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).
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1 Introduction
We denote by \(\mathbb{C}^{n\times n}\) and \(\mathbb{R}^{n\times n}\) the sets of \(n \times n\) complex matrices and \(n \times n\) real matrices, respectively. For a positive integer n, let \(S_{n}\) be the set of all n! permutations of \(\{1, 2, \ldots, n\}\). If \(A=(a_{i,j}) \in\mathbb{C}^{n\times n}\) and \(\sigma\in S_{n}\), then the sequence \(a_{1,\sigma(1)}, a_{2,\sigma(2)}, \ldots, a_{n,\sigma (n)}\) is called the transversal of A [2]. Let \(A\in \mathbb{C}^{n\times n}\), \(1\leq i_{1} \leq i_{2}\leq\cdots\leq i_{k}\leq n\), and \(1\leq j_{1} \leq j_{2}\leq\cdots\leq j_{s}\leq n\). We denote by \(A[i_{1}, i_{2}, \ldots, i_{k}|j_{1}, j_{2}, \ldots, j_{s}]\) the \(k\times s\) submatrix of A that lies in the rows \(i_{1}, i_{2}, \ldots, i_{k}\) and columns \(j_{1}, j_{2}, \ldots, j_{s}\). Denote by \(A(i_{1}, i_{2}, \ldots, i_{k}|j_{1}, j_{2}, \ldots, j_{s})\) the \((n-k)\times(n-s)\) submatrix of A obtained by deleting the rows \(i_{1}, i_{2}, \ldots, i_{k}\) and columns \(j_{1}, j_{2}, \ldots, j_{s}\). A matrix \(A=(a_{i,j})\in\mathbb{C}^{n\times n}\) is called diagonally magic if
for all \(\sigma, \pi\in S_{n}\).
Obviously, the zero matrix \(0_{n\times n}\) and \(J=[1]_{n\times n}\), the matrix of all ones, are diagonally magic matrices. In [1], we prove that
and the Henkel matrix
are diagonally magic matrices. So, there are a lot of diagonally magic matrices. The nonnegative matrices \(B_{n}\) and \(C_{n}\) have been a hot research area [3, 4].
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
Let \(C =(c_{i,j})\in\mathbb{R}^{n\times n}\) be a positive matrix with
for all \(\gamma, \tau\in S_{n}\). Then there exist positive diagonal matrices \(X=\operatorname {diag}(x_{1},x_{2},\ldots,x_{n})\) and \(Y= \operatorname {diag}(y_{1},y_{2},\ldots,y_{n})\) such that
Proof
Let B be a \(k\times k\) submatrix of C. Then there are row and column indices \(\alpha=(i_{1},i_{2},\ldots,i_{k})\) and \(\beta=(j_{1},j_{2},\ldots,j_{k})\) such that \(B=C[\alpha|\beta]\). Note that the union of a transversal of B and a transversal of \(C(\alpha|\beta)\) is a transversal of C. Choose an arbitrary but fixed transversal T of the square matrix \(C(\alpha|\beta)\). For any \(\sigma,\pi\in S_{k}\), \(c_{i_{1},j_{\sigma(1)}},\ldots ,c_{i_{k},j_{\sigma(k)}}\) and the entries in T constitute a transversal of C, whereas \(c_{i_{1},j_{\pi(1)}},\ldots,c_{i_{k},j_{\pi(k)}}\) and the entries in T also constitute a transversal of C. Let b be the product of the entries in T. Obviously, \(b>0\). Since
for all \(\gamma, \tau\in S_{n}\), we have
which yields
Particularly, this shows that any \(2\times2\) submatrix
of C satisfies
For any \(x_{1}>0\), let
for \(j=1,2,\ldots,n\) and
for \(i=2,3,\ldots,n\). According to (1), (2), and (3), we have
for all \(i,j=1,2,\ldots,n\). Let \(X=\operatorname {diag}(x_{1},x_{2},\ldots,x_{n})\) and \(Y= \operatorname {diag}(y_{1},y_{2},\ldots,y_{n})\). Obviously, X and Y are positive diagonal matrices, and we have
This completes the proof. □
We are now ready to present our proof of Theorem 2.1.
Proof of Theorem 2.1
First, let A be real. Let A be a diagonally magic matrix. Then the elementwise exponential \(C =\exp (A):=(c_{i,j})\in\mathbb{R}^{n\times n}\) is a positive matrix with all permutation products equal. Hence, by Theorem 2.2 it is diagonally equivalent to J, the all-1s matrix, that is,
for suitable positive vectors \(x=(x_{1}, x_{2}, \ldots, x_{n})^{T}\) and \(y=(y_{1}, y_{2}, \ldots, y_{n})^{T}\). Hence, if \(q = (\log (x_{1}), \log (x_{2}), \ldots, \log (x_{n}) )^{T}\) and \(r= (\log (y_{1}), \log(y_{2}), \ldots, \log (y_{n}) )^{T}\), then
Hence,
where \(e_{n}=(\underbrace{1, 1,\ldots, 1}_{n})^{T}\). Thus, A is the sum of two matrices of rank 1 and, hence, at most of rank 2.
Now let A be complex, so that
Since A is a diagonally magic matrix, so are B and C. Hence, both B and C are of the form
and hence
The matrix A has rank at most 2. This completes the proof. □
According to (4), we obtain that a diagonally magic matrix A can be presented in the form
If \(A=(a_{i,j}) \in\mathbb{C}^{n\times n}\) is a diagonally magic matrix, for any \(x_{1}\in\mathbb{C}\), let
for \(j=1,2,\ldots,n\) and
for \(i=2,3,\ldots,n\). By (5) we have
for all \(i,j=1,2,\ldots,n\). For example,
and
By (5) we can get the characteristic polynomial, the eigenvalues, and the eigenvectors of A. In fact, the characteristic polynomial of A is
From (6) we can see that the algebraic multiplicity of the eigenvalue 0 of the diagonally magic matrix A is at least \(n-2\).
References
Zhou, D, Chen, G, Cai, Q, Chen, X: The rank inequality for diagonally magic matrices. J. Inequal. Appl. 2015, 318 (2015)
Zhan, X: Matrix Theory. Am. Math. Soc., Providence (2013)
Ding, J, Rhee, N: When a matrix and its inverse are nonnegative. Mo. J. Math. Sci. 26, 98-103 (2014)
Zhan, X: Extremal numbers of positive entries of imprimitive nonnegative matrices. Linear Algebra Appl. 424, 132-138 (2007)
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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All authors conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.
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Zhou, D., Cai, Q. & Chen, X. A note on the rank inequality for diagonally magic matrices. J Inequal Appl 2016, 66 (2016). https://doi.org/10.1186/s13660-016-1013-4
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DOI: https://doi.org/10.1186/s13660-016-1013-4