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A new S-type upper bound for the largest singular value of nonnegative rectangular tensors
Journal of Inequalities and Applications volume 2017, Article number: 105 (2017)
Abstract
By breaking \(N=\{1,2,\ldots,n\}\) into disjoint subsets S and its complement, a new S-type upper bound for the largest singular value of nonnegative rectangular tensors is given and proved to be better than some existing ones. Numerical examples are given to verify the theoretical results.
1 Introduction
Singular value problems of rectangular tensors have become an important topic in applied mathematics and numerical multilinear algebra, and it has a wide range of practical applications, such as the strong ellipticity condition problem in solid mechanics [1, 2] and the entanglement problem in quantum physics [3, 4].
Let \(\mathbb{R}\) (respectively, \(\mathbb{C}\)) be the real (respectively, complex) field. Assume that \(p,q,m,n\) are positive integers, \(m,n\geq2\), \(l=p+q\), and \(N=\{1,2,\ldots,n\}\). A real \((p,q)\)th order \(m\times n\) dimensional rectangular tensor (or simply a real rectangular tensor) \(\mathcal{A}\) is defined as follows:
A real rectangular tensor \(\mathcal{A}\) is called nonnegative if \(a_{i_{1}\cdots i_{p}j_{1}\cdots j_{q}}\geq0\) for \(i_{k}=1,\ldots,m, k=1,\ldots, p\), and \(j_{v}=1,\ldots, n,v=1,\ldots,q\).
For vectors \(x=(x_{1},\ldots, x_{m})^{\textrm{T}}\), \(y=(y_{1},\ldots,y_{n})^{\textrm{T}}\) and a real number α, let \(x^{[\alpha]}=(x_{1}^{\alpha},x_{2}^{\alpha},\ldots, x_{m}^{\alpha})^{\textrm{T}}\), \(y^{[\alpha]}=(y_{1}^{\alpha},y_{2}^{\alpha},\ldots,y_{n}^{\alpha})^{\textrm{T}}\), \(\mathcal{A}x^{p-1}y^{q}\) be an m dimension real vector whose ith component is
and \(\mathcal{A}x^{p}y^{q-1}\) be an n dimension real vector whose jth component is
If \(\lambda\in\mathbb{C}\), \(x\in\mathbb{C}^{m}\backslash\{0\}\), and \(y\in \mathbb{C}^{n}\backslash\{0\}\) are solutions of
then we say that λ is a singular value of \(\mathcal{A}\), x and y are a left and a right eigenvectors of \(\mathcal{A}\), associated with λ. If \(\lambda\in\mathbb{R}, x\in\mathbb{R}^{m}\), and \(y\in\mathbb{R}^{n}\), then we say that λ is an H-singular value of \(\mathcal{A}\), x and y are a left and a right H-eigenvectors of \(\mathcal{A}\), associated with H-singular value λ [5]. Here,
is called the largest singular value [6].
The definition of singular values for tensors was first introduced in [7]. Note here that when l is even, the definitions in [5] is the same as in [7], and when l is odd, the definition in [5] is slightly different from that in [7], but parallel to the definition of eigenvalues of square matrices [8]; see [5] for details.
Recently, many people focus on bounding the largest singular value for nonnegative rectangular tensors [6, 9, 10]. For convenience, we first give some notation. Given a nonempty proper subset S of N, we denote
and then
This implies that, for a nonnegative rectangular tensor \(\mathcal {A}=(a_{i_{1}\cdots i_{p}j_{1}\cdots j_{q}})\), we have, for \(i,j\in S\),
where
and
In [6], Yang and Yang gave the following bound for the largest singular value of a nonnegative rectangular tensor \(\mathcal{A}\).
Theorem 1
[6], Theorem 4
Let \(\mathcal{A}\) be a \((p,q)\) th order \(m\times n\) dimensional nonnegative rectangular tensor. Then
where
When \(m=n\), He et al. [9] have given an upper bound which is lower than that in Theorem 1.
Theorem 2
[9], Theorem 1.3
Let \(\mathcal{A}\) be a \((p,q)\) th order \(n\times n\) dimensional nonnegative rectangular tensor. Then
where
Similarly, under the condition of \(m=n\), by breaking \(N=\{1,2,\ldots,n\} \) into disjoint subsets S and its complement SÌ„, Zhao and Sang [10] provided an S-type upper bound for the largest singular value of nonnegative rectangular tensors.
Theorem 3
[10], Theorem 2.2
Let \(\mathcal{A}\) be a \((p,q)\) th order \(n\times n\) dimensional nonnegative rectangular tensor, S be a nonempty proper subset of N, SÌ„ be the complement of S in N. Then
where
In this paper, we continue this research, and give a new S-type upper bound for the largest singular value of nonnegative rectangular tensors. It is proved that the new upper bound is better than those in Theorems 1-3.
2 Main results
Theorem 4
Let \(\mathcal{A}\) be a \((p,q)\) th order \(n\times n\) dimensional nonnegative rectangular tensor, S be a nonempty proper subset of N, SÌ„ be the complement of S in N. Then
where
Proof
Because \(\lambda_{0}\) is the largest singular value of \(\mathcal{A}\), from Theorem 2 in [6], there are nonnegative nonzero vectors \(x=(x_{1},x_{2},\ldots,x_{n})^{\mathrm{T}}\) and \(y=(y_{1},y_{2},\ldots,y_{n})^{\mathrm{T}}\), such that
Let
Then at least one of \(x_{t}\) and \(x_{h}\) is nonzero, and at least one of \(y_{f}\) and \(y_{g}\) is nonzero. We next divide into four cases to prove.
Case I: If \(w_{S}=x_{t}, w_{\bar{S}}=x_{h}\), then \(x_{t}\geq y_{t}, x_{h}\geq y_{h}\).
(i) If \(x_{h}\geq x_{t}\), then \(x_{h}=\max\{w_{i}:i\in N\}\). From (3) of Theorem 2.2 in [10], we have
If \(x_{t}=0\), by \(x_{h}>0\), we have \(\lambda_{0} -a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}(\mathcal{A})\leq0\). Then \(\lambda_{0}\leq a_{h\cdots hh\cdots h}+r_{h}^{\overline{\Delta^{S}}}(\mathcal {A})\leq\Psi_{1}^{S}(\mathcal{A})\). Otherwise, \(x_{t}>0\). From (1), we have
i.e.,
If \(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}(\mathcal{A})\leq 0\), then \(\lambda_{0}\leq a_{t\cdots tt\cdots t}+r_{t}^{\Delta^{S}}(\mathcal{A})\leq \Psi_{1}^{S}(\mathcal{A})\). If \(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}(\mathcal{A})> 0\), multiplying (3) with (4) and noting that \(x_{t}^{l-1}x_{h}^{l-1}>0\), we have
Solving \(\lambda_{0}\) in (5) gives
(ii) If \(x_{t}\geq x_{h}\), similar to the proof of (i), we have
and
Case II: Assume that \(w_{S}=y_{f}, w_{\bar{S}}=y_{g}\). If \(y_{g}\geq y_{f}\), similar to the proof of (i), we have
and
If \(y_{f}\geq y_{g}\), similarly, we have
and
Case III: Assume that \(w_{S}=x_{t}, w_{\bar{S}}=y_{g}\). If \(y_{g}\geq x_{t}\), similar to the proof of (i), we have
and
If \(x_{t}\geq y_{g}\), similarly, we have
and
Case IV: Assume that \(w_{S}=y_{f}, w_{\bar{S}}=x_{h}\). If \(x_{h}\geq y_{f}\), similar to the proof of (i), we have
and
If \(y_{f}\geq x_{h}\), similarly, we have
and
The conclusion follows from Cases I, II, III and IV. □
We next give the following comparison theorem for these upper bounds in Theorems 1-4.
Theorem 5
Let \(\mathcal{A}\) be a \((p,q)\) th order \(n\times n\) dimensional nonnegative rectangular tensor, SÂ be a nonempty proper subset of N, SÌ„ be the complement of S in N. Then
Proof
I. By Remark 2.2 in [9], \(\Phi(\mathcal{A})\leq\max_{i,j\in N}\{R_{i}(\mathcal{A}),C_{j}(\mathcal{A})\}\) holds.
II. Next, we prove \(U^{S}(\mathcal{A})\leq\Phi(\mathcal{A})\). Here, we only prove \(U_{1}^{S}(\mathcal{A})\leq\Phi(\mathcal{A})\). Similarly, we can prove \(U_{1}^{\bar{S}}(\mathcal{A}),U_{2}^{S}(\mathcal{A}), U_{2}^{\bar{S}}(\mathcal{A})\leq\Phi(\mathcal{A})\), respectively.
(i) Suppose that
From the proof of Theorem 2.2 in [10], we can see that the bound \(U_{1}^{S}(\mathcal{A})\) is obtained by solving \(\lambda_{0}\) from
From the proof of Theorem 1.3 in [9], we can see that the bound
is obtained by solving \(\lambda_{0}\) from
Taking \(i\in S\), \(j\in\bar{S}\) in (7), by the proof of Theorem 6 in [11], we know that if \(\lambda_{0}\) satisfies (6), then \(\lambda _{0}\) satisfies (7), which implies that
Obviously, \(U_{1}^{S}(\mathcal{A})\leq\Phi(\mathcal{A})\).
(ii) Suppose that
Similar to the proof of (i), we can obtain \(U_{1}^{S}(\mathcal{A})\leq\Phi _{3}(\mathcal{A})\leq\Phi(\mathcal{A})\).
III. Finally, we prove that \(\Psi^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\). Here, we only prove \(\Psi_{1}^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\). Similarly, we can prove \(\Psi_{1}^{\bar{S}}(\mathcal{A}),\Psi_{2}^{S}(\mathcal {A}),\Psi_{2}^{\bar{S}}(\mathcal{A}),\Psi_{3}^{S}(\mathcal{A}),\Psi_{3}^{\bar {S}}(\mathcal{A}),\Psi_{4}^{S}(\mathcal{A}),\Psi_{4}^{\bar{S}}(\mathcal {A})\leq U^{S}(\mathcal{A})\), respectively.
Let \(i\in S\) and \(j\in\bar{S}\). From the proof of Theorem 4, we can see that the bound \(\Psi_{1}^{S}(\mathcal{A})\) is obtained by solving \(\lambda_{0}\) from
(i) Suppose that \(r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})r_{j}^{\Delta ^{S}}(\mathcal{A})=0\). If \(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}(\mathcal{A})>0\), i.e., \(\lambda_{0}>a_{i\cdots ii\cdots i}+r_{i}^{\Delta^{S}}(\mathcal{A})\), then \(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\leq0\), and for any \(i\in S\),
That is to say, if \(\lambda_{0}\) satisfies (8), then \(\lambda _{0}\) satisfies (6), which implies that \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\).
If \(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}(\mathcal{A})\leq0\), then \(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta ^{S}}}(\mathcal{A})\geq0\), i.e., \(\lambda_{0} \geq a_{j\cdots jj\cdots j}+r_{j}^{\overline {\Delta^{S}}}(\mathcal{A})\). From (3), we can obtain \(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\leq r_{j}^{\Delta^{S}}(\mathcal{A})\), i.e.,
By \(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}(\mathcal{A})\leq 0\leq r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})\), i.e., \(\lambda_{0}-a_{i\cdots ii\cdots i}\leq r_{i}(\mathcal{A})\), we have
Multiplying (9) with (10), we can obtain
which implies that if \(\lambda_{0}\) satisfies (8), then \(\lambda_{0}\) satisfies (6), consequently, \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{\bar{S}}(\mathcal{A})\leq U^{S}(\mathcal{A})\).
(ii) Suppose that \(r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})r_{j}^{\Delta ^{S}}(\mathcal{A})>0\). Then dividing (8) by \(r_{i}^{\overline{\Delta^{S}}}(\mathcal {A})r_{j}^{\Delta^{S}}(\mathcal{A})\), we have
Furthermore, if \(\frac{\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A})}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})}\geq1\), then by Lemma 2.3 in [12] and (12), we have
Thus, (6) holds, which implies that if \(\lambda_{0}\) satisfies (8), then \(\lambda_{0}\) satisfies (6), consequently, \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{S}(\mathcal{A})\). And if \(\frac{\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A})}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})}\leq1\), then (10) holds, which leads to (11) from (9). This implies that if \(\lambda_{0}\) satisfies (8), then \(\lambda_{0}\) satisfies (6), consequently, \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{\bar{S}}(\mathcal{A})\leq U^{S}(\mathcal{A})\). The conclusion follows immediately from what we have proved. □
3 Numerical examples
Example 1
Let \(\mathcal{A}=(a_{ijkl})\) be a \((2,2)\)th order \(3\times3\) dimensional nonnegative rectangular tensor with entries defined as follows:
By Theorem 1, we have
By Theorem 2, we have
Taking \(S=\{1,2\},\bar{S}=\{3\}\), by Theorem 3, we have
by Theorem 4, we have
In fact, \(\lambda_{0}=29.8830\). This example shows that the upper bound in Theorem 4 is smaller than those in Theorems 1-3.
Example 2
Let \(\mathcal{A}=(a_{ijkl})\) be a \((2,2)\)th order \(2\times2\) dimensional nonnegative rectangular tensor with entries defined as follows:
the other \(a_{ijkl}=0\). By Theorem 4, we have
In fact, \(\lambda_{0}=3\). This example shows that the upper bound in Theorem 4 is sharp.
4 Conclusions
In this paper, a new S-type upper bound \(\Psi^{S}(\mathcal{A})\) of the largest singular value for a nonnegative rectangular tensor \(\mathcal {A}\) with \(m=n\) is obtained by breaking N into disjoint subsets S and its complement. It is proved that the bound \(\Psi^{S}(\mathcal{A})\) is better than those in [6, 9, 10].
Note here that when \(n=2\), \(\Phi(\mathcal{A})=U^{S}(\mathcal{A})=\Psi ^{S}(\mathcal{A})\), and when \(n\geq3\), \(\Phi(\mathcal{A})\geq U^{S}(\mathcal{A})\geq\Psi ^{S}(\mathcal{A})\) always holds. How to pick S to make \(\Psi^{S}(\mathcal{A})\) as small as possible is an interesting problem, but difficult when n is large. We will research this problem in the future.
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Acknowledgements
The authors are very indebted to the reviewers for their valuable comments and corrections, which improved the original manuscript of this paper. This work is supported by Natural Science Programs of Education Department of Guizhou Province (Grant No. [2016]066), Foundation of Guizhou Science and Technology Department (Grant No. [2015]2073) and National Natural Science Foundation of China (No. 11501141).
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Zhao, J., Sang, C. A new S-type upper bound for the largest singular value of nonnegative rectangular tensors. J Inequal Appl 2017, 105 (2017). https://doi.org/10.1186/s13660-017-1382-3
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DOI: https://doi.org/10.1186/s13660-017-1382-3