Two new lower bounds for the minimum eigenvalue of M-tensors

Abstract

Two new lower bounds for the minimum eigenvalue of an irreducible M-tensor are given. It is proved that the new lower bounds improve the corresponding bounds obtained by He and Huang (J. Inequal. Appl. 2014:114, 2014). Numerical examples are given to verify the theoretical results.

1 Introduction

Let $\mathbb{C} (\mathbb{R})$ be the set of all complex (real) numbers, n be positive integer, $n\geq2$, and $N=\{1,2,\ldots,n\}$. We call $\mathcal{A}=(a_{i_{1}\cdots i_{m}})$ a complex (real) tensor of order m and dimension n, if

$$a_{i_{1}\cdots i_{m}}\in \mathbb{C} (\mathbb{R}), $$

where $i_{j}\in N$ for $j=1,\ldots,m$. Obviously, a vector is a tensor of order 1 and a matrix is a tensor of order 2. We call $\mathcal{A}$ nonnegative if $\mathcal{A}$ is real and each of its entries $a_{i_{1}\cdots i_{m}}\geq0$. Let $\mathbb{R}^{[m, n]}$ denote the space of real-valued tensors with order m and dimension n.

A tensor $\mathcal{A}=(a_{i_{1}\cdots i_{m}})$ of order m and dimension n is called reducible if there exists a nonempty proper index subset $\alpha\subset N$ such that

$$a_{i_{1}i_{2}\cdots i_{m}}=0,\quad \forall i_{1}\in\alpha, \forall i_{2},\ldots,i_{m} \notin\alpha. $$

If $\mathcal{A}$ is not reducible, then we call $\mathcal{A}$ irreducible [2].

For a complex tensor $\mathcal{A}=(a_{i_{1}\cdots i_{m}})$ of order m and dimension n, if there are 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:

$$\mathcal{A}x^{m-1}=\lambda x^{[m-1]}, $$

then λ is called an eigenvalue of $\mathcal{A}$ and x an eigenvector of $\mathcal{A}$ associated with λ, where $\mathcal{A}x^{m-1}$ and $x^{[m-1]}$ are vectors, whose ith component are

$$\bigl(\mathcal{A}x^{m-1} \bigr)_{i}=\sum _{i_{2}, \ldots, i_{m}\in N} a_{ii_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}, $$

and

$$\bigl(x^{[m-1]} \bigr)_{i}=x_{i}^{m-1}, $$

respectively. This definition was introduced by Qi in [3] where he assumed that $\mathcal{A}$ is an order m and dimension n supersymmetric tensor and m is even. Independently, in [4], Lim gave such a definition but restricted x to be a real vector and λ to be a real number. In this case, we call λ an H-eigenvalue of $\mathcal{A}$ and x an H-eigenvector of $\mathcal{A}$ associated with λ.

Moreover, the spectral radius $\rho(\mathcal{A})$ of the tensor $\mathcal{A}$ is defined as

$$\rho(\mathcal{A})=\max\bigl\{ \vert \lambda \vert :\lambda\in\sigma(\mathcal {A})\bigr\} , $$

where $\sigma(\mathcal{A})$ is the spectrum of $\mathcal{A}$, that is, $\sigma(\mathcal{A})=\{\lambda:\lambda\mbox{ is an eigenvalue of } \mathcal{A}\}$; see [2, 5].

The class of M-tensors introduced in [6, 7] is related to nonnegative tensors, which is a generalization of M-matrices [8].

Definition 1

[6, 7]

Let $\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in\mathbb{R}^{[m, n]}$. $\mathcal {A}$ is called:

• (i) a Z-tensor if all of its off-diagonal entries are non-positive, that is, $a_{i_{1}\cdots i_{m}}\leq0$ for $i_{j}\in N$, $j=1,\ldots, m$;

• (ii) an M-tensor if $\mathcal{A}$ is a Z-tensor with the from $\mathcal{A}=c\mathcal{I}-\mathcal{B}$ such that $\mathcal{B}$ is a nonnegative tensor and $c>\rho(\mathcal{B})$, where $\rho(\mathcal{B})$ is the spectral radius of $\mathcal{B}$, and $\mathcal{I}$ is called the unit tensor with its entries

  $$\delta_{i_{1}\cdots i_{m}}= \textstyle\begin{cases} 1, &i_{1}=\cdots =i_{m},\\ 0,&\text{otherwise}. \end{cases} $$

Theorem 1

[1, 6]

Let $\mathcal{A}$ be an M-tensor and denote by $\tau(\mathcal{A})$ the minimal value of the real part of all eigenvalues of $\mathcal{A}$. Then $\tau(\mathcal{A})>0$ is an eigenvalue of $\mathcal{A}$ with a nonnegative eigenvector. If $\mathcal{A}$ is irreducible, then $\tau(\mathcal{A})$ is the unique eigenvalue with a positive eigenvector.

The minimum eigenvalue $\tau(\mathcal{A})$ of M-tensors has many applications and these are studied in [1, 6–13]. Very recently, He and Huang [1] provided some inequalities on $\tau(\mathcal{A})$ for an irreducible M-tensor $\mathcal{A}$ as follows.

Theorem 2

[1]

Let $\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in\mathbb{R}^{[m, n]}$ be an irreducible M-tensor. Then

$$0< \tau(\mathcal{A})\leq\min_{i\in N}a_{ii\cdots i},\quad \textit{and}\quad \tau(\mathcal{A})\geq\min_{i\in N}R_{i}( \mathcal{A}), $$

where $R_{i}(\mathcal{A})=\sum_{i_{2}, \ldots, i_{m}\in N} a_{ii_{2}\cdots i_{m}}$.

Theorem 3

[1]

Let $\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in\mathbb{R}^{[m, n]}$ be an irreducible M-tensor. Then

$$\tau(\mathcal{A})\geq\min_{i,j\in N,\atop j\neq i}\frac{1}{2} \bigl\{ a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}(\mathcal{A})- \bigl[\bigl(a_{i\cdots i}-a_{j\cdots j}+r_{i}^{j}( \mathcal {A})\bigr)^{2}-4a_{ij\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , $$

where

$$r_{i}(\mathcal{A})=\sum_{i_{2}, \ldots, i_{m}\in N, \atop \delta_{ii_{2}\cdots i_{m}}=0}\vert a_{ii_{2}\cdots i_{m}}\vert ,\qquad r_{i}^{j}(\mathcal{A})=r_{i}( \mathcal{A})-\vert a_{ij\cdots j}\vert =\sum_{\delta_{ii_{2}\cdots i_{m}}=0,\atop \delta_{ji_{2}\cdots i_{m}}=0} \vert a_{ii_{2}\cdots i_{m}}\vert . $$

In this paper, we continue to research the problem of estimating the minimum eigenvalue of M-tensors, give two new lower bounds for the minimum eigenvalue, and prove that the two new lower bounds are better than that in Theorem 2 and one of the two bounds is the correction of Theorem 3. Finally, some numerical examples are given to verify the results obtained.

2 Main results

In this section, we give two new lower bounds for the minimum eigenvalue of an irreducible M-tensor.

Theorem 4

Let $\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in\mathbb{R}^{[m, n]}$ be an irreducible M-tensor. Then

$$\tau(\mathcal{A})\geq\min_{i,j\in N,\atop j\neq i} L_{ij}(\mathcal{A}), $$

where

$$\begin{aligned} L_{ij}(\mathcal{A})=\frac{1}{2} \bigl\{ a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A})- \bigl[ \bigl(a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr)^{2}-4a_{ij\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

Proof

Because $\tau(\mathcal{A})$ is an eigenvalue of $\mathcal{A}$, from Theorem 2.1 in [14], there are $i,j\in N$, $j\neq i$, such that

$$\bigl(\bigl\vert \tau(\mathcal{A})-a_{i\cdots i}\bigr\vert -r_{i}^{j}( \mathcal{A})\bigr) \bigl(\bigl\vert \tau(\mathcal{A})-a_{j\cdots j }\bigr\vert \bigr)\leq \vert a_{ij\cdots j}\vert r_{j}(\mathcal{A}). $$

From Theorem 2, we can get

$$\bigl(a_{i\cdots i}-\tau(\mathcal{A})-r_{i}^{j}( \mathcal{A})\bigr) \bigl(a_{j\cdots j}-\tau(\mathcal{A})\bigr) \leq-a_{ij\cdots j}r_{j}(\mathcal{A}), $$

equivalently,

$$\begin{aligned} \tau(\mathcal{A})^{2}- \bigl(a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr)\tau(\mathcal{A})+a_{j\cdots j} \bigl(a_{i\cdots i}-r_{i}^{j}( \mathcal{A}) \bigr)+a_{ij\cdots j}r_{j}(\mathcal{A})\leq0. \end{aligned}$$

(1)

Solving for $\tau(\mathcal{A})$ gives

$$\begin{aligned} \tau(\mathcal{A}) \geq&\frac{1}{2} \bigl\{ a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A})- \bigl[ \bigl(a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr)^{2} \\ &{}-4 \bigl(a_{j\cdots j} \bigl(a_{i\cdots i}-r_{i}^{j}( \mathcal{A}) \bigr)+a_{ij\cdots j}r_{j}(\mathcal{A}) \bigr) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\frac{1}{2} \bigl\{ a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A})- \bigl[ \bigl(a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr)^{2}-4a_{ij\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \geq&\min_{i,j\in N,\atop j\neq i}\frac{1}{2} \bigl\{ a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A})- \bigl[ \bigl(a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr)^{2}-4a_{ij\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

The proof is completed. □

Remark 1

Note here that the bound in Theorem 4 is the correction of the bound in Theorem 3. Because the bound in Theorem 3 is obtained by solving for $\tau(\mathcal{A})$ from inequality (1); for details, see the proof of Theorem 2.2 in [1]. However, solving for $\tau(\mathcal{A})$ by inequality (1) gives the bound in Theorem 4.

In the following, a counterexample is given to show that the result in Theorem 3 is false. Consider the tensor $\mathcal{A}=(a_{ijkl})$ of order 4 and dimension 2 with entries defined as follows:

$$\begin{aligned}& \mathcal{A}(1,1,:,:)=\left( \begin{matrix} 21&-4\\ -3&-3 \end{matrix} \right),\qquad \mathcal{A}(1,2,:,:)= \left( \begin{matrix}-4&-2\\ -2&-1 \end{matrix} \right), \\& \mathcal{A}(2,1,:,:)=\left( \begin{matrix} -3&-3\\ -1&-1 \end{matrix} \right),\qquad \mathcal{A}(2,2,:,:)=\left( \begin{matrix} -3&-1\\ -1&27 \end{matrix} \right). \end{aligned}$$

By Theorem 3, we have $\tau(\mathcal{A})\geq8$. By Theorem 4, we have $\tau(\mathcal{A})\geq2.4700$. In fact, $\tau(\mathcal{A})= 6.9711$.

We now give the following comparison theorem for Theorem 2 and Theorem 4.

Theorem 5

Let $\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in\mathbb{R}^{[m, n]}$ be an irreducible M-tensor. Then

$$\min_{i,j\in N,\atop j\neq i}L_{ij}(\mathcal{A})\geq\min _{i\in N} R_{i}(\mathcal{A}). $$

Proof

(i) For any $i,j\in N$, $j\neq i$, if $R_{i}(\mathcal{A})\leq R_{j}(\mathcal{A})$, i.e., $a_{ii\cdots i}+a_{ij\cdots j}-r_{i}^{j}(\mathcal{A})\leq a_{jj\cdots j}-r_{j}(\mathcal{A})$, then

$$\begin{aligned} 0\leq r_{j}(\mathcal{A})\leq-a_{ij\cdots j}- \bigl(a_{ii\cdots i}-a_{jj\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr). \end{aligned}$$

(2)

Hence,

$$\begin{aligned}& \bigl[a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr]^{2}-4a_{ij\cdots j}r_{j}(\mathcal{A}) \\& \quad \leq \bigl[a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr]^{2}-4a_{ij\cdots j} \bigl[-a_{ij\cdots j}- \bigl(a_{ii\cdots i}-a_{jj\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr) \bigr] \\& \quad = \bigl[a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr]^{2}+4a_{ij\cdots j} \bigl[a_{ii\cdots i}-a_{jj\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr]+4a_{ij\cdots j}^{2} \\& \quad = \bigl[-2a_{ij\cdots j}- \bigl(a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr) \bigr]^{2}. \end{aligned}$$

From (2), we have

$$-2a_{ij\cdots j}-\bigl(a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A})\bigr)\geq r_{j}(\mathcal{A})\geq0. $$

Thus,

$$\begin{aligned} L_{ij}(\mathcal{A}) =&\frac{1}{2} \bigl\{ a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A})- \bigl[ \bigl(a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr)^{2}-4a_{ij\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \geq& \frac{1}{2} \bigl\{ a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A})- \bigl[-2a_{ij\cdots j}- \bigl(a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr) \bigr] \bigr\} \\ =&\frac{1}{2} \bigl\{ 2a_{i\cdots i}+2a_{ij\cdots j}-2r_{i}^{j}( \mathcal{A}) \bigr\} \\ =&R_{i}(\mathcal{A}), \end{aligned}$$

which implies

$$\min_{i,j\in N,\atop j\neq i} L_{ij}(\mathcal{A}) \geq\min _{i\in N} R_{i}(\mathcal{A}). $$

(ii) For any $i,j\in N$, $j\neq i$, if $R_{j}(\mathcal{A})\leq R_{i}(\mathcal{A})$, i.e., $a_{jj\cdots j}-r_{j}(\mathcal{A})\leq a_{ii\cdots i}+a_{ij\cdots j}-r_{i}^{j}(\mathcal{A})$, then

$$0\leq-a_{ij\cdots j}\leq r_{j}(\mathcal{A})-\bigl(a_{jj\cdots j}-a_{ii\cdots i}+r_{i}^{j}( \mathcal{A})\bigr). $$

Similar to the proof of (i), we have $L_{ij}(\mathcal{A})\geq R_{j}(\mathcal{A})$. Hence,

$$\min_{i,j\in N,\atop j\neq i} L_{ij}(\mathcal{A}) \geq\min _{j\in N} R_{j}(\mathcal{A}). $$

The conclusion follows. □

Theorem 5 shows the lower bound in Theorem 4 is better than that in Theorem 2. To obtain a better lower bound, we give the following theorem by breaking N into disjoint subsets S and its complement S̅.

Theorem 6

Let $\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in\mathbb{R}^{[m, n]}$ be an irreducible M-tensor, S be a nonempty proper subset of N, S̅ be the complement of S in N. Then

$$\begin{aligned} \tau(\mathcal{A})\geq\min \Bigl\{ \min_{i\in S}\max _{j\in\overline{S}}L_{ij}(\mathcal{A}), \min_{i\in\overline{S}} \max_{j\in S}L_{ij}(\mathcal{A}) \Bigr\} . \end{aligned}$$

Proof

Let $x=(x_{1},x_{2},\ldots,x_{n})^{T}$ be an associated positive eigenvector of $\mathcal{A}$ corresponding to $\tau(\mathcal{A})$, i.e.,

$$\begin{aligned} \mathcal{A}x^{m-1}=\tau(\mathcal{A}) x^{[m-1]}. \end{aligned}$$

(3)

Let $x_{p}=\max\{x_{i}:i\in S\}$ and $x_{q}=\max\{x_{j}: j\in\overline{S}\}$. We next distinguish two cases to prove.

Case I: If $x_{p}\geq x_{q}$, then $x_{p}=\max\{x_{i}:i\in N\}$. For $p\in S$ and any $j\in\overline{S}$, we have by (3)

$$\tau(\mathcal{A})x_{p}^{m-1}=\sum _{\delta_{pi_{2}\cdots i_{m}}=0,\atop \delta_{ji_{2}\cdots i_{m}}=0}a_{pi_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}+a_{p\cdots p}x_{p}^{m-1}+a_{pj\cdots j}x_{j}^{m-1} $$

and

$$\tau(\mathcal{A})x_{j}^{m-1}=\sum _{\delta_{ji_{2}\cdots i_{m}}=0,\atop \delta_{pi_{2}\cdots i_{m}}=0}a_{ji_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}+a_{j\cdots j}x_{j}^{m-1}+a_{jp\cdots p}x_{p}^{m-1}, $$

equivalently,

$$\begin{aligned} \bigl(\tau(\mathcal{A})-a_{p\cdots p} \bigr)x_{p}^{m-1}-a_{pj\cdots j}x_{j}^{m-1}= \sum_{\delta_{pi_{2}\cdots i_{m}}=0,\atop \delta_{ji_{2} \cdots i_{m}}=0}a_{pi_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}} \end{aligned}$$

(4)

and

$$\begin{aligned} \bigl(\tau(\mathcal{A})-a_{j\cdots j} \bigr)x_{j}^{m-1}-a_{jp\cdots p}x_{p}^{m-1}= \sum_{\delta_{ji_{2}\cdots i_{m}}=0,\atop \delta_{pi_{2} \cdots i_{m}}=0}a_{ji_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}}. \end{aligned}$$

(5)

Solving $x_{p}^{m-1}$ by (4) and (5), we obtain

$$\begin{aligned}& \bigl( \bigl(\tau(\mathcal{A})-a_{p\cdots p} \bigr) \bigl(\tau( \mathcal{A})-a_{j\cdots j} \bigr)-a_{pj\cdots j}a_{jp\cdots p} \bigr)x_{p}^{m-1} \\& \quad = \bigl(\tau(\mathcal{A})-a_{j\cdots j} \bigr)\sum _{\delta_{pi_{2} \cdots i_{m}}=0,\atop \delta_{ji_{2} \cdots i_{m}}=0}a_{pi_{2}\cdots i_{m}}x_{i_{2}}\cdots x_{i_{m}} +a_{pj\cdots j}\sum_{\delta_{ji_{2} \cdots i_{m}}=0,\atop \delta_{pi_{2} \cdots i_{m}}=0}a_{ji_{2}\cdots i_{m}}x_{i_{2}} \cdots x_{i_{m}}. \end{aligned}$$

Since $\tau(\mathcal{A})\leq\min_{i\in N}a_{i\cdots i}$ by Theorem 2 and $\mathcal{A}$ is a Z-tensor, we have

$$\begin{aligned}& \bigl( \bigl(a_{p\cdots p}-\tau(\mathcal{A}) \bigr) \bigl(a_{j\cdots j}- \tau(\mathcal{A}) \bigr)-a_{pj\cdots j}a_{jp\cdots p} \bigr)x_{p}^{m-1} \\& \quad = \bigl(a_{j\cdots j}-\tau(\mathcal{A}) \bigr)\sum _{\delta_{pi_{2} \cdots i_{m}}=0,\atop \delta_{ji_{2} \cdots i_{m}}=0}\vert a_{pi_{2}\cdots i_{m}}\vert x_{i_{2}}\cdots x_{i_{m}} +\vert a_{pj\cdots j}\vert \sum _{\delta_{ji_{2} \cdots i_{m}}=0,\atop \delta_{pi_{2} \cdots i_{m}}=0}\vert a_{ji_{2}\cdots i_{m}}\vert x_{i_{2}}\cdots x_{i_{m}}. \end{aligned}$$

Hence,

$$\begin{aligned}& \bigl( \bigl(a_{p\cdots p}-\tau(\mathcal{A}) \bigr) \bigl(a_{j\cdots j}- \tau(\mathcal{A}) \bigr)-\vert a_{pj\cdots j}\vert \vert a_{jp\cdots p} \vert \bigr)x_{p}^{m-1} \\& \quad \leq \bigl(a_{j\cdots j}-\tau( \mathcal{A}) \bigr)\sum _{\delta_{pi_{2} \cdots i_{m}}=0,\atop \delta_{ji_{2} \cdots i_{m}}=0}\vert a_{pi_{2}\cdots i_{m}} \vert x_{p}^{m-1} +\vert a_{pj\cdots j}\vert \sum_{\delta_{ji_{2} \cdots i_{m}}=0,\atop \delta_{pi_{2} \cdots i_{m}}=0}\vert a_{ji_{2}\cdots i_{m}}\vert x_{p}^{m-1}. \end{aligned}$$

Note that $x_{p}>0$, then

$$\begin{aligned}& \bigl(a_{p\cdots p}-\tau(\mathcal{A}) \bigr) \bigl(a_{j\cdots j}-\tau( \mathcal{A}) \bigr)-\vert a_{pj\cdots j}\vert \vert a_{jp\cdots p} \vert \\& \quad \leq \bigl(a_{j\cdots j}-\tau(\mathcal{A}) \bigr)\sum _{\delta_{pi_{2} \cdots i_{m}}=0,\atop \delta_{ji_{2} \cdots i_{m}}=0}\vert a_{pi_{2}\cdots i_{m}}\vert +\vert a_{pj\cdots j} \vert \sum_{\delta_{ji_{2} \cdots i_{m}}=0,\atop \delta_{pi_{2} \cdots i_{m}}=0}\vert a_{ji_{2}\cdots i_{m}}\vert , \end{aligned}$$

equivalently,

$$\bigl(a_{p\cdots p}-\tau(\mathcal{A})\bigr) \bigl(a_{j\cdots j}-\tau( \mathcal{A})\bigr)-\vert a_{pj\cdots j}\vert \vert a_{jp\cdots p}\vert \leq\bigl(a_{j\cdots j}-\tau(\mathcal{A})\bigr)r_{p}^{j}( \mathcal{A})+\vert a_{pj\cdots j}\vert r_{j}^{p}( \mathcal{A}). $$

This implies

$$\bigl(a_{p\cdots p}-\tau(\mathcal{A})\bigr) \bigl(a_{j\cdots j}-\tau( \mathcal{A})\bigr)-\bigl(a_{j\cdots j}-\tau(\mathcal{A})\bigr)r_{p}^{j}( \mathcal{A})-\vert a_{pj\cdots j}\vert r_{j}(\mathcal{A})\leq0, $$

that is,

$$\tau(\mathcal{A})^{2}-\bigl(a_{p\cdots p}+a_{j\cdots j}-r_{p}^{j}( \mathcal{A})\bigr)\tau(\mathcal{A})+a_{p\cdots p}a_{j\cdots j}-a_{j\cdots j}r_{p}^{j}( \mathcal{A})-\vert a_{pj\cdots j}\vert r_{j}(\mathcal{A})\leq0. $$

Solving for $\tau(\mathcal{A})$ gives

$$\tau(\mathcal{A})\geq\frac{1}{2} \bigl\{ a_{p\cdots p}+a_{j\cdots j}-r_{p}^{j}( \mathcal{A})-\bigl[\bigl(a_{p\cdots p}-a_{j\cdots j}-r_{p}^{j}( \mathcal{A})\bigr)^{2}+4\vert a_{pj\cdots j}\vert r_{j}( \mathcal {A})\bigr]^{\frac{1}{2}} \bigr\} . $$

This must also be true for any $j\in\overline{S}$. Therefore,

$$\begin{aligned} \tau(\mathcal{A}) \geq& \max_{j\in\overline{S}}\frac{1}{2} \bigl\{ a_{p\cdots p}+a_{j\cdots j}-r_{p}^{j}(\mathcal{A})- \bigl[ \bigl(a_{p\cdots p}-a_{j\cdots j}-r_{p}^{j}( \mathcal{A}) \bigr)^{2}+4\vert a_{pj\cdots j}\vert r_{j}( \mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\max_{j\in\overline{S}}\frac{1}{2} \bigl\{ a_{p\cdots p}+a_{j\cdots j}-r_{p}^{j}( \mathcal{A})- \bigl[ \bigl(a_{p\cdots p}-a_{j\cdots j}-r_{p}^{j}( \mathcal{A}) \bigr)^{2}-4a_{pj\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

Since this could be true for some $p\in S$, we finally have

$$\begin{aligned} \tau(\mathcal{A})\geq\min_{i\in S}\max_{j\in\overline{S}} \frac{1}{2} \bigl\{ a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A})- \bigl[ \bigl(a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr)^{2}-4a_{ij\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

Case II: If $x_{q}\geq x_{p}$, then $x_{q}=\max\{x_{i}:i\in N\}$. Similar to the proof of Case I, we can easily prove that

$$\begin{aligned} \tau(\mathcal{A})\geq\min_{i\in\overline{S}}\max_{j\in S} \frac{1}{2} \bigl\{ a_{i\cdots i}+a_{j\cdots j}-r_{i}^{j}( \mathcal{A})- \bigl[ \bigl(a_{i\cdots i}-a_{j\cdots j}-r_{i}^{j}( \mathcal{A}) \bigr)^{2}-4a_{ij\cdots j}r_{j}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$

The conclusion follows from Cases I and II. □

By Theorem 4, Theorem 5, and Theorem 6, the following comparison theorem is obtained easily.

Theorem 7

Let $\mathcal{A}=(a_{i_{1}\cdots i_{m}})\in\mathbb{R}^{[m, n]}$ be an irreducible M-tensor. Then

$$\min\Bigl\{ \min_{i\in S}\max_{j\in\overline{S}}L_{ij}( \mathcal{A}), \min_{i\in\overline{S}}\max_{j\in S}L_{ij}( \mathcal{A}) \Bigr\} \geq\min_{i,j\in N,\atop j\neq i}L_{ij}(\mathcal{A}) \geq\min_{i\in N} R_{i}(\mathcal{A}). $$

Remark 2

• (i) Theorem 7 shows that the bound in Theorem 6 is better than those in Theorem 2 and Theorem 4, respectively.

• (ii) For an M-tensor $\mathcal{A}$ of order m and dimension n, as regards Theorem 4 and Theorem 6 we need to compute $n(n-1)$ and $2\vert S\vert (n-\vert S\vert )$ $L_{ij}(\mathcal {A})$ to obtain their lower bound for $\tau(\mathcal{A})$, respectively, where $\vert S\vert$ is the cardinality of S. When n is very large, one needs more computations to obtain these lower bounds by Theorem 4 and Theorem 6 than Theorem 2.

• (iii) Note that $\vert S\vert < n$. When $n=2$, then $\vert S\vert =1$ and $n(n-1)=2\vert S\vert (n-\vert S\vert )=2$, which implies that

  $$\min\Bigl\{ \min_{i\in S}\max_{j\in\overline{S}}L_{ij}( \mathcal{A}), \min_{i\in\overline{S}}\max_{j\in S}L_{ij}( \mathcal{A}) \Bigr\} =\min_{i,j\in N,\atop j\neq i}L_{ij}(\mathcal{A}). $$

  When $n\geq3$, then $2\vert S\vert (n-\vert S\vert )< n(n-1)$ and

  $$\min\Bigl\{ \min_{i\in S}\max_{j\in\overline{S}}L_{ij}( \mathcal{A}), \min_{i\in\overline{S}}\max_{j\in S}L_{ij}( \mathcal{A}) \Bigr\} \geq\min_{i,j\in N,\atop j\neq i}L_{ij}(\mathcal{A}). $$

3 Numerical examples

In this section, two numerical examples are given to verify the theoretical results.

Example 1

Let $\mathcal{A}=(a_{ijk})\in\mathbb{R}^{[3, 4]}$ be an irreducible M-tensor with elements defined as follows:

$$\begin{aligned}& \mathcal{A}(1,:,:)=\left( \begin{matrix} 37&-2&-1&-4\\ -1&-3&-3&-2\\ -1&-1&-3&-2\\ -2&-3&-3&-3 \end{matrix} \right),\qquad \mathcal{A}(2,:,:)=\left( \begin{matrix} -2&-4&-2&-3\\ -1&39&-2&-1\\ -3&-3&-4&-2\\ -2&-3&-1&-4 \end{matrix} \right), \\& \mathcal{A}(3,:,:)=\left( \begin{matrix} -4&-1&-1&-1\\ -1&-2&-2&-3\\ -1&-1&62&-1\\ -2&-2&-4&-3 \end{matrix} \right),\qquad \mathcal{A}(4,:,:)=\left( \begin{matrix} -2&-4&-3&-1\\ -4&-4&-2&-4\\ -3&-3&-3&-3\\ -3&-3&-4&55 \end{matrix} \right). \end{aligned}$$

Let $S=\{1,2\}$. Obviously $\overline{S}=\{3,4\}$. By Theorem 2, we have

$$\tau(\mathcal{A})\geq2. $$

By Theorem 4, we have

$$\tau(\mathcal{A})\geq 2.0541. $$

By Theorem 6, we have

$$\tau(\mathcal{A})\geq 4. $$

In fact, $\tau(\mathcal{A})=9.9363$. Hence, this example verifies Theorem 7, that is, the bound in Theorem 6 is better than those in Theorem 2 and Theorem 4, respectively.

Example 2

Let $\mathcal{A}=(a_{ijkl})\in\mathbb{R}^{[4, 2]}$ be an irreducible M-tensor with elements defined as follows:

$$a_{1111}=5,\qquad a_{1222}=-1,\qquad a_{2111}=-2,\qquad a_{2222}=4, $$

the other $a_{ijkl}=0$. By Theorem 2, we have

$$\tau(\mathcal{A})\geq2. $$

By Theorem 4, we have

$$\tau(\mathcal{A})\geq 3. $$

In fact, $\tau(\mathcal{A})=3$. Hence, the lower bound in Theorem 4 is tight and sharper than that in Theorem 2.

4 Further work

In this paper, we give an S-type lower bound

$$\Delta^{S}(\mathcal{A})=\min\Bigl\{ \min_{i\in S}\max _{j\in\overline {S}}L_{ij}(\mathcal{A}), \min_{i\in\overline{S}} \max_{j\in S}L_{ij}(\mathcal{A}) \Bigr\} $$

for the minimum eigenvalue of an irreducible M-tensor $\mathcal{A}$ by breaking N into disjoint subsets S and its complement S̅. Then an interesting problem is how to pick S to make $\Delta^{S}(\mathcal{A})$ as big as possible. But it is difficult when the dimension of the tensor $\mathcal{A}$ is large. We will continue to study this problem in the future.