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By a Zero Line-Sum (ZLS) matrix I mean matrices with the property, that each row sum and each column sum equals zero:

$$A\in\mathbb{R}^{m\times n}:\ \sum_{i=1}^{n}a_{ij}=\sum_{j=1}^{m}a_{ij}=0$$

These can be thought of as being the difference of two "ordinary" doubly stochastic matrices.

ZLS matrices obviously don't have full rank but, as not all rank-deficient matrices have the property of being ZLS, I wonder if ZLS matrices are also special in different aspects and also w.r.t. the mappings they define.

Question:
Have ZLS matrices already been investigated, and what are non-trivial special properties of them that have been identified? I am looking for information on matrices, that equal the difference of two doubly stochastic matrices and, on the special properties of the transformations they define.

Clarification in response to Jochen Glueck's correct remarks: I use the term "projection" in a formally not correct way, namely meaning any mapping to a lower-dimensional space.

Remark: I have edited this question to replace the former "0-Stochastic projection matrix" with the "Zero Line-Sum" matrix following the suggestion of Gerry Myerson; "line" in this context is the common term for row and column.

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  • $\begingroup$ Could you please specify what you mean by "of the projections they define"? There are matrices whose rows and columns all sum up to $0$, but which are not projecctions. $\endgroup$ Commented Feb 15, 2018 at 12:34
  • $\begingroup$ Do you have any constraint on the value or sign of their elements? $\endgroup$ Commented Feb 15, 2018 at 12:36
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    $\begingroup$ @Manfred Weis: Let us consider the matrix $M =\begin{pmatrix} 1 & -1 \\ -1 &1 \end{pmatrix}$ The columns and rows sum up to $0$, but $M^2 \not= M$. What projection associated to $M$ do you have in mind? $\endgroup$ Commented Feb 15, 2018 at 13:05
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    $\begingroup$ So these "doubly stochastic projection matrices" are neither doubly stochastic nor projection matrices. $\endgroup$ Commented Feb 15, 2018 at 21:54
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    $\begingroup$ How about "zero line-sum matrices"? $\endgroup$ Commented Feb 16, 2018 at 6:13

1 Answer 1

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Below find six suggestions.

Case of general ZLS matrices.

Examples of properties are:

  1. All cofactors of a square ZLS are equal.
  2. Each square ZLS matrix of dimension $n$ has eigenvector $1^{n\times 1}$ with eigenvalue $0$.
  3. If an $n\times n$ ZLS $A$ is moreover symmetric, and if $\mathrm{Sp}(\cdot)$ denotes the spectrum of a matrix, then each cofactor of $A$ equals $\frac{1}{n}\prod_{\lambda\in\mathrm{Sp(A)\setminus\{0\}}}\lambda$.
  4. Rectangular ZLS matrices over a field $K$ form a $K$-vector space in the obvious way, and according to this thread its dimension is $(m-1)(n-1)$, and if $m=n$, then according to that thread its orthogonal complement w.r.t. the Frobenius inner product consists solely of sums of two rank-one matrices, more precisely, this orthogonal complement consists solely of matrices of the form $v\cdot 1^{1\times n} + 1^{n\times 1}\cdot w^{\text{t}}$.
  5. Some relevant Lie theoretic consideration can be found in

    Andreas Boukas, Philip Feinsilver, Anargyros Felouris, On the Lie structure of zero sum and related matrices. Random Oper. Stoch. Equ. 23, No. 4, 209-218 (2015)

Case of matrices which are differences of two doubly-stochastic matrices

  1. If $A$ is the sum of two symmetric stochastic matrices, you may find it profitable to look in the direction of Horn's conjecture. Starting points could be R Bhatia: Algebraic Geometry Solves an Old Matrix Theorem. Resonance. December 1999, or this MO thread.
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  • $\begingroup$ Why do you need $m=n$ for the statement about the orthogonal complement? It does not hold if $m \ne n$? $\endgroup$
    – valle
    Commented Apr 24 at 15:21

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