Positivity of GCD tensors and their determinants
Source
Linear Algebra and Its Applications
ISSN
00243795
Date Issued
2026-02-15
Author(s)
Panigrahy, Krushnachandra
Abstract
Let S={s<inf>1</inf>,s<inf>2</inf>,…,s<inf>n</inf>} be an ordered set of n distinct positive integers. The mth-order n-dimensional tensor T<inf>[S]</inf>=(t<inf>i<inf>1</inf>i<inf>2</inf>…i<inf>m</inf></inf>), where t<inf>i<inf>1</inf>i<inf>2</inf>…i<inf>m</inf></inf>=GCD(s<inf>i<inf>1</inf></inf>,s<inf>i<inf>2</inf></inf>,…,s<inf>i<inf>m</inf></inf>), the greatest common divisor (GCD) of s<inf>i<inf>1</inf></inf>,s<inf>i<inf>2</inf></inf>,…, and s<inf>i<inf>m</inf></inf> is called the GCD tensor on S. The earliest result on GCD tensors goes back to Smith [Proc. Lond. Math. Soc., 1976], who computed the determinant of GCD matrix on S={1,2,…,n} using the Euler's totient function, followed by Beslin–Ligh [Linear Algebra Appl., 1989] who showed all GCD matrices are positive definite. In this note, we study the positivity of higher-order tensors in the k-mode product. We show that all GCD tensors are strongly completely positive (CP). We then show that GCD tensors are infinite divisible. In fact, we prove that for every positive real number r, the tensor T<inf>[S]</inf><sup>∘r</sup>=(t<inf>i<inf>1</inf>i<inf>2</inf>…i<inf>m</inf></inf><sup>r</sup>) is strongly CP. Finally, we obtain an interesting decomposition of GCD tensors using Euler's totient function Φ. Using this decomposition, we show that the determinant (also called hyperdeterminant) of the mth-order GCD tensor T<inf>[S]</inf> on a factor-closed set S={s<inf>1</inf>,…,s<inf>n</inf>} is ∏i=1nΦ(s<inf>i</inf>)<sup>(m−1)<sup>(n−1)</sup></sup>.
Keywords
Completely positive tensor | Determinant of tensor | GCD tensor | K-mode product | Tensor decomposition
