1 Introduction
For a positive integer
n, let
\(S_{n}\) be the set of all
n! permutations of
\(\{1, 2, \ldots, n\}\). We denote by
\(\mathbb{C}^{n\times n}\) and
\(\mathbb{R}^{n\times n}\) the set of
\(n \times n\) complex matrices and the set of
\(n \times n\) real matrices, respectively. 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 a
transversal of
A [
1]. In 2012, Professor Xingzhi Zhan defined the following new concept at a seminar and suggested studying its properties.
Obviously, the zero matrix
\(0_{n\times n}\) and
\(J=[1]_{n\times n}\), the matrix of all ones, are diagonally magic matrices. Denote
$$ B_{n}= \begin{pmatrix} 1&2 & \cdots& n\\ n+1&n+2 & \cdots& 2n\\ \vdots&\vdots& \ddots& \vdots\\ (n-1)n+1&(n-1)n+2 & \cdots& n^{2} \end{pmatrix} $$
(2)
and
$$ C_{n}= \begin{pmatrix} 1&2 & \cdots& n\\ 2&3 & \cdots& n+1\\ \vdots&\vdots& \ddots& \vdots\\ n&n+1 & \cdots& 2n-1 \end{pmatrix} . $$
(3)
We will show that
\(B_{n}\) and
\(C_{n}\) are diagonally magic matrices. So, there are a lot of diagonally magic matrices.
\(C_{n}\) is a Hankel matrix.
\(B_{n}\) and
\(C_{n}\) are nonnegative matrices which have been a hot research area [
2,
3].
For matrix
\(D=(d_{ij})\in\mathbb{C}^{m\times n}\), let the columns of
D be
\(d_{1},d_{2},\ldots,d_{n}\).
\(\operatorname{vec}(D)\) is a vector defined by
\(\operatorname{vec}(D)=(d^{T}_{1},d^{T}_{2},\ldots,d^{T}_{n})^{T}\), where the superscript
T
denotes the transpose. The matrix
\(E_{i,j}^{n}\) denotes the Type 1
elementary matrix [
4], p.8, which is simply the identity matrix
\(I_{n}\) of order
n, the
i,
i and
j,
j entries replaced by 0 and the
i,
j entry (respectively
j,
i entry) replaced by 1 (respectively 1). Given two matrices
A and
B, their direct sum is written as
\(A\oplus B\). Given a sequence of matrices
\(A_{i}\), for
\(i=1, \ldots, k\), one may write their direct sum as
$$A=\bigoplus_{i=1}^{k}A_{i}= \operatorname{diag} (A_{1}, \ldots, A_{k}). $$
Each
\(A_{i}\) is called a direct summand of
A. Let
\(e_{n}=(\underbrace{1, \ldots, 1} _{n})^{T}\) and
\(\widehat{e}_{n}=(\underbrace{0, \ldots, 0} _{n-1}, 1)^{T}\).
2 Main results
Let
\(A=(a_{i,j})\in\mathbb{C}^{n\times n}\) be a diagonally magic matrix with
\(n\geq2\). Assume that the sum of every transversal is
c. From the definition of diagonally magic matrices (
1), we have a system of linear equations
$$ \widetilde{A}_{n} \operatorname{vec} \bigl(A^{T} \bigr)=ce_{n!}, $$
(4)
where
\(\widetilde{A}_{n}=(\widetilde{a}_{1}, \widetilde{a}_{2}, \ldots, \widetilde{a}_{n^{2}}) \in\mathbb{R}^{n!\times n^{2}}\) is the
coefficient matrix. If
\(n=2\), from the definition of the diagonally magic matrices and (
4), the coefficient matrix
\(\widetilde {A}_{2}\) can be chosen to be the
\(2 \times4\) matrix
$$\widetilde{A}_{2}= \begin{pmatrix} 1 & 0 & 0 & 1\\ 0 & 1 & 1 & 0 \end{pmatrix} =(\widetilde{a}_{1}, \widetilde{a}_{2}, \widetilde{a}_{3}, \widetilde{a}_{4}), $$
and the augmented matrix is
$$ \widehat{A}_{2}= \left ( \textstyle\begin{array}{c@{\hspace{3pt}}c@{\quad}c@{\quad}c|@{\quad}c} 1 & 0 & 0 & 1& c\\ 0 & 1 & 1 & 0& c \end{array}\displaystyle \right ). $$
(5)
This augmented matrix is the row-reduced echelon form.
Suppose
\(n\geq2\) and that, for
\(n = 2, \ldots, k\), we have constructed the coefficient matrix
$$\widetilde{A}_{k}=(\widetilde{a}_{1}, \widetilde{a}_{2}, \ldots, \widetilde{a}_{k^{2}}). $$
Let
\(n=k+1\). We use the following method to construct the coefficient matrix
\(\widetilde{A}_{k+1}\). Firstly, let
$$\begin{aligned}& C_{1,1}^{k+1}= (e_{k!}, 0_{k!\times k} ), \\& C_{1, m+1}^{k+1}= (0_{k!\times1},\widetilde{a}_{(m-1)k+1}, \widetilde{a}_{(m-1)k+2}, \ldots, \widetilde{a}_{mk} ) \end{aligned}$$
for
\(1\leq m \leq k\). Secondly, construct
$$C_{i,j}^{k+1}=C_{i-1,j}^{k+1}E_{i-1,i}^{k+1} $$
for
\(i=2,3,\ldots,k+1\),
\(j=1,2,3,\ldots,k+1\), where
\(E_{i,j}^{n}\) denotes the Type 1
elementary matrix [
4], p.8, which is simply the identity matrix
\(I_{n}\) of order
n, the
i,
i and
j,
j entries replaced by 0 and the
i,
j entry (respectively
j,
i entry) replaced by 1 (respectively 1). Then we get the coefficient matrix
$$ \widetilde{A}_{k+1}= \begin{pmatrix} C_{1,1}^{k+1}& C_{1,2}^{k+1}& \cdots& C_{1,k+1}^{k+1}\\ C_{2,1}^{k+1}& C_{2,2}^{k+1}& \cdots& C_{2,k+1}^{k+1}\\ \vdots& \vdots& \ddots& \vdots\\ C_{k+1,1}^{k+1}& C_{k+1,2}^{k+1}& \cdots& C_{k+1,k+1}^{k+1} \end{pmatrix}. $$
For example, if
\(n=3\), according to the constructing method and
\(\widetilde{A}_{2}\), we have
$$\begin{aligned} \widetilde{A}_{3}&= \begin{pmatrix} C_{1,1}^{3}& C_{1,2}^{3}& C_{1,3}^{3}\\ C_{2,1}^{3}& C_{2,2}^{3}& C_{2,3}^{3}\\ C_{3,1}^{3}& C_{3,2}^{3}& C_{3,3}^{3} \end{pmatrix} \\ &=\left ( \textstyle\begin{array}{ccc|ccc|ccc} 1& 0& 0 &0&1&0 &0&0&1\\ 1& 0& 0 &0&0&1 &0&1&0\\ \hline 0&1& 0 &1&0&0 &0&0&1\\ 0&1& 0 &0&0&1 &1&0&0\\ \hline 0& 0&1 &1&0&0 &0&1&0\\ 0& 0&1 &0&1&0 &1&0&0 \end{array}\displaystyle \right ) \\ &=(\widetilde{a}_{1}, \widetilde{a}_{2}, \widetilde{a}_{3}, \widetilde {a}_{4},\widetilde{a}_{5}, \widetilde{a}_{6},\widetilde{a}_{7}, \widetilde {a}_{8}, \widetilde{a}_{9}). \end{aligned}$$
Assume
$$C_{j}= \bigl(C_{j,1}^{k+1}, C_{j,2}^{k+1}, \ldots, C_{j,k+1}^{k+1}, ce_{k!} \bigr) $$
for
\(j=1, 2, \ldots, k+1\). Consequently, the augmented matrix of
\(\widetilde{A}_{k+1}\) is
$$\bigl(C_{1}^{T}, C_{2}^{T}, \ldots, C_{k+1}^{T} \bigr)^{T}. $$
Let
$$\begin{aligned}& D_{n}= \begin{pmatrix} 0 & \cdots& 0 & 1\\ \vdots& \ddots& \vdots& \vdots\\ 0 & \cdots& 0 & 1\\ 0 & \cdots& 0 & 1 \end{pmatrix} \in \mathbb{R}^{n\times n}, \\& F_{(n-1)\times n}= (\textstyle\begin{array}{@{}c@{\quad}c@{}} I_{n-1} & -e_{n-1} \end{array}\displaystyle ) \in\mathbb{R}^{(n-1)\times n}, \\& G_{n}= \begin{pmatrix} 0 & 1& \cdots& 1 & 3-n\\ 1 & 0& \cdots& 1 & 3-n\\ \vdots&\vdots& \ddots&\vdots&\vdots\\ 1 & 1& \cdots& 0 & 3-n\\ 1 & 1& \cdots& 1 & 2-n \end{pmatrix} \in \mathbb{R}^{n\times n}. \end{aligned}$$
We claim that the row-reduced echelon form of the augmented matrix in the system of linear equations (
4) has the following form:
$$ \textstyle\begin{array}{@{}c@{}} \textstyle\begin{array}{@{\hspace{-30pt}}l} \overbrace{\hphantom{\hspace{43mm}}}^{(n-2)n} \end{array}\displaystyle \\ \widehat{A}_{n}=\left ( \textstyle\begin{array}{cccccc|c} I_{n}&D_{n} &D_{n} &\cdots&D_{n}&G_{n}&ce_{n}\\ & F_{(n-1)\times n} & 0 &\cdots& 0& -F_{(n-1)\times n}&0 \\ & &F_{(n-1)\times n}& \cdots&0&-F_{(n-1)\times n}&0 \\ & & & \ddots&\vdots&\vdots&\vdots\\ & & & & F_{(n-1)\times n}&-F_{(n-1)\times n}&0\\ &&{\Huge0}&& &0&0 \end{array}\displaystyle \right ). \end{array} $$
(6)
We prove this by induction on
n. For example,
\((\widetilde{A}_{2}, ce_{2!})\) is row-equivalent to
$$ \widehat{A}_{2}=\left ( \textstyle\begin{array}{cccc|c} 1 & 0 & 0 & 1& c\\ 0 & 1 & 1 & 0& c \end{array}\displaystyle \right ). $$
The row-reduced echelon form of the augmented matrix
\((\widetilde{A}_{3}, ce_{3!})\) has the following form:
$$\begin{aligned} \widehat{A}_{3}&=\left ( \textstyle\begin{array}{ccc|c} I_{3}& D_{3}& G_{3}&ce_{3}\\ 0_{2\times3}& F_{2\times3}& -F_{2\times3}&0_{3\times1}\\ 0_{1\times3}& 0_{1\times3}& 0_{1\times3}&0 \end{array}\displaystyle \right ) \\ &=\left ( \textstyle\begin{array}{ccc|ccc|ccc|c} 1& 0& 0 &0&0&1 &0&1&0 &1\\ 0& 1& 0 &0&0&1 &1&0&0 &1\\ 0& 0 &1 &0&0&1 &1&1&-1 &1\\ \hline 0&0&0 &1&0& -1 &-1&0&1 &0\\ 0&0&0 &0& 1&-1 &0&-1&1 &0\\ \hline 0& 0&0 &0&0&0 &0&0&0 &0 \end{array}\displaystyle \right ). \end{aligned}$$
Obviously, if
\(n=2\),
\(\widehat{A}_{2}\) in (
5) has the form (
6). Suppose
\(n\geq2\) and that, for
\(n = 2, \ldots, k\), the assertion has been proved for
n!-by-
\(n^{2}\) matrix
\(\widetilde {A}_{n}\). Let
\(n=k+1\),
$$\begin{aligned}& H_{k\times(k+1)}= (e_{k}, 0_{k\times k} ), \qquad J_{k\times (k+1)}= (0_{k\times1}, I_{k} ), \\& \widetilde{D}_{k\times(k+1)}= (0_{k\times1}, D_{k} ),\qquad \widetilde{G}_{k\times(k+1)}= (0_{k\times1}, G_{k} ), \\& \widetilde{F}_{(k-1)\times(k+1)}= (0_{k\times1}, F_{(k-1)\times k} ). \end{aligned}$$
By the inductive hypothesis, we can obtain the following matrix after a sequence of elementary operations for
\(C_{1}\):
$$ {{\begin{aligned} P_{1} &=\left ( \textstyle\begin{array}{ccccccc|c} H_{k\times(k+1)}&J_{k\times(k+1)}&\widetilde{D}_{k\times(k+1)} &\widetilde{D}_{k\times(k+1)} &\cdots&\widetilde{D}_{k\times (k+1)}&\widetilde{G}_{k\times(k+1)}&ce_{k}\\ && \widetilde{F}_{(k-1)\times(k+1)} & 0 &\cdots& 0& -\widetilde {F}_{(k-1)\times(k+1)}&0 \\ && &\widetilde{F}_{(k-1)\times(k+1)}& \cdots&0&-\widetilde {F}_{(k-1)\times(k+1)}&0 \\ && & & \ddots&\vdots&\vdots&\vdots\\ & & & & & \widetilde{F}_{(k-1)\times(k+1)}&-\widetilde{F}_{(k-1)\times (k+1)}&0\\ &&\Huge0&&& &0&0 \end{array}\displaystyle \right ) \\ &\equiv \begin{pmatrix} P_{1,1} \\ P_{1,2} \\ P_{1,3} \\ \vdots\\ P_{1,k-1} \\ 0 \end{pmatrix} , \end{aligned}}} $$
where
$$\begin{aligned}& P_{1,1}=\left ( \textstyle\begin{array}{ccccccc|c} H_{k\times(k+1)}&J_{k\times(k+1)}&\widetilde{D}_{k\times(k+1)} &\widetilde{D}_{k\times(k+1)} &\cdots&\widetilde{D}_{k\times (k+1)}&\widetilde{G}_{k\times(k+1)}&ce_{k} \end{array}\displaystyle \right ),\\& \textstyle\begin{array}{@{}c@{}} \textstyle\begin{array}{@{\hspace{-30pt}}c} \overbrace{\hphantom{\hspace{15mm}}}^{(k-i-1)(k+1)\ \mathrm{columns}} \end{array}\displaystyle \\ P_{1,i}=\left ( \textstyle\begin{array}{cccccccc|c} 0&\cdots& 0&\widetilde{F}_{(k-1)\times(k+1)} & 0 &\cdots& 0& -\widetilde{F}_{(k-1)\times(k+1)}&0 \end{array}\displaystyle \right )\in \mathbb{R}^{(k-1)\times((k+1)^{2}+1)} \end{array}\displaystyle \end{aligned}$$
for
\(i=2, 3, \ldots, k-1\).
\(C_{j}\) is row-equivalent to
$$P_{j}=P_{j-1} \Biggl( \Biggl(\bigoplus ^{k+1}_{i=1}E_{j-1,j}^{k+1} \Biggr) \oplus1 \Biggr)= \bigl(P_{j,1}^{T}, \ldots, P_{j,k-1}^{T}, 0^{T} \bigr)^{T} $$
for
\(j=2, \ldots, k+1\).
\(P_{j,1}\in\mathbb{R}^{k\times((k+1)^{2}+1)}\) is the first
k rows of
\(P_{j}\).
\(P_{j,i}\in\mathbb{R}^{(k-1)\times ((k+1)^{2}+1)}\) is the rows of
\(P_{j}\) from
\((k+(i-2)(k-1)+1)\)th to
\((k+(i-1)(k-1))\)th row,
\(i=2, \ldots, k-1\). In
\(P_{1,1}\), multiplying row
k by the scalar −1 and adding to row
j, for
\(j=1, \ldots, k-1\), then
\(P_{1,1}\) is row-equivalent to
$$\widehat{P}_{1,1}= (\widehat{H}_{k\times(k+1)}, \widehat {J}_{k\times(k+1)}, \widehat{\widetilde{D}}_{k\times(k+1)}, \widehat { \widetilde{D}}_{k\times(k+1)}, \ldots, \widehat{\widetilde {D}}_{k\times(k+1)}, \widehat{\widetilde{G}}_{k\times(k+1)}, c\widehat {e}_{k} ), $$
where
$$\begin{aligned}& \widehat{H}_{k\times(k+1)}= \begin{pmatrix} 0&0_{(k-1)\times k}\\ 1&0 \end{pmatrix} ,\qquad \widehat{J}_{k\times(k+1)}= \begin{pmatrix} 0& I_{k-1}&-e_{k-1}\\ 0&0&1 \end{pmatrix} , \\& \widehat{\widetilde{D}}_{k\times(k+1)}= \begin{pmatrix} 0_{(k-1)\times k}&0\\ 0&1 \end{pmatrix} ,\qquad \widehat{\widetilde{G}}_{k\times(k+1)}= \begin{pmatrix} 0 & -1& 0&\cdots& 0 & 1\\ 0 & 0& -1&\cdots& 0 & 1\\ \vdots&\vdots& \vdots&\ddots&\vdots&\vdots\\ 0 & 0& 0&\cdots& -1 & 1\\ 0 & 1&1& \cdots& 1 & 2-k \end{pmatrix} . \end{aligned}$$
Applying this method to
\(P_{j,1}\),
\(j=2, \ldots, k+1\), then
\(P_{j,1}\) is row-equivalent to
$$\widehat{P}_{j,1}=\widehat{P}_{j-1,1} \Biggl( \Biggl(\bigoplus ^{k+1}_{i=1}E_{j-1,j}^{k+1} \Biggr)\oplus1 \Biggr). $$
Multiplying row
\(k-1\) of
\(\widehat{P}_{1,1}\) by the scalar −1 and adding to row
k of
\(\widehat{P}_{k+1,1}\), and multiplying row
\(k-1\) of
\(P_{1,i}\) by the scalar −1 and adding to row
k of
\(\widehat {P}_{k+1,1}\) for
\(i=2, 3, \ldots, k-1\), then row
k of
\(\widehat {P}_{k+1,1}\) changes to
$$ \bigl(\underbrace{\widehat{e}_{k+1}^{T}, \ldots, \widehat{e}_{k+1}^{T}} _{k(k+1)}, \underbrace{1, \ldots, 1} _{k}, 1-k, c \bigr). $$
(7)
Picking row
k of
\(\widehat{P}_{j,1}\),
\(j=1, 2, \ldots, k\), and (
7), we have
$$ (I_{k+1}, \underbrace{D_{k+1}, D_{k+1}, \ldots, D_{k+1}} _{(k-1)(k+1)}, G_{k+1}, ce_{k+1} ). $$
(8)
Combining row 1 of
\(\widehat{P}_{2,1}\) and row
i of
\(\widehat {P}_{1,1}\), for
\(i=1, \ldots, k-1\), we have
$$ (0_{k\times(k+1)}, F_{k\times(k+1)}, \underbrace{0_{k\times (k+1)}, \ldots, 0_{k\times(k+1)}} _{(k-3)(k+1)}, -F_{k\times(k+1)}, 0 ). $$
(9)
Combining row 1 of
\(P_{2,i}\) and row
j of
\(P_{1,i}\), for
\(i, j=1, 2, \ldots, k-1\), we get
$$ \left ( \textstyle\begin{array}{ccccccc|c} 0_{k\times(k+1)}& 0_{k\times(k+1)}&F_{k\times(k+1)}&0_{k\times (k+1)}& \cdots&0_{k\times(k+1)}&-F_{k\times(k+1)}&0 \\ 0_{k\times(k+1)}& 0_{k\times(k+1)}&0_{k\times(k+1)}&F_{k\times (k+1)}& \cdots&0_{k\times(k+1)}&-F_{k\times(k+1)}&0 \\ & & & &\ddots&\vdots&\vdots&\vdots\\ &&\Huge0&& & F_{k\times (k+1)}&-F_{k\times(k+1)}&0 \end{array}\displaystyle \right ). $$
(10)
Combining (
8), (
9) and (
10), we have
$$ \left ( \textstyle\begin{array}{cccccc|c} I_{k+1}& D_{k+1}& D_{k+1}& \cdots& D_{k+1}& G_{k+1}& ce_{k+1}\\ & F_{k\times(k+1)} & 0 &\cdots& 0& -F_{k\times(k+1)}&0 \\ & &F_{k\times(k+1)}& \cdots&0&-F_{k\times(k+1)}&0 \\ & & & \ddots&\vdots&\vdots&\vdots\\ &&\Huge0& & F_{k\times (k+1)}&-F_{k\times(k+1)}&0 \end{array}\displaystyle \right ). $$
The other rows depend linearly on some rows of the above matrix. From the
row-reduced echelon form (
6), we get
\(\operatorname{rank}(\widetilde{A}_{n})=n^{2}-2n+2\).
\(a_{j,n}\),
\(a_{n,i}\), for
\(2\leq j\leq n\),
\(1\leq i\leq n-1\), are free variables. The other entries of matrix
\(A=(a_{i,j})\in\mathbb{C}^{n\times n}\) are the pivot variables. The pivot variables are completely determined in terms of free variables.
According to (
11), we know that the matrices
\(B_{n}\) in (
2) and
\(C_{n}\) in (
3) are diagonally magic matrices. It is easy to verify that
\(B_{n}\) is row-equivalent to
$$ B_{n}\longrightarrow \begin{pmatrix} 1&2 &3& \cdots& n\\ 0&1 &2& \cdots& n-1\\ 0&0 &0& \cdots& 0\\ \vdots&\vdots& \vdots&\ddots& \vdots\\ 0&0 &0& \cdots& 0 \end{pmatrix} . $$
\(C_{n}\) is row-equivalent to
$$ C_{n}\longrightarrow \begin{pmatrix} 1&2 &3& \cdots& n\\ 0&1 &2& \cdots& n-1\\ 0&0 &0& \cdots& 0\\ \vdots&\vdots& \vdots&\ddots& \vdots\\ 0&0 &0& \cdots& 0 \end{pmatrix} . $$
Now it is clear that 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
\(\operatorname{rank}(B_{n})=\operatorname{rank}(C_{n})=2\).
From (
13), we can see that
the algebraic multiplicity of the eigenvalue 0 of the diagonally magic matrix
A is
at least
\(n-2\).
Let \(A\in\mathbb{C}^{n\times n}\), \(1\leq i_{1} \leq i_{2}\leq\cdots\leq i_{k}\leq n\), \(1\leq j_{1} \leq j_{2}\leq\cdots\leq j_{s}\leq n\). 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 \(j_{1}, j_{2}, \ldots, j_{s}\).
Acknowledgements
This work is supported by the National Natural Science Foundation of China (No. 11501126, No. 11471122, No. 61502107), the Youth Natural Science Foundation of Jiangxi Province (No. 20151BAB211011, No. 20151BAB211014), the Education Department of Jiangxi Province (No. 15YB106), the Supporting the Development for Local Colleges and Universities Foundation of China - Applied Mathematics Innovative team building, the Graduate student innovation Foundation of Gannan Normal University (No. YCX14B002, No. YCX14B003), and the Natural Science Foundation of Anhui Province (No. 1508085MA12).
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
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.