Open Access

p-Norm SDD tensors and eigenvalue localization

Journal of Inequalities and Applications20162016:178

https://doi.org/10.1186/s13660-016-1119-8

Received: 1 May 2016

Accepted: 23 June 2016

Published: 8 July 2016

Abstract

We present a new class of nonsingular tensors (p-norm strictly diagonally dominant tensors), which is a subclass of strong \(\mathcal{H}\)-tensors. As applications of the results, we give a new eigenvalue inclusion set, which is tighter than those provided by Li et al. (Linear Multilinear Algebra 64:727-736, 2016) in some case. Based on this set, we give a checkable sufficient condition for the positive (semi)definiteness of an even-order symmetric tensor.

Keywords

p-norm SDD tensor strong \(\mathcal{H}\)-tensor positive (semi)definiteness eigenvalue localization

1 Introduction

Let \(\mathbb{C}(\mathbb{R})\) denote the set of all complex (real) numbers, and \([n]:=\{1,2,\ldots,n\}\). An mth-order n-dimensional complex (real) tensor, denoted by \(\mathcal{A}\in\mathbb{C}^{[m,n]}(\mathbb{R}^{[m,n]})\), is a multidimensional array of \(n^{m}\) elements of the form
$$ \mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}}),\quad a_{i_{1}\cdots i_{m}}\in\mathbb {C}(\mathbb{R}), i_{j} \in[n], j\in[m]. $$
When \(m=2\), \(\mathcal{A}\) is an n-by-n matrix. A tensor \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in\mathbb{R}^{[m,n]}\) is called nonnegative if each its entry is nonnegative, and it is called symmetric [2, 3] if
$$a_{i_{1}\cdots i_{m} }= a_{\pi(i_{1}\cdots i_{m} )},\quad \forall\pi\in\Pi_{m}, $$
where \(\Pi_{m}\) is the permutation group of m indices. Moreover, an mth-order n-dimensional tensor \(\mathcal{I}=(\delta_{i_{1}i_{2}\cdots i_{m}})\) is called the identity tensor [4] if
$$\delta_{i_{1}i_{2}\cdots i_{m}}= \left \{ \textstyle\begin{array}{l@{\quad}l} 1 &\mbox{if } i_{1}=i_{2}=\cdots=i_{m}, \\ 0 &\mbox{otherwise}. \end{array}\displaystyle \right . $$
For an n-dimensional vector \(x=(x_{1},x_{2},\ldots,x_{n})^{T}\), real or complex, we define the n-dimensional vector
$$\mathcal{A}x^{m-1}:= \biggl(\sum_{i_{2},\ldots,i_{m}\in[n]} a_{ii_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}} \biggr)_{1\leq i \leq n}, $$
and the n-dimensional vector
$$x^{[m-1]}:=\bigl(x_{i}^{m-1}\bigr)_{1\leq i \leq n}. $$

The following definition related to eigenvalues of tensors was first introduced and studied by Qi [3] and Lim [5].

Definition 1

[3, 5]

Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\). A pair \((\lambda,x)\in\mathbb{C}\times(\mathbb{C}^{n}\setminus\{0\})\) is called an eigenvalue-eigenvector (or simply eigenpair) of \(\mathcal{A}\) if satisfies the equation
$$ \mathcal{A}x^{m-1}=\lambda x^{[m-1]}. $$
(1)
We call \((\lambda,x)\) an H-eigenpair if they are both real.
In addition, the spectral radius of a tensor \(\mathcal{A}\) is defined as
$$\rho\mathcal{(A)} = \max\bigl\{ |\lambda|:\lambda \mbox{ is an eigenvalue of } \mathcal{A}\bigr\} . $$

Definition 2

A tensor \(\mathcal{A}\in\mathbb{C}^{[m,n]}\) is said to be nonsingular if zero is not an eigenvalue of \(\mathcal{A}\). Otherwise, it is called singular.

Tensor eigenvalue problems have gained special attention in the realm of numerical multilinear algebra, and they have a wide range in practice; see [3, 4, 616]. For instance, we can use the smallest H-eigenvalues of tensors to determine their positive (semi)definiteness, that is, for an even-order real symmetric tensor \(\mathcal{A}\), if its smallest H-eigenvalue is positive (nonnegative), then \(\mathcal{A}\) is positive (semi)definite; consequently, the multivariate homogeneous polynomial \(f(x)\) determined by \(\mathcal{A}\) is positive (semi)definite [3].

Most often, it is difficult to compute the smallest H-eigenvalue. Therefore, we always try to give a distribution range of eigenvalues of a given tensor in the complex plane. In particular, if this range is in the right-half complex plane, which means that the smallest H-eigenvalue is positive, then the corresponding tensor is positive definite.

Qi [3] generalized the Geršgorin eigenvalue inclusion theorem from matrices to real symmetric tensors, which can be easily extended to generic tensors; see [4, 17].

Theorem 1

Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\). Then
$$ \sigma(\mathcal{A})\subseteq\Gamma(\mathcal{A})=\bigcup _{i\in [n]}\Gamma_{i}(\mathcal{A}), $$
where \(\sigma(\mathcal{A})\) is the set of all the eigenvalues of \(\mathcal{A}\), and
$$ \Gamma_{i}(\mathcal{A})=\bigl\{ z\in\mathbb{C}:|z-a_{ii\cdots i}| \leq r_{i}(\mathcal{A})\bigr\} ,\quad r_{i}(\mathcal{A})=\sum _{\substack{i_{2},\ldots ,i_{m}\in[n],\\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|. $$
Recently, as an extension of the theory in [18], Li et al. [1, 17, 19] proposed three new Brauer-type eigenvalue localization sets for tensors and showed tighter bounds than \(\Gamma(\mathcal{A})\) of Theorem 1. We list the latest Brauer-type eigenvalue localization set as follows. For convenience, we denote
$$\begin{aligned}& \Delta_{i}=\bigl\{ (i_{2},i_{3},\ldots, i_{m}):i_{j}=i \mbox{ for some } j\in\{2,3,\ldots , m\}, \mbox{where } i,i_{2},\ldots,i_{m}\in[n]\bigr\} , \\& \overline{\Delta}_{i}=\bigl\{ (i_{2},i_{3}, \ldots, i_{m}):i_{j}\neq i \mbox{ for any } j\in\{2,3,\ldots, m\}, \mbox{where } i,i_{2},\ldots,i_{m}\in[n]\bigr\} , \end{aligned}$$
and
$$ r_{i}^{\Delta_{i}}(\mathcal{A})=\sum_{\substack{(i_{2},\ldots,i_{m})\in\Delta _{i},\\ \delta_{ii_{2}\cdots i_{m}}=0}} |a_{ii_{2}\cdots i_{m}}|,\qquad r_{i}^{\overline{\Delta}_{i}}(\mathcal{A})=\sum _{(i_{2},\ldots,i_{m})\in\overline{\Delta}_{i}} |a_{ii_{2}\cdots i_{m}}|. $$

Theorem 2

[1]

Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\). Then
$$ \sigma(\mathcal{A})\subseteq\Omega(\mathcal{A})= \biggl(\bigcup _{i\in [n]}\hat{\Omega}_{i}(\mathcal{A}) \biggr)\cup \biggl(\bigcup_{\substack {i,j\in[n],\\ i\neq j}} \Bigl(\hat{\Omega}_{i,j}( \mathcal{A})\cap \Gamma_{i}(\mathcal{A}) \Bigr) \biggr), $$
where
$$ \hat{\Omega}_{i}(\mathcal{A})= \bigl\{ z\in\mathbb{C}:|z-a_{i\cdots i}| \leq r_{i}^{\Delta_{i}}(\mathcal{A}) \bigr\} $$
and
$$ \hat{\Omega}_{i,j}(\mathcal{A})= \bigl\{ z\in\mathbb{C}: \bigl(|z-a_{i\cdots i}|-r_{i}^{\Delta_{i}}(\mathcal{A}) \bigr) \bigl(|z-a_{j\cdots j}|-r_{j}^{\overline{\Delta}_{i}}(\mathcal{A}) \bigr)\leq r_{i}^{\overline{\Delta}_{i}}(\mathcal{A})r_{j}^{\Delta_{i}}( \mathcal {A}) \bigr\} . $$

Li et al. [1] proved that the set \(\Omega(\mathcal{A})\) in Theorem 2 is tighter than \(\Theta(\mathcal{A})\) in [19] and \(\mathcal{K}(\mathcal{A})\) in [17]; for details, see Theorem 2.3 in [1].

In this paper, we continue this research on the eigenvalue localization problem for tensors. A class of strictly diagonally dominant tensors that involve a parameter p in the interval \([1,\infty]\), denoted by p-norm SDD tensor, is introduced in Section 2. In Section 3, we discuss the relationships between p-norm SDD tensors and strong \(\mathcal{H}\)-tensors. A new eigenvalue inclusion set for tensors based on p-norm SDD tensors is obtained in Section 4, and numerical results show that the new set is tighter than \(\Omega(\mathcal{A})\) in Theorem 2 in some case. Finally, in Section 5, we give a checkable sufficient condition for the positive (semi)definiteness of even-order symmetric tensors.

2 p-Norm SDD tensors

In this section, we propose a new class of nonsingular tensors, namely p-norm strictly diagonally dominant tensors. First, some notation and the definition of strictly diagonally dominant tensors are given.

Given a tensor \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\) and a real number \(p\in[1,\infty]\), denote
$$ r_{i}^{p}(\mathcal{A}):= \biggl(\sum _{\substack{i_{2},\ldots,i_{m}\in[n],\\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|^{p} \biggr)^{\frac {1}{p}} \quad \mbox{for all } i\in[n]. $$
In particular, if \(p=1\), then \(r_{i}^{1}(\mathcal{A})=r_{i}(\mathcal{A})\) for all \(i\in[n]\). If \(p=\infty\), then \(r_{i}^{\infty}(\mathcal{A})= \max_{\substack{i_{2},\ldots,i_{m}\in[n],\\ \delta{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|\) for all \(i\in[n]\). For a vector \(x=(x_{1},x_{2},\ldots, x_{n})^{T}\in\mathbb{C}^{n}\), the \(l_{q}\)-norm on \(\mathbb{C}^{n}\) is
$$\|x\|_{q}:= \biggl(\sum_{i\in[n]}|x_{i}|^{q} \biggr)^{\frac{1}{q}}. $$

Definition 3

[16]

A tensor \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\) is diagonally dominant if
$$ |a_{ii\cdots i}|\geq r_{i}(\mathcal{A})\quad \mbox{for all } i\in[n], $$
(2)
and \(\mathcal{A}\) is strictly diagonally dominant if the strict inequality holds in (2) for all i.

Remark 1

\(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\) is strictly diagonally dominant if and only if
$$ \max_{i\in[n]}\frac{r_{i}(\mathcal{A})}{|a_{ii\cdots i}|}< 1. $$
It is well known that strictly diagonally dominant tensors are nonsingular. An interesting problem arises: for a tensor \(\mathcal {A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\) satisfying
$$\max_{i\in[n]}\frac{r_{i}^{p}(\mathcal{A})}{|a_{ii\cdots i}|}< 1, $$
is \(\mathcal{A}\) nonsingular or not? Certainly, when \(p=1\), \(\mathcal {A}\) is a strictly diagonally dominant tensor, which means that \(\mathcal{A}\) is nonsingular, but when \(p>1\), \(\mathcal{A}\) may be singular as the following simple example shows.

Example 1

Let \(\mathcal{A}=(a_{ijk})\in\mathbb{R}^{[3,2]}\), where
$$a_{111}=a_{222}=-3,\quad \mbox{and the remaining}\quad a_{ijk}=1. $$
Then, since \(\mathcal{A}e^{2}=0\), where \(e=(1,1,1)^{T}\), this implies \(0\in\sigma (\mathcal{A})\). However, for every \(p>1\), we have
$$\max_{i\in[2]}\frac{r_{i}^{p}(\mathcal{A})}{|a_{iii}|}=3^{\frac{1-p}{p}}< 1. $$

Therefore, something needs to be added in order to obtain a nonsingular \(\mathcal{A}\) for a real number \(p\in(1,\infty]\). We provide an answer further, but we first introduce a class of strictly diagonally dominant tensors that involve a parameter p in the interval \([1,\infty]\).

Definition 4

Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\) and \(p\in [1,\infty]\), \(\mathcal{A}\) is called a p-norm strictly diagonally dominant tensor (or, shortly, p-norm SDD tensor) if
$$ \bigl\Vert \delta_{p}(\mathcal{A})\bigr\Vert _{q}< 1, $$
(3)
where
$$\delta_{p}(\mathcal{A}):=(\delta_{1},\delta_{2}, \ldots,\delta_{n})^{T}, \qquad \delta_{i}:= \biggl( \frac{r_{i}^{p}(\mathcal{A})}{|a_{i\cdots i}|} \biggr)^{\frac{1}{m-1}}\quad \mbox{for all } i\in[n], $$
and q is Hölder’s complement of p, that is, \(\frac{1}{p}+\frac{1}{q}=1\).

Remark 2

Definition 4 extends the concept of \(\operatorname{SDD}(p)\) matrix given in [20] to tensors. Clearly, the \(\operatorname{SDD}(p)\) matrix is a 2nd-order p-norm SDD tensor.

Remark 3

Taking \(p=1\), \(\mathcal{A}\) is a 1-norm SDD tensor if and only if
$$\bigl\Vert \delta_{1}(\mathcal{A})\bigr\Vert _{\infty}=\max _{i\in[n]} \biggl(\frac {r_{i}(\mathcal{A})}{|a_{ii\cdots i}|} \biggr)^{\frac{1}{m-1}}< 1, $$
that is,
$$\max_{i\in[n]}\frac{r_{i}(\mathcal{A})}{|a_{ii\cdots i}|}< 1, $$
which is equivalent to the fact that \(\mathcal{A}\) is a strictly diagonally dominant tensor. The other extreme case is \(p=\infty\). \(\mathcal{A}\) is a ∞-norm SDD tensor if and only if
$$ \bigl\Vert \delta_{\infty}(\mathcal{A})\bigr\Vert _{1}=\sum _{i\in[n]} \biggl(\frac{\max_{\substack{i_{2},\ldots,i_{m}\in[n],\\\delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|}{|a_{ii\cdots i}|} \biggr)^{\frac{1}{m-1}}< 1. $$

The p-norm SDD tensors can also be characterized in the following way.

Proposition 1

Let \(\mathcal{A}\in\mathbb{C}^{[m,n]}\) and \(p\in[1,\infty]\). Then \(\mathcal{A}\) is a p-norm SDD tensor if and only if there exists an entrywise positive vector \(x=(x_{1},x_{2},\ldots,x_{n})^{T}\in\mathbb{R}^{n}\) such that \(\|x\|_{q}\leq1\), where q is Hölder’s complement of p such that
$$ x_{i}^{m-1}|a_{i\cdots i}|> r_{i}^{p}(\mathcal{A})\quad \textit{for all } i\in[n]. $$
(4)

Proof

Necessity. Suppose that \(\mathcal{A}\) is a p-norm SDD tensor. It follows from inequality (3) of Definition 4 that there exists a sufficiently small \(\varepsilon>0\) such that, for \(x_{i}:=\delta_{i}+\varepsilon>0\), where \(i\in[n]\), \(\|x\|_{q}\leq1\). Thus, \(x_{i}^{m-1}>\delta_{i}^{m-1}=\frac{r_{i}^{p}(\mathcal{A})}{|a_{i\cdots i}|}\), which implies inequality (4).

Sufficiency. Suppose that there exists an entrywise positive vector \(x>0\) such that \(\|x\|_{q}\leq1\) and inequality (4) holds. By inequality (4) we have
$$x_{i}^{m-1}>\frac{r_{i}^{p}(\mathcal{A})}{|a_{i\cdots i}|}=\delta_{i}^{m-1} \quad \mbox{for all } i\in[n], $$
which implies \(x_{i}>\delta_{i}\) for all \(i\in[n]\), which, together with \(\| x\|_{q}\leq1\), yields
$$\bigl\Vert \delta_{p}(\mathcal{A})\bigr\Vert _{q}< \|x \|_{q}\leq1. $$
Thus, \(\mathcal{A}\) is a p-norm SDD tensor. The proof is completed. □

The following result proves the nonsingular of p-norm SDD tensors.

Theorem 3

Let \(\mathcal{A}\in\mathbb{C}^{[m,n]}\) be a p-norm SDD tensor. Then \(\mathcal{A}\) is nonsingular.

Proof

Suppose that \(\mathcal{A}\) is singular, that is, \(0\in\sigma(\mathcal {A})\). It follows from equality (1) that there exists \(y\in \mathbb{C}^{n}\setminus\{0\}\), such that
$$ \mathcal{A}y^{m-1}=0. $$
(5)
Without the loss of generality, we can assume that \(\|y\|_{q}=1\). Then, equality (5) yields
$$a_{i\cdots i}y_{i}^{m-1}=-\sum _{\substack{i_{2},\ldots,i_{m}\in[n],\\ \delta _{ii_{2}\cdots i_{m}}=0}}a_{ii_{2}\cdots i_{m}}y_{i_{2}}\cdots y_{i_{m}} \quad \mbox{for all } i\in[n], $$
which implies that
$$ |a_{i\cdots i}||y_{i}|^{m-1}=\biggl\vert \sum_{\substack{i_{2},\ldots,i_{m}\in[n],\\ \delta_{ii_{2}\cdots i_{m}}=0}}a_{ii_{2}\cdots i_{m}}y_{i_{2}}\cdots y_{i_{m}}\biggr\vert \quad \mbox{for all } i\in[n]. $$
(6)
Then, applying the Hölder inequality to the right-hand side of equality (6), we obtain
$$\begin{aligned} |a_{i\cdots i}||y_{i}|^{m-1} \leq& \biggl( \sum_{\substack{i_{2},\ldots,i_{m}\in [n], \\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|^{p} \biggr)^{\frac {1}{p}} \biggl(\sum_{\substack{i_{2},\ldots,i_{m}\in[n], \\ \delta _{ii_{2}\cdots i_{m}}=0}}|y_{i_{2}} \cdots y_{i_{m}}|^{q} \biggr)^{\frac {1}{q}} \\ =&r_{i}^{p}(\mathcal{A}) \Biggl[ \Biggl(\sum _{j=1}^{n}|y_{j}|^{q} \Biggr)^{m-1}-|y_{i}|^{(m-1)q} \Biggr]^{\frac{1}{q}} \\ =&r_{i}^{p}(\mathcal{A}) \bigl(\|y\|_{q}^{(m-1)q}-|y_{i}|^{(m-1)q} \bigr)^{\frac{1}{q}} \\ \leq&r_{i}^{p}(\mathcal{A})\|y\|_{q}^{m-1} \\ =&r_{i}^{p}(\mathcal{A}) \quad \mbox{for all } i\in[n]. \end{aligned}$$
(7)
Since \(\mathcal{A}\) is a p-norm SDD tenor, there exists an entrywise positive vector \(x>0\) such that \(\|x\|_{q}\leq1\) and inequality (4) holds. Combining inequality (4) with (7), we obtain
$$ |a_{i\cdots i}||y_{i}|^{m-1}\leq r_{i}^{p}( \mathcal {A})< x_{i}^{m-1}|a_{i\cdots i}|\quad \mbox{for all } i\in[n], $$
which means that
$$ |y_{i}|< x_{i} \quad \mbox{for all } i\in[n]. $$
Thus, \(\|x\|_{q}>\|y\|_{q}=1\), which contradicts \(\|x\|_{q}\leq1\). The proof is completed. □

3 Relationships between p-norm SDD tensors and strong \(\mathcal{H}\)-tensors

The following lemma shows that the strong \(\mathcal{H}\)-tensors play an important role in identifying the positive definiteness of even-order real symmetric tensors.

Lemma 1

[21]

Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in\mathbb{R}^{[m,n]}\) be an even-order real symmetric tensor with \(a_{i\cdots i}>0\) for all \(i\in[n]\). If \(\mathcal{A}\) is a strong \(\mathcal{H}\)-tensor, then \(\mathcal{A}\) is positive definite.

It is known that the strictly diagonally dominant tensors are a subclass of strong \(\mathcal{H}\)-tensors. An interesting problem arises: whether the class of p-norm SDD tensors is a subclass of strong \(\mathcal{H}\)-tensors for an arbitrary \(p\in[1,\infty]\). In this section, we discuss this problem. We first recall the definition of strong \(\mathcal{H}\)-tensors.

Definition 5

[16]

A tenor \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{R}^{[m,n]}\) is called an \(\mathcal {M}\)-tensor if there exist a nonnegative tensor \(\mathcal{B}\) and a positive real number \(\eta\geq\rho\mathcal{(B)}\) such that \(\mathcal{A}=\eta\mathcal{I}\)-\(\mathcal{B}\). If \(\eta>\rho\mathcal{(B)}\), then \(\mathcal{A}\) is called a strong \(\mathcal{M}\)-tensor.

Definition 6

[9]

Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\). We call another tensor \(\mathcal {M(A)}=(m_{i_{1}i_{2}\cdots i_{m}})\) the comparison tensor of \(\mathcal {A}\) if
$$m_{i_{1}i_{2}\cdots i_{m}}=\left \{ \textstyle\begin{array}{l@{\quad}l} +|a_{i_{1}i_{2}\cdots i_{m}}| &\mbox{if } (i_{2},i_{3},\ldots, i_{m})=(i_{1},i_{1},\ldots, i_{1}), \\ -|a_{i_{1}i_{2}\cdots i_{m}}| &\mbox{if } (i_{2},i_{3},\ldots, i_{m})\neq (i_{1},i_{1},\ldots, i_{1}). \end{array}\displaystyle \right . $$

Definition 7

[9]

We call a tensor an \(\mathcal{H}\)-tensor if its comparison tensor is an \(\mathcal{M}\)-tensor. We call it a strong \(\mathcal{H}\)-tensor if its comparison tensor is a strong \(\mathcal{M}\)-tensor.

Note that Li et al. [21] also provided an equivalent definition of strong \(\mathcal{H}\)-tensors; for details, see [21].

In [22], the multiplication of matrices has been extended to tensors. In the following, we state these results for reference.

Definition 8

[22]

Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\) and \(\mathcal {B}=(b_{i_{1}i_{2}\cdots i_{k}})\) be n-dimensional tensors of orders \(m\geq 2\) and \(k\geq1\), respectively. The product \(\mathcal{AB}\) is the following n-dimensional tensor \(\mathcal{C}\) of order \((m-1)(k-1)+1\) with entries
$$ c_{i\alpha_{1}\alpha_{2}\cdots\alpha_{m-1}}=\sum_{i_{2},\ldots,i_{m}\in [n]}a_{ii_{2}\cdots i_{m}}b_{i_{2}\alpha_{1}} \cdots b_{i_{m}\alpha_{m-1}}, $$
where \(i\in[n]\) and \(\alpha_{1},\ldots,\alpha_{m-1}\in\{j_{2}j_{3}\cdots j_{k}:j_{l}\in[n],l=2,3,\ldots, k\}\).

Remark 4

When \(m=2\) and \(\mathcal{A}=(a_{ij})\) is a matrix of dimension n, then \(\mathcal{AB}\) is an mth-order n-dimensional tensor, and we have
$$(\mathcal{AB})_{i_{1}i_{2}\cdots i_{m}}=\sum_{l_{2}\in [n]}a_{i_{1}l_{2}}b_{l_{2}i_{2}\cdots i_{m}}, \quad i_{j}\in[n], j\in[m]. $$
In particular, the product of a diagonal matrix \(X=\operatorname{diag}(x_{1},x_{2},\ldots, x_{n})\) and the tensor \(\mathcal{A}\) is given by
$$(X\mathcal{A})_{i_{1}i_{2}\cdots i_{m}}=x_{i_{1}}a_{i_{1}i_{2}\cdots i_{m}}, \quad i_{j}\in [n], j\in[m]. $$

Remark 5

Given an n-by-n matrix X and two mth order n-dimensional tensors \(\mathcal{A}\), \(\mathcal{B}\), we have the right distributive law for tensors [22], that is,
$$ X\cdot\mathcal{A}+X\cdot\mathcal{B}=X\cdot(\mathcal{A}+\mathcal{B}). $$

Based on this multiplication of tensors, Kannan, Shaked-Monderer, and Berman [23] established a necessary and sufficient condition for a tensor to be a strong \(\mathcal{H}\)-tensor.

Lemma 2

[23]

Let \(\mathcal{A}\in\mathbb{C}^{[m,n]}\). Then \(\mathcal{A}\) is a strong \(\mathcal{H}\)-tensor if and only if \(a_{i\cdots i}\neq0\) for all \(i\in[n]\) and
$$\rho\bigl(\mathcal{I}-D_{\mathcal{M(A)}}^{-1}\mathcal{M(A)}\bigr)< 1, $$
where \(D_{\mathcal{M(A)}}\) is the diagonal matrix with the same diagonal entries as \(\mathcal{M(A)}\).

The following lemma is given by Qi [3].

Lemma 3

[3]

Let \(\mathcal{A}\in\mathbb{C}^{[m,n]}\) and \(\mathcal{B}=a(\mathcal {A}+b\mathcal{I})\), where a and b are two complex numbers. Then μ is an eigenvalue of \(\mathcal{B}\) if and only if \(\mu=a(\lambda+b)\) and λ is an eigenvalue of \(\mathcal{A}\). In this case, they have the same eigenvectors.

Next, we present an equivalence condition for singular tensors.

Lemma 4

Let \(\mathcal{A}\in\mathbb{C}^{[m,n]}\). Then \(\mathcal{A}\) is singular if and only if \(D\mathcal{A}\) is singular, where \(D=\operatorname{diag}(d_{1},d_{2},\ldots,d_{n})\) is a positive diagonal matrix.

Proof

Suppose that \(D\mathcal{A}\) is singular, that is, \(0\in\sigma (D\mathcal{A})\). Then there exists a vector \(x=(x_{1},x_{2},\ldots,x_{n})^{T}\neq0\) such that
$$\sum_{i_{2},\ldots,i_{m}\in[n]}d_{i}a_{ii_{2}\cdots i_{m}}x_{i_{2}} \cdots x_{i_{m}}=0\quad \mbox{for all } i\in[n], $$
which is equivalent to
$$\sum_{i_{2},\ldots,i_{m}\in[n]}a_{ii_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}=0\quad \mbox{for all } i\in[n], $$
which implies \(0\in\sigma(\mathcal{A})\), that is, \(\mathcal{A}\) is singular. The proof is completed. □

By applying Lemmas 2, 3, and 4, we can now reveal the relationship of p-norm SDD tensors and strong \(\mathcal {H}\)-tensors.

Theorem 4

Let \(\mathcal{A}\in\mathbb{C}^{[m,n]}\). If \(\mathcal{A}\) is a p-norm SDD tensor, then \(\mathcal{A}\) is a strong \(\mathcal{H}\)-tensor.

Proof

By Lemma 2 the theorem will be proved if we can show that \(\rho (\mathcal{I}-D_{\mathcal{M(A)}}^{-1}\mathcal{M(A)})< 1\). Assume, on the contrary, that there exists \(\lambda\in\sigma(\mathcal{I}-D_{\mathcal {M(A)}}^{-1}\mathcal{M(A)})\) such that \(|\lambda|\geq1\). Then, by Lemma 3,
$$0\in\sigma\bigl(\lambda\mathcal{I}-\mathcal{I}+D_{\mathcal {M(A)}}^{-1} \mathcal{M(A)}\bigr). $$
According to the right distributive law for tensors, we have
$$ \lambda\mathcal{I}-\mathcal{I}+D_{\mathcal{M(A)}}^{-1}\mathcal {M(A)}=D_{\mathcal{M(A)}}^{-1} \bigl((\lambda-1)D_{\mathcal {M(A)}} \mathcal{I}+\mathcal{M(A)} \bigr), $$
which, together with Lemma 4, yields
$$ 0\in\sigma \bigl((\lambda-1)D_{\mathcal{M(A)}}\mathcal{I}+\mathcal {M(A)} \bigr). $$
(8)
However, there exists an entrywise positive vector \(x>0\) such that \(\| x\|_{q}\leq1\), and inequality (4) holds because \(\mathcal{A}\) is a p-norm SDD tensor. By inequality (4) we have
$$\begin{aligned} x_{i}^{m-1}\bigl\vert (\lambda-1)|a_{i\cdots i}|+|a_{i\cdots i}| \bigr\vert =&x_{i}^{m-1}|\lambda| |a_{i\cdots i}| \\ \geq&x_{i}^{m-1}|a_{i\cdots i}| \\ >&r_{i}^{p}(\mathcal{A}) \\ =&r_{i}^{p} \bigl((\lambda-1)D_{\mathcal{M(A)}}\mathcal{I}+ \mathcal {M(A)} \bigr), \end{aligned}$$
which implies that \((\lambda-1)D_{\mathcal{M(A)}}\mathcal{I}+\mathcal {M(A)}\) is a p-norm SDD tensor. By Theorem 3 we have
$$0\notin\sigma \bigl((\lambda-1)D_{\mathcal{M(A)}}\mathcal{I}+\mathcal {M(A)} \bigr), $$
which contradicts (8). The proof is completed. □

4 Eigenvalue localization

Similarly to matrices, a nonsingular class of tensors can lead to an eigenvalue localization result. In this section, we illustrate this fact with the class of p-norm SDD tensors.

Theorem 5

Let \(\mathcal{A}\in\mathbb{C}^{[m,n]}\) and \(p\in[1,\infty]\). Then
$$ \sigma(\mathcal{A})\subseteq\Phi^{p}(\mathcal{A}). $$
When \(p=1\), \(\Phi^{1}(\mathcal{A})=\Gamma(\mathcal{A})\). When \(p>1\),
$$\Phi^{p}(\mathcal{A})= \biggl\{ z\in\mathbb{C}:\sum _{i\in[n]} \biggl[\frac {r_{i}^{p}(\mathcal{A})}{|z-a_{i\cdots i}|} \biggr]^{\frac {p}{(m-1)(p-1)}}\geq1 \biggr\} . $$

Proof

Clearly, if \(p=1\), \(\sigma(\mathcal{A})\subseteq\Gamma(\mathcal{A})\) can be easily obtained from Theorem 1. If \(p>1\), suppose that there exists \(\lambda\in\sigma(\mathcal{A})\) such that \(\lambda\notin \Phi^{p}(\mathcal{A})\), that is,
$$ \sum_{i\in[n]} \biggl[\frac{r_{i}^{p}(\mathcal{A})}{|\lambda-a_{i\cdots i}|} \biggr]^{\frac{p}{(m-1)(p-1)}}< 1. $$
(9)
Let \(\mathcal{B}:=\lambda\mathcal{I}-\mathcal{A}=(b_{i_{1}i_{2}\cdots i_{m}})\). Since \(\lambda\in\sigma(\mathcal{A})\), this, together with Lemma 3, yields that \(\mathcal{B}\) is surely singular. On the other hand, by the definition of \(\mathcal{B}\) we obtain \(r_{i}^{p}(\mathcal{B})=r_{i}^{p}(\mathcal{A})\) and \(|b_{i\cdots i}|=|\lambda-a_{i\cdots i}|\) for all \(i\in[n]\), so that (9) becomes
$$\sum_{i\in[n]} \biggl[\frac{r_{i}^{p}(\mathcal{B})}{|b_{i\cdots i}|} \biggr]^{\frac{p}{(m-1)(p-1)}}< 1, $$
which implies
$$\bigl\Vert \delta_{p}(\mathcal{B})\bigr\Vert _{q}< 1, $$
where q is Hölder’s complement of p, which means that \(\mathcal{B}\) is a p-norm SDD tensor. By Theorem 3, \(\mathcal{B}\) is nonsingular. This leads to a contradiction. □

Remark 6

When \(m=2\), Theorem 5 reduces to the result of [20].

Remark 7

In particular, taking \(p=\infty\), we have
$$\Phi^{\infty}(\mathcal{A})= \biggl\{ z\in\mathbb{C}:\sum _{i\in[n]} \biggl[\frac{\max_{\substack{i_{2},\ldots,i_{m}\in[n],\\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|}{|z-a_{i\cdots i}|} \biggr]^{\frac{1}{(m-1)}}\geq1 \biggr\} . $$
It follows from Theorem 5 that
$$ \sigma(\mathcal{A})\subseteq\Phi^{p}(\mathcal{A}), $$
but since this conclusion holds for any \(p\in[1,\infty]\), we immediately have the following theorem.

Theorem 6

Let \(\mathcal{A}\in\mathbb{C}^{[m,n]}\). Then
$$ \sigma(\mathcal{A})\subseteq\bigcap_{p\in[1,\infty]} \Phi^{p}(\mathcal{A}), $$
where \(\Phi^{p}(\mathcal{A})\) is defined as in Theorem  5.

The following example shows that \(\Phi^{\infty}(\mathcal{A})\) is tighter than \(\Omega(\mathcal{A})\) in some case.

Example 2

Let \(\mathcal{A}=(a_{ijk})\in\mathbb{R}^{[3,2]}\) and \(\mathcal {B}=(b_{ijk})\in\mathbb{R}^{[3,2]}\) with elements defined as follows:
$$\begin{aligned}& a_{111}=1.96, \qquad a_{222}=16, \quad \mbox{and the remaining}\quad a_{ijk}=1, \\& b_{111}=18,\qquad b_{222}=16,\quad \mbox{and the remaining}\quad b_{ijk}=1, \end{aligned}$$
respectively. The eigenvalue inclusion regions \(\Omega(\mathcal{A})\) (\(\Omega (\mathcal{B})\)), \(\Phi^{\infty}(\mathcal{A})\) (\(\Phi^{\infty}(\mathcal{B})\)) and the exact eigenvalues of \(\mathcal{A}\) (\(\mathcal{B}\)) are drawn in Figure 1A (Figure 1B), where \(\Omega (\mathcal{A})\) (\(\Omega(\mathcal{B})\)), \(\Phi^{\infty}(\mathcal{A})\) (\(\Phi^{\infty}(\mathcal{B})\)) and the exact eigenvalues of \(\mathcal{A}\) (\(\mathcal{B}\)) are respectively denoted by the blue area, the green area, and red asterisks. In addition, by Corollary 7.8 in [10] we have
$$\sigma(\mathcal{A})=\{ 15.6288, 16.7844, 1.7534+0.4687i, 1.7534-0.4687i\} $$
and
$$\sigma (\mathcal{B})=\{20.1811, 17.3673, 15.2258+0.5245i, 15.2258-0.5245i\}. $$
It is easy to see that \(\Phi^{\infty}(\mathcal{A})\subseteq\Omega (\mathcal{A})\), but \(\Omega(\mathcal{B})\subseteq\Phi^{\infty}(\mathcal{B})\).
Figure 1

The eigenvalue inclusion sets of tensors \(\pmb{\mathcal {A}}\) and \(\pmb{\mathcal{B}}\) for Example 2 .

5 Determining the positive (semi)definiteness for an even-order real symmetric tensor

By applying the results obtained in Sections 3 and 4 we give a sufficient condition for the positive (semi)definiteness of an even-order real symmetric tensor.

Theorem 7

Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{R}^{[m,n]}\) be an even-order symmetric tensor with \(a_{i\cdots i}>0\) for all \(i\in[n]\). If \(\mathcal{A}\) is a p-norm SDD tensor, then \(\mathcal{A}\) is positive definite.

Proof

The theorem follows immediately from Lemma 1 and Theorem 4. □

Theorem 8

Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{R}^{[m,n]}\) be an even-order symmetric tensor with \(a_{i\cdots i}>0\) for all \(i\in[n]\), and \(p\in[1,\infty]\). If
$$ \bigl\Vert \delta_{p}(\mathcal{A})\bigr\Vert _{q}\leq1, $$
where q is Hölder’s complement of p. Then \(\mathcal{A}\) is positive semidefinite.

Proof

If \(p=1\), then
$$\bigl\Vert \delta_{1}(\mathcal{A})\bigr\Vert _{\infty}=\max _{i\in[n]} \biggl(\frac {r_{i}(\mathcal{A})}{|a_{ii\cdots i}|} \biggr)^{\frac{1}{m-1}}\leq1, $$
which implies that \(\mathcal{A}\) is diagonally dominant. By Theorem 3 of [24] it follows that \(\mathcal{A}\) is positive semidefinite.
If \(p>1\), then let λ be an H-eigenvalue of \(\mathcal{A}\), and \(\lambda<0\). By Theorem 5 we have \(\lambda\in\Phi^{p}(\mathcal {A})\), which implies that
$$\sum_{i\in[n]} \biggl[\frac{r_{i}^{p}(\mathcal{A})}{|\lambda-a_{i\cdots i}|} \biggr]^{\frac{p}{(m-1)(p-1)}}\geq1. $$
However, it follows from \(a_{ii\cdots i}>0\) for all \(i\in[n]\) that
$$\sum_{i\in[n]} \biggl[\frac{r_{i}^{p}(\mathcal{A})}{|a_{i\cdots i}|} \biggr]^{\frac{p}{(m-1)(p-1)}}> \sum_{i\in[n]} \biggl[ \frac{r_{i}^{p}(\mathcal{A})}{|\lambda-a_{i\cdots i}|} \biggr]^{\frac{p}{(m-1)(p-1)}}\geq1, $$
which implies
$$\bigl\Vert \delta_{p}(\mathcal{A})\bigr\Vert _{q}>1, $$
which contradicts \(\|\delta_{p}(\mathcal{A})\|_{q}\leq1\). This completes the proof. □

Example 3

Let \(\mathcal{A}\in\mathbb{R}^{[4,2]}\) be a symmetric tensor with elements defined as follows:
$$a_{1111}=21,\qquad a_{1222}=a_{2122}=a_{2212}=a_{2221}=-3, \qquad a_{2222}=8, $$
and the remaining \(a_{i_{1}i_{2}i_{3}i_{4}}=0\). By computation,
$$|a_{2222}|=8< 9=r_{2}^{\Delta_{2}}(\mathcal{A})=\sum _{\substack {(i_{2},i_{3},i_{4})\in\Delta_{2},\\ \delta_{2i_{2}i_{3} i_{4}}=0}}|a_{2i_{2}i_{3}i_{4}}|, $$
which means that the statement (I) of Proposition 2.4 in [1] does not hold, and hence we cannot use Proposition 2.4 in [1] to determine the positive definiteness of \(\mathcal{A}\). However, it is easy to verify that \(\mathcal{A}\) is a ∞-norm SDD tensor. By Theorem 7, \(\mathcal{A}\) is positive definite.

6 Conclusions

In this paper, we proposed a new class of nonsingular tensors (p-norm SDD tensors) and proved that the class of p-norm SDD tensors is a subclass of strong \(\mathcal{H}\)-tensors. Furthermore, we presented a new eigenvalue inclusion set, which is tighter than those provided by Li et al.[1] in some case. Based on this set, we presented a checkable sufficient condition for the positive (semi)definiteness of an even-order symmetric tensor.

Declarations

Acknowledgements

This work was supported by the National Natural Science Foundations of China (11361074, 11501141), Natural Science Foundations of Yunnan Province (2013FD002) and Foundation of Guizhou Science and Technology Department ([2015]2073).

Open Access This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.

Authors’ Affiliations

(1)
School of Mathematics and Statistics, Yunnan University

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