Does Anyone Know Anything about the Determinant and/or Inverse of this Matrix?

Question

The matrix I am inquiring about here is the $n \times n$ matrix where the entry $A_{ij}$ is $\frac{1}{(i+j-1)^2}$. The $2 \times 2$ matrix looks like $$ \begin{pmatrix} 1 & 1/4 \\ 1/4 & 1/9 \end{pmatrix}. $$ The $3 \times 3$ case looks like $$ \begin{pmatrix} 1 & 1/4 & 1/9 \\ 1/4 & 1/9 & 1/16 \\ 1/9 & 1/16 & 1/25 \end{pmatrix}. $$ The $n \times n$ matrix looks like $$ \begin{pmatrix} 1 & 1/4 & \cdots & 1/(n^2) \\ 1/4 & 1/9 & \cdots & 1/{(n+1)^2} \\ \vdots & \vdots & \cdots & \vdots \\ 1/(n^2) & 1/{(n+1)^2} & \cdots & 1/{(2n-1)^2} \end{pmatrix}. $$ If anyone has any information or knows of any papers talking about this matrix please let me know. Thanks!

Answer 1

Not an answer, but an amusing observation: The determinant of the matrix usually (but not always) has denominator a perfect square (empirically), while the numerator always seems to have a huge prime factor.

EDIT Actually, it seems that the expectation of $\log(P(n))/\log(n),$ where $P(n)$ is the largest prime factor of $n$ approaches the Golomb-Dickman constant, which is about 0.62. In view of this, the largest prime factor of the numerator is pretty much par for the course, or at least, not obviously NOT par for the course.

n=1: det = 1/1

n=2: det=7/(12^2)

n=3 det = 647/(2160^2)

n=4 det = (19 * 571)/(672000^2)

n=5 det = (179 * 179357)/(7*4233600000^2)

n=6 det = (97 * 157 * 384191938531)/(186313420339200000^2)

n=7 det = (23 * 1280587616051046200369)/(2067909047925770649600000^2)

n=8 det = (317 * 6337 * 25997 * 87403 * 511645991608091)/(365356847125734485878112256000000^2)

(and some more. Note that mysterious 7 which keeps popping up in the denominator, for n=5,9,11 )

n=9 det = (55091 * 7731550926975871647518813143593349)/(7 * 146968826339795671126721851844198400000000^2)

n=10 det = (257 * 47360083 * 530916328215423816923887043836865928533)/(15402297982638230438765209613012092908994560000000000^2)

n=11 det = (31 * 1193 * 2647 * 538580971 * 957346850101 * 71222443011485886519799225151)/(7 * 175251348661711183890804992735665222783492739007774720000000000^2))

Answer 2

The determinant formula for $\det (1/(x_i+y_j)^2)_{i,j}$ is due to Borchardt, see Krattenthaler's article given in Steve's answer, which contains a $(q)$-deformation of it as well. I want to mention, that Emmanuel Preissmann conjectured also a formula for the generalized Cauchy determinant identity $$\det \left(\frac{1}{(x_i+y_j)^k}\right)_{i,j}$$ for every $k\ge 1$, see http://ipg.epfl.ch/~leveque/Conjectures/cauchy.pdf.
Further generalizations of Cauchy's determinant identity are given in Okada's paper http://www.math.kobe-u.ac.jp/publications/rlm18/9.pdf, where $(1.6)$ is again Borchardt's identity.

Answer 3

See (3.9) and (3.10)/Theorem 25 in Krattenthaler's "Advanced determinant calculus", http://arxiv.org/abs/math/9902004

Answer 4

Just a couple observations, building upon and explaining why Igor Rivin found that the denominators are often perfect squares or close to being such.

Let $A$ be the matrix with components $A_{ij} = 1/(x_i + y_j)^2$. Expressing its determinant as a sum over permutations and putting everything on the same denominator gives $$ \det A = \frac{P(\{x_i\}, \{y_i\})}{\prod_{i,j} (x_i + y_j)^2} \,, $$ for some polynomial $P$ with integer coefficients. Also, if two $x_i$ or two $y_i$ are equal, then the corresponding rows/columns are equal, and the determinant vanishes. Hence $$ \det A = \frac{Q(\{x_i\}, \{y_i\}) \prod_{i\neq j} (x_i - x_j) (y_i - y_j)}{\prod_{i,j} (x_i + y_j)^2} \,. $$ Actually, as pointed out in other answers, the Cauchy formula provides $Q$ as the permanent of some matrix, but all I need here is to know $Q$ has integer coefficients.

In your particular case, the products give $$ \frac{\prod_{i\neq j} (i - j) ((i - 1) - (j - 1))}{\prod_{i,j} (i + j - 1)^2} = \frac{\prod_{i < j} (j - i)^4}{\prod_{i,j} (i + j - 1)^2} = \prod_{j=1}^N \left[(j - 1)!^4 \frac{(j - 1)!^2}{(j + N - 1)!^2} \right] = \frac{\prod_{j=0}^{N-1} j!^8}{\prod_{j=0}^{2N-1} j!^2} $$ Note that $j!^8$ divides $(2j)!(2j+1)!$, so the numerator divides the denominator. All in all, $$ \det A = \frac{Q}{\prod_{j=0}^{2N-1} j!^2 \bigg/ \prod_{j=0}^{N-1} j!^8} $$ for some integer $Q$. The denominator appearing in this formula is typically rather close to the ones given in Igor Rivin's answer, and are trivially squares: $1$, $12^2$, $2160^2$, $6048000^2 = 672000^2 \cdot 9^2$, etc. The discrepancy is of course accounted for by cancellations between $Q$ and the product of factorials. Again using Igor Rivin's data, I find (indices denote the size of the matrix) $$ \begin{aligned} Q_1 &= 1 \\ Q_2 &= 7 \\ Q_3 &= 647 = 2^3 \cdot 3^4 - 1 = 8 \cdot (16 \cdot (6 - 1) + 1) - 1 \\ Q_4 &= 878769 = 16 \cdot (12 \cdot (32 \cdot (144 - 1) + 1) - 1) + 1 \\ Q_5 &= 18203480001 = 40000 \cdot ( 48 \cdot ( 120 \cdot (80 - 1) + 1) - 1) + 1 \\ Q_6 &= 5850859031888599 \,. \end{aligned} $$ Here I've included some expressions which may or may not generalize. In any case, it is certain that $Q$ will not be expressed as a (simple) product, as it involves some very large prime factors.