# p-Norm SDD tensors and eigenvalue localization

## 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.

## 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 ofÂ $$\mathcal{A}$$. 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, 6â€“16]. 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 GerÅ¡gorin 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 TheoremÂ 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 TheoremÂ 2 is tighter than $$\Theta(\mathcal{A})$$ in [19] and $$\mathcal{K}(\mathcal{A})$$ in [17]; for details, see TheoremÂ 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 SectionÂ 2. In SectionÂ 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 SectionÂ 4, and numerical results show that the new set is tighter than $$\Omega(\mathcal{A})$$ in TheoremÂ 2 in some case. Finally, in SectionÂ 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 HÃ¶lderâ€™s complement of p, that is, $$\frac{1}{p}+\frac{1}{q}=1$$.

### Remark 2

DefinitionÂ 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 âˆž-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 HÃ¶lderâ€™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 DefinitionÂ 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.â€ƒâ–¡

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 HÃ¶lder 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.â€ƒâ–¡

## 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 Î¼ is an eigenvalue of $$\mathcal{B}$$ if and only if $$\mu=a(\lambda+b)$$ and Î» 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.â€ƒâ–¡

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 LemmaÂ 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 LemmaÂ 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 LemmaÂ 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 TheoremÂ 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.â€ƒâ–¡

## 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 TheoremÂ 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 LemmaÂ 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 HÃ¶lderâ€™s complement of p, which means that $$\mathcal{B}$$ is a p-norm SDD tensor. By TheoremÂ 3, $$\mathcal{B}$$ is nonsingular. This leads to a contradiction.â€ƒâ–¡

### Remark 6

When $$m=2$$, TheoremÂ 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 TheoremÂ 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 TheoremÂ  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 FigureÂ 1A (FigureÂ 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 CorollaryÂ 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 $$\Omega(\mathcal{B})\subseteq\Phi^{\infty}(\mathcal{B})$$.

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

By applying the results obtained in SectionsÂ 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 LemmaÂ 1 and TheoremÂ 4.â€ƒâ–¡

### 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 HÃ¶lderâ€™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 TheoremÂ 3 of [24] it follows that $$\mathcal{A}$$ is positive semidefinite.

If $$p>1$$, then let Î» be an H-eigenvalue of $$\mathcal{A}$$, and $$\lambda<0$$. By TheoremÂ 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.â€ƒâ–¡

### 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 PropositionÂ 2.4 in [1] does not hold, and hence we cannot use PropositionÂ 2.4 in [1] to determine the positive definiteness of $$\mathcal{A}$$. However, it is easy to verify that $$\mathcal{A}$$ is a âˆž-norm SDD tensor. By TheoremÂ 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.

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## 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).

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### Corresponding author

Correspondence to Yaotang Li.

### Competing interests

The authors declare that they have no competing interests.

### Authorsâ€™ contributions

Both authors contributed equally to this work. Both authors read and approved the final manuscript.

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Liu, Q., Li, Y. p-Norm SDD tensors and eigenvalue localization. J Inequal Appl 2016, 178 (2016). https://doi.org/10.1186/s13660-016-1119-8