A note on the rank inequality for diagonally magic matrices
- Duanmei Zhou^{1}Email author,
- Qingyou Cai^{1} and
- Xiaoyan Chen^{2}
https://doi.org/10.1186/s13660-016-1013-4
© Zhou et al. 2016
Received: 18 December 2015
Accepted: 5 February 2016
Published: 17 February 2016
Abstract
We prove that a positive matrix with all permutation products equal is diagonally equivalent to J, the all-1s matrix. Then we give a simple proof of the rank inequality for diagonally magic matrices (J. Inequal. Appl. 2015:318, 2015).
Keywords
MSC
1 Introduction
2 Main result
The rank inequality for diagonally magic matrices can be stated as follows.
Theorem 2.1
([1], Theorem 2.1)
Let \(A \in\mathbb{C}^{n\times n}\) be a diagonally magic matrix. Then \(\operatorname {rank}(A)\leq2\).
There are diagonally magic matrices of ranks 0, 1, 2. Indeed, \(\operatorname {rank}(0_{n\times n} )=0\), \(\operatorname {rank}([1]_{n\times n} )=1\), and \(\operatorname {rank}(B_{n})=\operatorname {rank}(C_{n})=2\).
The purpose of this note is to give a simple proof of Theorem 2.1. Our proof depends only on the following fact.
Theorem 2.2
Proof
We are now ready to present our proof of Theorem 2.1.
Proof of Theorem 2.1
Declarations
Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 11501126, No. 11471122), the Youth Natural Science Foundation of Jiangxi Province (No. 20151BAB211011), the Science Foundation of Jiangxi Provincial Department of Education (No. GJJ150979), the Supporting the Development for Local Colleges and Universities Foundation of China - Applied Mathematics Innovative team building, and the Education Department of Jiangxi Province (No. 15YB106).
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Authors’ Affiliations
References
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