Abstract
Some new criteria for \(\mathcal{H}\)-tensors are obtained. As an application, some sufficient conditions of the positive definiteness for an even-order real symmetric tensor are given. The advantages of the results obtained are illustrated by numerical examples.
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1 Introduction
Let \(\mathbb{C}(\mathbb{R})\) be the complex (real) field and \(N=\{1,2,\ldots,n\}\). We call \(\mathcal{A}=(a_{i_{1}i_{2} \cdots i_{m}})\) a complex (real) order m dimension n tensor, if
where \(i_{j}=1,\ldots,n\) for \(j=1,\ldots,m\) [1, 2]. A tensor \(\mathcal{A}=(a_{i_{1}i_{2} \cdots i_{m}})\) is called symmetric [3], if
where \(\Pi_{m}\) is the permutation group of m indices. Furthermore, an order m dimension n tensor \(\mathcal{I}=(\delta_{i_{1}i_{2}\cdots i_{m}})\) is called the unit tensor [4], if its entries
Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\) be an order m dimension n complex tensor. If there exist a complex number λ and a nonzero complex vector \(x=(x_{1},x_{2},\ldots,x_{n})^{T}\) that are solutions of the following homogeneous polynomial equations:
then we call λ an eigenvalue of \(\mathcal{A}\) and x the eigenvector of \(\mathcal{A}\) associated with λ [5–7], \(\mathcal{A} x^{m-1}\), and \(\lambda x^{[m-1]}\) are vectors, whose ith components are
and
respectively. If the eigenvalue λ and the eigenvector x are real, then λ is called an H-eigenvalue of \(\mathcal{A}\) and x is its corresponding H-eigenvector [1].
Throughout this paper, we will use the following definitions.
Definition 1
[8]
Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\) be a tensor of order m dimension n. \(\mathcal{A}\) is called a diagonally dominant tensor if
If all inequalities in (1) hold, then we call \(\mathcal{A}\) a strictly diagonally dominant tensor.
Definition 2
[9]
Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) be an order m dimension n complex tensor. \(\mathcal{A}\) is called an \(\mathcal{H}\)-tensor if there is a positive vector \(x=(x_{1},x_{2},\ldots,x_{n})^{T} \in{\mathbb {R}}^{n}\) such that
Definition 3
[10]
Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\) be a tensor of order m dimension n, \(X=\operatorname{diag}(x_{1}, x_{2}, \ldots, x_{n})\). Denote
we call \(\mathcal{B}\) the product of the tensor \(\mathcal{A}\) and the matrix X.
Definition 4
[11]
A complex tensor \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) of order m dimension n is called reducible, if there exists a nonempty proper index subset \(I\subset N\) such that
If \(\mathcal{A}\) is not reducible, then we call \(\mathcal{A}\) irreducible.
Definition 5
Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) be an order m dimension n complex tensor, for \(i,j\in N\) (\(i\neq j\)), if there exist indices \(k_{1},k_{2},\ldots, k_{r}\) with
where \(k_{0}=i\), \(k_{r+1}=j\), we call there is a nonzero elements chain from i to j.
For an mth-degree homogeneous polynomial of n variables \(f(x)\) can be denoted
where \(x=(x_{1},x_{2},\ldots,x_{n})\in{\mathbb{R}}^{n}\). The homogeneous polynomial \(f(x)\) in (2) is equivalent to the tensor product of an order m dimensional n symmetric tensor \(\mathcal{A}\) and \(x^{m}\) defined by
where \(x=(x_{1},x_{2},\ldots,x_{n})\in{\mathbb{R}}^{n}\) [1].
The positive definiteness of homogeneous polynomials have applications in automatic control [12, 13], polynomial problems [14], magnetic resonance imaging [15, 16], and spectral hypergraph theory [17, 18]. However, for \(n> 3\) and \(m> 4\), it is a hard problem to identify the positive definiteness of such a multivariate form. For solving this problem, Qi [1] pointed out that \(f(x)\) defined by (3) is positive definite if and only if the real symmetric tensor \(\mathcal{A}\) is positive definite, and Qi provided an eigenvalue method to verify the positive definiteness of \(\mathcal{A}\) when m is even (see Theorem 1).
Theorem 1
[1]
Let \(\mathcal{A}\) be an even-order real symmetric tensor, then \(\mathcal{A}\) is positive definite if and only if all of its H-eigenvalues are positive.
Although from Theorem 1 we can verify the positive definiteness of an even-order symmetric tensor \(\mathcal{A}\) (the positive definiteness of the mth-degree homogeneous polynomial \(f(x)\)) by computing the H-eigenvalues of \(\mathcal{A}\), it is difficult to compute all these H-eigenvalues when m and n are large. Recently, by introducing the definition of \(\mathcal{H}\)-tensor, Li et al. [9] provided a practical sufficient condition for identifying the positive definiteness of an even-order symmetric tensor (see Theorem 2).
Theorem 2
[9]
Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\) be an even-order real symmetric tensor of order m dimension n with \(a_{k\cdots k}>0\) for all \(k \in N\). If \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor, then \(\mathcal{A}\) is positive definite.
Theorem 2 provides a method for identifying the positive definiteness of an even-order symmetric tensor by determining \(\mathcal{H}\)-tensors. Thus the identification of \(\mathcal{H}\)-tensors is useful in checking the positive definiteness of homogeneous polynomials. In this paper, some new criteria for identifying \(\mathcal{H}\)-tensors are presented, which is easy to calculate since it only depends on the entries of tensors. As an application of these criteria, some sufficient conditions of the positive definiteness for an even-order real symmetric tensor are obtained. Numerical examples are also given to verify the corresponding results.
2 Main results
In this section, we give some new criteria for \(\mathcal {H}\)-tensors. First of all, we give some notation and lemmas.
Let S be a nonempty subset of N and let \(N\setminus S\) be the complement of S in N. Given an order m dimension n complex tensor \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\), we denote
It is obvious that if \(N_{1}= \emptyset\), then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor. It is known that, for an \(\mathcal {H}\)-tensor \(\mathcal{A}\), \(N_{2}\neq\emptyset\) [9]. So we always assume that both \(N_{1}\) and \(N_{2}\) are not empty. Otherwise, we assume that \(\mathcal{A}\) satisfies: \(a_{ii\cdots i}\neq0\), \(R_{i}(\mathcal{A}) \neq0\), \(\forall i\in N\).
Lemma 1
[8]
If \(\mathcal{A}\) is a strictly diagonally dominant tensor, then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor.
Lemma 2
[10]
Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) be a complex tensor of order m dimension n. If there exists a positive diagonal matrix X such that \(\mathcal{A}X^{m-1}\) is an \(\mathcal{H}\)-tensor, then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor.
Lemma 3
[9]
Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) be a complex tensor of order m dimension n. If \(\mathcal{A}\) is irreducible,
and strictly inequality holds for at least one i, then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor.
Lemma 4
Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) be an order m dimension n complex tensor. If
-
(i)
\(|a_{ii\cdots i}|\geq R_{i}(\mathcal{A})\), \(\forall i\in N\),
-
(ii)
\(J(\mathcal{A})=\{i\in N: |a_{ii\cdots i}|> R_{i}(\mathcal{A}) \}\neq\emptyset\),
-
(iii)
for any \(i\notin J(\mathcal{A})\), there exists a nonzero elements chain from i to j, such that \(j\in J(\mathcal{A})\),
then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor.
Proof
It is evident that the result holds with \(J(\mathcal{A})=N\). Next, we assume that \(J(\mathcal{A})\neq N\). Suppose \(J(\mathcal{A})=\{k+1, \ldots, n\}\), \(N\setminus J(\mathcal{A})=\{1, \ldots, k\}\), \(1\leq k< n\). By hypothesis,
By the condition (iii), there exist indices \(k_{1},k_{2},\ldots, k_{r}\) such that
where \(k_{0}=k\), \(k_{r+1}=j\), \(j\in J(\mathcal{A})\). Then
Further, without loss of generality, we assume that \(k_{1}, \ldots, k_{r}\notin J(\mathcal{A})\), that is, \(1\leq k_{1}, \ldots, k_{r}< k\). From \(j\in J(\mathcal{A})\), we have \(|a_{jj\cdots j}|> R_{j}(\mathcal{A})\), so there exists \(0<\varepsilon<1\) such that \(\varepsilon |a_{jj\cdots j}|> R_{j}(\mathcal{A})\).
Construct a positive diagonal matrix \(X_{k_{r}}=\operatorname{diag}(x_{1}, \ldots, x_{n})\), where
Let \(\mathcal{A}_{k_{r}}=[a_{i_{1}i_{2}\cdots i_{m}}^{(k_{r})}]=\mathcal{A}X_{k_{r}}^{m-1}\). Then
Obviously, \(\mathcal{A}_{k_{r}}\) is also a diagonally dominant tensor, and \(J(\mathcal{A}_{k_{r}})=J(\mathcal{A})\cup\{k_{r}\}\).
If \(J(\mathcal{A}_{k_{r}})=N\), then \(\mathcal{A}_{k_{r}}\) is strictly diagonally dominant. By Lemma 2, \(\mathcal{A}\) is an \(\mathcal {H}\)-tensor.
If \(N\setminus J(\mathcal{A}_{k_{r}})\neq\emptyset\), then \(\mathcal{A}_{k_{r}}\) also satisfies the conditions of the lemma, that is, for any \(i\in N\setminus J(\mathcal{A}_{k_{r}})\), there exist indices \(l_{1},l_{2},\ldots, l_{s}\), such that
where \(l_{0}=i\), \(l_{s+1}=j\), \(j\in J(\mathcal{A}_{k_{r}})\). Then
Similar to the above argument, for \(\mathcal{A}_{k_{r}}\), there exists a positive diagonal matrix \(X_{l_{s}}\) such that \(\mathcal{A}_{l_{s}}=\mathcal{A}_{k_{r}}X_{l_{s}}^{m-1}\) is diagonally dominant, and \(J(\mathcal{A}_{l_{s}})=J(\mathcal{A}_{k_{r}})\cup \{l_{s}\}\).
If \(J(\mathcal{A}_{l_{s}})=N\), then \(\mathcal{A}_{l_{s}}\) is strictly diagonally dominant. By Lemma 2, \(\mathcal{A}\) is an \(\mathcal {H}\)-tensor.
If \(N\setminus J(\mathcal{A}_{l_{s}})\neq\emptyset\), then \(\mathcal{A}_{l_{s}}\) also satisfies the conditions of the lemma. Similarly as the above argument, for \(\mathcal{A}_{l_{s}}\), there exist at most k positive diagonal matrices \(X_{k_{r}}, X_{l_{s}}, \ldots, X_{p_{q}}\) such that \(\mathcal{B}\) is strictly diagonally dominant, where \(\mathcal{B}=\mathcal{A}(X_{k_{r}}X_{l_{s}}\cdots X_{p_{q}})^{m-1}\). Hence, \(\mathcal{B}\) is an \(\mathcal{H}\)-tensor, and by Lemma 2, \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor. The proof is completed. □
Theorem 3
Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) be an order m dimension n complex tensor. If
then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor.
Proof
Let
If \(\sum_{i_{2}i_{3}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|=0\), we denote \(M_{i}=+\infty\). From inequality (4), we obtain \(M_{i}>0\) (\(i\in N_{1}\)). Hence, there exists a positive number \(\varepsilon> 0\) such that
Let the matrix \(X=\operatorname{diag}(x_{1},x_{2},\ldots,x_{n})\), where
By inequality (6), we have \((\varepsilon+t_{i} )^{\frac{1}{m-1}}<1\) (\(i\in N_{2}\)). As \(\varepsilon\neq+\infty\), so \(x_{i}\neq+\infty\), which implies that X is a diagonal matrix with positive entries. Let \(\mathcal {B}=(b_{i_{1}i_{2}\cdots i_{m}})=\mathcal{A}X^{m-1}\). Next, we will prove that \(\mathcal{B}\) is strictly diagonally dominant.
For all \(i\in N_{1}\), if \(\sum_{i_{2}i_{3}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|=0\), then by inequality (4), we have
If \(\sum_{i_{2}i_{3}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|\neq0\), then by inequalities (5) and (6), we obtain
Now, we consider \(i\in N_{2}\). Since \(|a_{ii\cdots i}|>R_{i}(\mathcal {A})\), we have
and
By inequalities (9), (10), and \(\varepsilon>0\), we get
From inequality (11), for any \(i\in N_{2}\), we obtain
Therefore, from inequalities (7), (8), and (12), we obtain \(|b_{ii\cdots i}|>R_{i}(\mathcal{B})\) for all \(i\in N\), that is, \(\mathcal{B}\) is strictly diagonally dominant. By Lemma 1 and Lemma 2, \(\mathcal{A}\) is an \(\mathcal {H}\)-tensor. The proof is completed. □
Theorem 4
Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) be an order m dimension n complex tensor. If \(\mathcal{A}\) is irreducible and
and a strict inequality holds for at least one \(i \in N_{1}\), then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor.
Proof
Let the matrix \(X=\operatorname{diag}(x_{1},x_{2},\ldots,x_{n})\), where
By the irreducibility of \(\mathcal{A}\), we have \(x_{i}\neq+\infty\), then X is a diagonal matrix with positive diagonal entries. Let \(\mathcal{B}=[b_{i_{1}\cdots i_{m}}]=\mathcal{A}X^{m-1}\).
Adopting the same procedure as in the proof of Theorem 3, we can obtain \(|b_{ii\cdots i}|\geq R_{i}(\mathcal{B})\) (\(\forall i\in N\)), and there exists at least an \(i_{0} \in N_{1}\) such that \(|b_{i_{0}i_{0}\cdots i_{0}}|> R_{i_{0}}(\mathcal{B})\).
On the other hand, since \(\mathcal{A}\) is irreducible and so is \(\mathcal{B}\). Then by Lemma 3, we see that \(\mathcal{B}\) is an \(\mathcal{H}\)-tensor. By Lemma 2, \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor. The proof is completed. □
Let
Theorem 5
Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) be an order m dimension n complex tensor. If
\(J_{1}\cup J_{2} \neq\emptyset\), and for \(\forall i\in(N_{1}\setminus J_{1})\cup(N_{2}\setminus J_{2})\), there exists a nonzero elements chain from i to j such that \(j\in J_{1}\cup J_{2}\), then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor.
Proof
Let the matrix \(X=\operatorname{diag}(x_{1},x_{2},\ldots,x_{n})\), where
Obviously \(x_{i}\neq+\infty\), then X is a diagonal matrix with positive diagonal entries. Let \(\mathcal{B}=[b_{i_{1}\cdots i_{m}}]=\mathcal{A}X^{m-1}\). Similarly as in the proof of Theorem 3, we can obtain \(|b_{ii\cdots i}|\geq R_{i}(\mathcal{B})\) (\(\forall i\in N\)). From \(J_{1}\cup J_{2} \neq\emptyset\), there exists at least an \(i_{0} \in N\) such that \(|b_{i_{0}i_{0}\cdots i_{0}}|> R_{i_{0}}(\mathcal {B})\).
On the other hand, if \(|b_{ii\cdots i}|= R_{i}(\mathcal{B})\), then \(i\in(N_{1}\setminus J_{1})\cup(N_{2}\setminus J_{2})\), by the assumption, we know that there exists a nonzero elements chain from i to j of \(\mathcal{A}\) such that \(j\in J_{1}\cup J_{2}\). Then there exists a nonzero elements chain from i to j of \(\mathcal {B}\) with j satisfying \(|b_{jj\cdots j}|> R_{j}(\mathcal{B})\).
Based on above analysis, we conclude that the tensor \(\mathcal{B}\) satisfies the conditions of Lemma 4, so \(\mathcal{B}\) is an \(\mathcal{H}\)-tensor. By Lemma 2, \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor. The proof is completed. □
Theorem 6
Let \(\mathcal{A}=(a_{i_{1}i_{2}\cdots i_{m}})\) be a complex tensor of order m dimension n. If
and
then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor.
Proof
By inequality (15), for each \(i\in N_{1}\), there exists a positive number \(K_{i}>1\), such that
Let \(K\equiv\max_{i\in N_{1}}\{K_{i}\}\). By inequality (17), we obtain
Since \(|a_{ii\cdots i}|\leq R_{i}(\mathcal{A})\) (\(i\in N_{1}\)) and inequality (15), so
For any \(i\in N_{1}\), denote
From inequalities (18) and (19), we have \(T_{i}>0\). Therefore there exists a positive number \(\varepsilon>0\) such that
Let the matrix \(X=\operatorname{diag}(x_{1},x_{2},\ldots,x_{n})\), where
Mark \(\mathcal{B}=\mathcal {A}X^{m-1}\). Similarly as in the proof of Theorem 3, we can prove that \(\mathcal{B}\) is strictly diagonally dominant. By Lemma 1 and Lemma 2, \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor. The proof is completed. □
There is no inclusion relation between the conditions of Theorem 3 and the conditions of Theorem 6. This can be seen from the following examples.
Example 1
Consider a tensor \(\mathcal{A}=(a_{ijk})\) of order 3 dimension 3 defined as follows:
Obviously,
so \(N_{1}=\{1\}\), \(N_{2}=\{2, 3\}\). By calculation, we have
Since
we know that \(\mathcal{A}\) satisfies the conditions of Theorem 3, then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor. But
so \(\mathcal{A}\) does not satisfy the conditions of Theorem 6.
Example 2
Consider a tensor \(\mathcal{A}=(a_{ijk})\) of order 3 dimension 3 defined as follows:
Obviously,
so \(N_{1}=\{1\}\), \(N_{2}=\{2, 3\}\). By calculation, we have
Since
and
we see that \(\mathcal{A}\) satisfies the conditions of Theorem 6, then \(\mathcal{A}\) is an \(\mathcal{H}\)-tensor. But
so \(\mathcal{A}\) does not satisfy the conditions of Theorem 3.
3 An application
In this section, based on the criteria for \(\mathcal{H}\)-tensors in Section 2, we present some criteria for identifying the positive definiteness of an even-order real symmetric tensor (the positive definiteness of a multivariate form).
From Theorems 2-6, we obtain easily the following result.
Theorem 7
Let \(\mathcal{A}=(a_{i_{1}\cdots i_{m}})\) be an even-order real symmetric tensor of order m dimension n with \(a_{kk\cdots k}>0\) for all \(k \in N\). If \(\mathcal{A}\) satisfies one of the following conditions, then \(\mathcal{A}\) is positive definite:
-
(i)
all the conditions of Theorem 3;
-
(ii)
all the conditions of Theorem 4;
-
(iii)
all the conditions of Theorem 5;
-
(iv)
all the conditions of Theorem 6.
Example 3
Let \(f(x)=\mathcal {A}x^{4}=11x_{1}^{4}+18x_{2}^{4}+18x_{3}^{4}+12x_{4}^{4}+12x_{1}^{2}x_{2}x_{3}-24x_{1}x_{2}x_{3}x_{4}\) be a 4th-degree homogeneous polynomial. We can get an order 4 dimension 4 real symmetric tensor \(\mathcal{A}=(a_{i_{1}i_{2}i_{3} i_{4}})\), where
and other \(a_{i_{1}i_{2}i_{3} i_{4}}=0\). It can be verified that \(\mathcal{A}\) satisfies all the conditions of Theorem 3. Thus, from Theorem 7, we see that \(\mathcal{A}\) is positive definite, that is, \(f(x)\) is positive definite.
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Acknowledgements
The authors are very indebted to the referees for their valuable comments and corrections, which improved the original manuscript of this paper. This work was supported by the National Natural Science Foundation of China (11361074), the Foundation of Science and Technology Department of Guizhou Province ([2015]2073, [2015]7206), the Natural Science Programs of Education Department of Guizhou Province ([2015]420), and the Research Foundation of Guizhou Minzu University (15XRY004).
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Wang, F., Sun, Ds. New criteria for \(\mathcal{H}\)-tensors and an application. J Inequal Appl 2016, 96 (2016). https://doi.org/10.1186/s13660-016-1041-0
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DOI: https://doi.org/10.1186/s13660-016-1041-0