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p-Norm SDD tensors and eigenvalue localization
Journal of Inequalities and Applications volume 2016, Article number: 178 (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.
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
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
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
For an n-dimensional vector \(x=(x_{1},x_{2},\ldots,x_{n})^{T}\), real or complex, we define the n-dimensional vector
and the n-dimensional vector
The following definition related to eigenvalues of tensors was first introduced and studied by Qi [3] and Lim [5].
Definition 1
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
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
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, 6–16]. 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 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
where \(\sigma(\mathcal{A})\) is the set of all the eigenvalues of \(\mathcal{A}\), and
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
and
Theorem 2
[1]
Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\). Then
where
and
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
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
Definition 3
[16]
A tensor \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\in\mathbb{C}^{[m,n]}\) is diagonally dominant if
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
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
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
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
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
where
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
that is,
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
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
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
which implies \(x_{i}>\delta_{i}\) for all \(i\in[n]\), which, together with \(\| x\|_{q}\leq1\), yields
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
Without the loss of generality, we can assume that \(\|y\|_{q}=1\). Then, equality (5) yields
which implies that
Then, applying the Hölder inequality to the right-hand side of equality (6), we obtain
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
which means that
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
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
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
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
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,
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
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
which is equivalent to
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,
According to the right distributive law for tensors, we have
which, together with Lemma 4, yields
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
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
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
When \(p=1\), \(\Phi^{1}(\mathcal{A})=\Gamma(\mathcal{A})\). When \(p>1\),
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,
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
which implies
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
It follows from Theorem 5 that
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
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:
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
and
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 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
where q is Hölder’s complement of p. Then \(\mathcal{A}\) is positive semidefinite.
Proof
If \(p=1\), then
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
However, it follows from \(a_{ii\cdots i}>0\) for all \(i\in[n]\) that
which implies
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:
and the remaining \(a_{i_{1}i_{2}i_{3}i_{4}}=0\). By computation,
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.
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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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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
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DOI: https://doi.org/10.1186/s13660-016-1119-8