Positive semi-definiteness and sum-of-squares property of fourth order four dimensional Hankel tensors

详细信息    查看全文

• 作者：Yannan Chen^a ; ^{ynchen@zzu.edu.cn" class="auth_mail" title="E-mail the corresponding author} ; Liqun Qi^b ; ^{maqilq@polyu.edu.hk" class="auth_mail" title="E-mail the corresponding author} ; Qun Wang^b ; ^{wangqun876@gmail.com" class="auth_mail" title="E-mail the corresponding author}
• 关键词：15A18 ; 15A69
• 刊名：Journal of Computational and Applied Mathematics
• 出版年：2016
• 出版时间：15 August 2016
• 年：2016
• 卷：302
• 期：Complete
• 页码：356-368
• 全文大小：924 K

文摘

A symmetric positive semi-definite (PSD) tensor, which is not sum-of-squares (SOS), is called a PSD non-SOS (PNS) tensor. Is there a fourth order four dimensional PNS Hankel tensor? The answer for this question has both theoretical and practical significance. Under the assumptions that the generating vector $v$ of a Hankel tensor A$\mathcal{A}$ is symmetric and the fifth element v₄$v_4$ of $v$ is fixed at 1$1$, we show that there are two surfaces M₀$M_0$ and N₀$N_0$ with the elements v₂,v₆,v₁,v₃,v₅$v_2,v_6,v_1,v_3,v_5$ of $v$ as variables, such that M₀≥N₀$M_0\ge N_0$, A$\mathcal{A}$ is SOS if and only if v₀≥M₀$v_0\ge M_0$, and A$\mathcal{A}$ is PSD if and only if v₀≥N₀$v_0\ge N_0$, where v₀$v_0$ is the first element of $v$. If M₀=N₀$M_0=N_0$ for a point P=(v₂,v₆,v₁,v₃,v₅)^⊤$P={(v_2,v_6,v_1,v_3,v_5)}^{\top }$, there are no fourth order four dimensional PNS Hankel tensors with symmetric generating vectors for such v₂,v₆,v₁,v₃,v₅$v_2,v_6,v_1,v_3,v_5$. Then, we call such P$P$ a PNS-free point. We prove that a 45$45$-degree planar closed convex cone, a segment, a ray and an additional point are PNS-free. Numerical tests check various grid points and report that they are all PNS-free.