# 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.

## 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 .

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  where he assumed that $$\mathcal{A}$$ is an order m and dimension n supersymmetric tensor and m is even. Independently, in , 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 .

### Definition 1

[6, 7]

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

1. (i)

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

2. (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, 613]. Very recently, He and Huang  provided some inequalities on $$\tau(\mathcal{A})$$ for an irreducible M-tensor $$\mathcal{A}$$ as follows.

### Theorem 2



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



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.

## 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 , 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 . 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 .

### 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, 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

1. (i)

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

2. (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.

3. (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

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

## 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.

## 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 . 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.

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## Acknowledgements

This work is supported by the National Natural Science Foundation of China (Nos. 11361074, 11501141), the Natural Science Programs of Education Department of Guizhou Province (Grant No. 066), and the Foundation of Guizhou Science and Technology Department (Grant No. 2073).

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Correspondence to Jianxing Zhao.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

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

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