - and I recall having seen it in a computational setting before, maybe numerical integration, though I'm not sure...

Anyhow, is there any technique I can use to show this has non-zero determinant for any size I pick?

• Observation: $A$ is symmetric. Although I don't see how that can be used yet
– fwd
Aug 25, 2021 at 20:56
• Also, why do you have zero entries in your examples?
– fwd
Aug 25, 2021 at 20:58
• It might be a bit of extra algebra, but for the first few cases, use row reduction to diagonalize. You may find a common pattern due to the common structure. This way, you could show all diagonal elements are nonzero and thus $\det(A)=\prod_k \lambda_k$ is nonzero. Aug 25, 2021 at 21:07
• Extra hint/observation: your second example contains a principal submatrix equal to your first example. So I believe you will have a straightforward proof by induction (using also my prior comment). Aug 25, 2021 at 21:09
• Outline of another solution: $A$ is half the Grammian matrix of the linearly independent vectors $1, x, x^2, \ldots, x^{n-1}$ in $L^2[-1, 1]$ and is therefore positive definite. (So, to unfold the standard proof of Grammian matrices being positive definite: for a column vector $b$, $b^T A b = \frac{1}{2} \int_{-1}^1 (b_0 + b_1 x + \cdots + b_{n-1} x^{n-1})^2\,dx$.) Aug 25, 2021 at 22:27

• Let $$C:=(\frac1{i+j-1})_{1\le i,j\le n}$$. This is a symmetric matrix known as a Cauchy matrix. By Sylvester's criterion, it is positive-definite.

• Let $$D$$ be the $$n\times n$$ diagonal matrix whose $$i$$-th diagonal element is $$(-1)^{i-1}$$. We can easily check by matrix multiplication that $$A=(DCD+C)/2$$.

• We observe that $$DCD=(D\sqrt C)(D\sqrt C)^T$$ where $$\det(D\sqrt C)\neq0$$, so $$DCD$$ is also a positive-definite matrix. Hence $$A$$ is the sum of two positive-definite matrices, and therefore also positive-definite. In particular $$\det(A)>0$$.

• [+1] for the astute method of solution, but I disagree with the term "Cauchy matrix" for $C$ : it is a Hilbert matrix Aug 26, 2021 at 7:59
• @Jean Marie: you're right, I kind of mixed up the two. A Hilbert matrix is a special case of a Cauchy matrix. Aug 26, 2021 at 8:01
• I agree with the qualification "special case" but it is not worth the value to refer to a Cauchy matrix. Aug 26, 2021 at 9:54