The inverse of a symmetric nonnegative matrix can be copositive

IF 0.7 4区 数学 Q2 Mathematics
Robert Reams
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引用次数: 0

Abstract

Let $A$ be an $n\times n$ symmetric matrix. We first show that if $A$ and its pseudoinverse are strictly copositive, then $A$ is positive semidefinite, which extends a similar result of Han and Mangasarian. Suppose $A$ is invertible, as well as being symmetric. We showed in an earlier paper that if $A^{-1}$ is nonnegative with $n$ zero diagonal entries, then $A$ can be copositive (for instance, this happens with the Horn matrix), and when $A$ is copositive, it cannot be of form $P+N$, where $P$ is positive semidefinite and $N$ is nonnegative and symmetric. Here, we show that if $A^{-1}$ is nonnegative with $n-1$ zero diagonal entries and one positive diagonal entry, then $A$ can be of the form $P+N$, and we show how to construct $A$. We also show that if $A^{-1}$ is nonnegative with one zero diagonal entry and $n-1$ positive diagonal entries, then $A$ cannot be copositive.
对称非负矩阵的逆可以是共积的
设A是一个n乘以n的对称矩阵。我们首先证明了如果$A$和它的伪逆是严格合的,则$A$是正半定的,推广了Han和Mangasarian的类似结果。假设A是可逆的,并且是对称的。我们在之前的一篇论文中证明了如果$A^{-1}$是非负的且$n$零对角线项,那么$A$可以是共积的(例如,这发生在Horn矩阵中),当$A$是共积的时,它不可能是$P+ n$的形式,其中$P$是半正定的,而$n$是非负对称的。在这里,我们证明了如果$A^{-1}$是非负的且有$ N -1$ 0个对角线项和1个正对角线项,那么$A$可以是$P+N$的形式,并且我们证明了如何构造$A$。我们还证明了如果$A^{-1}$是非负的,且有1个对角线项为零,且有$n-1个对角线项为正,则$A$不可能是共生的。
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来源期刊
CiteScore
1.20
自引率
14.30%
发文量
45
审稿时长
6-12 weeks
期刊介绍: The journal is essentially unlimited by size. Therefore, we have no restrictions on length of articles. Articles are submitted electronically. Refereeing of articles is conventional and of high standards. Posting of articles is immediate following acceptance, processing and final production approval.
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