Determinant Inequalities for Positive Definite Matrices Via Diananda’s Result for Arithmetic and Geometric Weighted Means

In this paper we prove among others that, if (A_j)_j=1,...,m are positive definite matrices of order n ≥ 2 and q_j ≥ 0, j = 1, ..., m with $\sum_{j = 1}^{m} q_{j} = 1$ $$\sum\nolimits_{j = 1}^m {{q_j} = 1} $$, then $\begin{array}{l} 0 & \leq \frac{1}{1 - \min_{i \in {1, \dots, m}} {q_{i}}} \\ \times & [{\sum^{}}_{i = 1}^{m} q_{i} (1 - q_{i}) {[\det (A_{i})]}^{- 1} - 2^{n + 1} \underset{1 \leq i < j \leq m}{\sum^{}} q_{i} q_{j} {[\det (A_{i} + A_{j})]}^{- 1}] \\ \leq & {\sum^{}}_{i = 1}^{m} q_{i} {[\det (A_{i})]}^{- 1} - {[\det ({\sum^{}}_{i = 1}^{m} q_{i} A_{i})]}^{- 1} \\ \leq & \frac{1}{\min_{i \in {1, \dots, m}} {q_{i}}} \\ \times & [{\sum^{}}_{i = 1}^{m} q_{i} (1 - q_{i}) {[\det (A_{i})]}^{- 1} - 2^{n + 1} \underset{1 \leq i < j \leq m}{\sum^{}} q_{i} q_{j} {[\det (A_{i} + A_{j})]}^{- 1}] . \end{array}$

Language:: English

Publication timeframe:: Volume Open
Journal Subjects:: Mathematics, General Mathematics

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Determinant Inequalities for Positive Definite Matrices Via Diananda’s Result for Arithmetic and Geometric Weighted Means

Silvestru Sever Dragomir

Published Online: May 03, 2023

Page range: 21 - 34

DOI: https://doi.org/10.2478/awutm-2023-0003

KeywordsPositive definite matrices, Determinants, Inequalities

© 2023 Silvestru Sever Dragomir, published by Sciendo

This work is licensed under the Creative Commons Attribution-NonCommercial-NoDerivatives 3.0 License.

Keywords
Positive definite matrices, Determinants, Inequalities