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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?

Thanks in advance!

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    $\begingroup$ Observation: $A$ is symmetric. Although I don't see how that can be used yet $\endgroup$
    – fwd
    Aug 25, 2021 at 20:56
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    $\begingroup$ Also, why do you have zero entries in your examples? $\endgroup$
    – fwd
    Aug 25, 2021 at 20:58
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    $\begingroup$ 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. $\endgroup$
    – Jacob A
    Aug 25, 2021 at 21:07
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    $\begingroup$ 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). $\endgroup$
    – Jacob A
    Aug 25, 2021 at 21:09
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    $\begingroup$ 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 $b$, $b^T A b = \frac{1}{2} \int_{-1}^1 (b_0 + b_1 x + \cdots + b_{n-1} x^{n-1})^2\,dx$.) $\endgroup$
    – Daniel Schepler
    Aug 25, 2021 at 22:27

1 Answer 1

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  • 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 $\det(A)>0$.

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

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