Hankel Tensors: Associated Hankel Matrices and Vandermonde Decomposition

Hankel tensors arise from applications such as signal processing. In this paper, we make an initial study on Hankel tensors. For each Hankel tensor, we associate it with a Hankel matrix and a higher order two-dimensional symmetric tensor, which we call the associated plane tensor. If the associated Hankel matrix is positive semi-definite, we call such a Hankel tensor a strong Hankel tensor. We show that an $m$ order $n$-dimensional tensor is a Hankel tensor if and only if it has a Vandermonde decomposition. We call a Hankel tensor a complete Hankel tensor if it has a Vandermonde decomposition with positive coefficients. We prove that if a Hankel tensor is copositive or an even order Hankel tensor is positive semi-definite, then the associated plane tensor is copositive or positive semi-definite, respectively. We show that even order strong and complete Hankel tensors are positive semi-definite, the Hadamard product of two strong Hankel tensors is a strong Hankel tensor, and the Hadamard product of two complete Hankel tensors is a complete Hankel tensor. We show that all the H-eigenvalue of a complete Hankel tensors (maybe of odd order) are nonnegative. We give some upper bounds and lower bounds for the smallest and the largest Z-eigenvalues of a Hankel tensor, respectively. Further questions on Hankel tensors are raised

The necessary and sufficient conditions of copositive tensors

In this paper, it is proved that (strict) copositivity of a symmetric tensor $\mathcal{A}$ is equivalent to the fact that every principal sub-tensor of $\mathcal{A}$ has no a (non-positive) negative $H^{++}$-eigenvalue. The necessary and sufficient conditions are also given in terms of the $Z^{++}$-eigenvalue of the principal sub-tensor of the given tensor. This presents a method of testing (strict) copositivity of a symmetric tensor by means of the lower dimensional tensors. Also the equivalent definition of strictly copositive tensors is given on entire space $\mathbb{R}^n$.Comment: 13 pages. arXiv admin note: text overlap with arXiv:1302.608

Convergence of a Second Order Markov Chain

In this paper, we consider convergence properties of a second order Markov chain. Similar to a column stochastic matrix is associated to a Markov chain, a so called {\em transition probability tensor} $P$ of order 3 and dimension $n$ is associated to a second order Markov chain with $n$ states. For this $P$, define $F_P$ as $F_P(x):=Px^{2}$ on the $n-1$ dimensional standard simplex $\Delta_n$. If 1 is not an eigenvalue of $\nabla F_P$ on $\Delta_n$ and $P$ is irreducible, then there exists a unique fixed point of $F_P$ on $\Delta_n$. In particular, if every entry of $P$ is greater than $\frac{1}{2n}$, then 1 is not an eigenvalue of $\nabla F_P$ on $\Delta_n$. Under the latter condition, we further show that the second order power method for finding the unique fixed point of $F_P$ on $\Delta_n$ is globally linearly convergent and the corresponding second order Markov process is globally $R$-linearly convergent.Comment: 16 pages, 3 figure

Infinite and finite dimensional Hilbert tensors

For an $m$-order $n-$dimensional Hilbert tensor (hypermatrix) $\mathcal{H}_n=(\mathcal{H}_{i_1i_2\cdots i_m})$, $$\mathcal{H}_{i_1i_2\cdots i_m}=\frac1{i_1+i_2+\cdots+i_m-m+1},\ i_1,\cdots, i_m=1,2,\cdots,n$$ its spectral radius is not larger than $n^{m-1}\sin\frac{\pi}{n}$, and an upper bound of its $E$-spectral radius is $n^{\frac{m}2}\sin\frac{\pi}{n}$. Moreover, its spectral radius is strictly increasing and its $E$-spectral radius is nondecreasing with respect to the dimension $n$. When the order is even, both infinite and finite dimensional Hilbert tensors are positive definite. We also show that the $m$-order infinite dimensional Hilbert tensor (hypermatrix) $\mathcal{H}_\infty=(\mathcal{H}_{i_1i_2\cdots i_m})$ defines a bounded and positively $(m-1)$-homogeneous operator from $l^1$ into $l^p$ ($1<p<\infty$), and the norm of corresponding positively homogeneous operator is smaller than or equal to $\frac{\pi}{\sqrt6}$