New criteria for $\mathcal{H}$-tensors and an application

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.

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

$$a_{i_{1}i_{2}\cdots i_{m}}\in \mathbb{C}(\mathbb{R}), $$

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

$$a_{i_{1}i_{2}\cdots i_{m}}=a_{\pi(i_{1}i_{2}\cdots i_{m})},\quad \forall\pi\in \Pi_{m}, $$

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

$$\delta_{i_{1}i_{2}\cdots i_{m}}=\left \{ \textstyle\begin{array}{l@{\quad}l} 1, &\text{if } i_{1}=\cdots=i_{m}, \\ 0,&\text{otherwise}. \end{array}\displaystyle \right . $$

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:

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

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

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

and

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

respectively. 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

$$ |a_{ii\cdots i}|\geq\sum_{\substack{i_{2},\ldots ,i_{m}\in N\\ \delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2}\cdots i_{m}}|,\quad \forall i\in N. $$

(1)

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

$$\vert {a_{ii\cdots i}} \vert x_{i}^{m-1}>\sum _{\substack{ i_{2},\ldots,i_{m}\in N\\ \delta_{ii_{2}\cdots i_{m}}=0}}\vert { a_{ii_{2}\cdots i_{m}}}\vert x_{i_{2}}\cdots x_{i_{m}},\quad i=1,2,\ldots,n. $$

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

$$ \mathcal{B}=(b_{i_{1}\cdots i_{m}})=\mathcal{A}X^{m-1},\qquad b_{i_{1}i_{2}\cdots i_{m}}=a_{i_{1}i_{2}\cdots i_{m}}x_{i_{2}}x_{i_{3}}\cdots x_{i_{m}}, \quad i_{j}\in N, j\in N, $$

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

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

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

$$\begin{aligned} \sum_{\substack{i_{2},\ldots,i_{m}\in N \\ \delta_{k_{s}i_{2}\cdots i_{m}}=0 \\ k_{s+1}\in{\{i_{2},\ldots, i_{m}\}} }} \vert { a_{k_{s}i_{2}\cdots i_{m}}}\vert \neq0, \quad s=0,1,\ldots, r, \end{aligned}$$

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

$$ f(x)=\sum_{i_{1},i_{2},\ldots,i_{m}\in N} a_{i_{1}i_{2}\cdots i_{m}}x_{i_{1}}x_{i_{2}} \cdots x_{i_{m}}, $$

(2)

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

$$ f(x)\equiv\mathcal{A}x^{m}=\sum _{i_{1},i_{2},\ldots,i_{m}\in N} a_{i_{1}i_{2}\cdots i_{m}}x_{i_{1}}x_{i_{2}}\cdots x_{i_{m}}, $$

(3)

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

$$\begin{aligned}& R_{i}(\mathcal{A}) = \sum_{\substack{i_{2},\ldots,i_{m}\in N\\ \delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2}\cdots i_{m}}|=\sum_{i_{2},\ldots,i_{m}\in N} |a_{ii_{2}\cdots i_{m}}|-|a_{ii\cdots i}|, \\& N_{1} = \bigl\{ i\in N: 0< |a_{ii\cdots i}|\leq R_{i}( \mathcal{A})\bigr\} ,\qquad N_{2}=\bigl\{ i\in N: |a_{ii\cdots i}|> R_{i}(\mathcal {A})\bigr\} , \\& s_{i} = \frac{|a_{ii\cdots i}|}{R_{i}(\mathcal{A})},\qquad t_{i}=\frac{R_{i}(\mathcal{A})}{|a_{ii\cdots i}|}, \qquad r=\max \Bigl\{ \max_{i\in N_{1}} s_{i}, \max _{i\in N_{2}} t_{i} \Bigr\} , \\& S^{m-1} = \{i_{2}i_{3}\cdots i_{m}: i_{j} \in S, j=2, 3, \ldots, m \}, \\& N^{m-1}\setminus S^{m-1} = \bigl\{ i_{2}i_{3} \cdots i_{m}: i_{2}i_{3}\cdots i_{m}\in N^{m-1} \text{ and } i_{2}i_{3}\cdots i_{m} \notin S^{m-1} \bigr\} . \end{aligned}$$

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,

$$|a_{i\cdots i}|\geq R_{i}(\mathcal{A}),\quad \forall i \in N, $$

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,

$$ |a_{kk\cdots k}|= R_{k}(\mathcal{A}). $$

By the condition (iii), there exist indices $k_{1},k_{2},\ldots, k_{r}$ such that

$$ \sum_{\substack{i_{2},\ldots,i_{m}\in N \\ \delta_{k_{s}i_{2}\cdots i_{m}}=0 \\ k_{s+1}\in{\{i_{2},\ldots, i_{m}\}} }} \vert { a_{k_{s}i_{2}\cdots i_{m}}}\vert \neq0, \quad s=0,1,\ldots, r, $$

where $k_{0}=k$, $k_{r+1}=j$, $j\in J(\mathcal{A})$. Then

$$ \sum_{\substack{i_{2},\ldots,i_{m}\in N \\ \delta_{k_{r}i_{2}\cdots i_{m}}=0 \\ j\in{\{i_{2},\ldots, i_{m}\}} }} \vert { a_{k_{r}i_{2}\cdots i_{m}}}\vert \neq0. $$

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

$$ x_{i}=\left \{ \textstyle\begin{array}{l@{\quad}l} \varepsilon^{\frac{1}{m-1}}, & i=j, \\ 1, & i\neq j. \end{array}\displaystyle \right . $$

Let $\mathcal{A}_{k_{r}}=[a_{i_{1}i_{2}\cdots i_{m}}^{(k_{r})}]=\mathcal{A}X_{k_{r}}^{m-1}$. Then

$$\begin{aligned}& \bigl\vert a_{ii\cdots i}^{(k_{r})}\bigr\vert = \vert a_{ii\cdots i}\vert = R_{i}(\mathcal{A})\geq R_{i}( \mathcal{A}_{k_{r}}),\quad 1\leq i\leq k, i\neq k_{r}, \\& \bigl\vert a_{k_{r}k_{r}\cdots k_{r}}^{(k_{r})}\bigr\vert = \vert a_{k_{r}k_{r}\cdots k_{r}}\vert = R_{k_{r}}(\mathcal{A})> R_{k_{r}}( \mathcal{A}_{k_{r}}), \\& \bigl\vert a_{ii\cdots i}^{(k_{r})}\bigr\vert = \vert a_{ii\cdots i}\vert > R_{i}(\mathcal{A})\geq R_{k_{r}}( \mathcal{A}_{k_{r}}),\quad i\in J(\mathcal{A}), i\neq j, \\& \bigl\vert a_{jj\cdots j}^{(k_{r})}\bigr\vert = \varepsilon \vert a_{jj\cdots j}\vert > R_{j}(\mathcal{A})\geq R_{k_{r}}( \mathcal{A}_{k_{r}}). \end{aligned}$$

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

$$ \sum_{\substack{i_{2},\ldots,i_{m}\in N, \\ \delta_{l_{t}i_{2}\cdots i_{m}}=0, \\ l_{t+1}\in{\{i_{2},\ldots, i_{m}\}} }} \vert { a_{l_{t}i_{2}\cdots i_{m}}}\vert \neq0, \quad t=0,1,\ldots, s, $$

where $l_{0}=i$, $l_{s+1}=j$, $j\in J(\mathcal{A}_{k_{r}})$. Then

$$ \sum_{\substack{i_{2},\ldots,i_{m}\in N, \\ \delta_{l_{s}i_{2}\cdots i_{m}}=0, \\ j\in{\{i_{2},\ldots, i_{m}\}} }} \vert { a_{l_{s}i_{2}\cdots i_{m}}}\vert \neq0. $$

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

$$ |a_{ii\cdots i}| s_{i}> r \sum _{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}\\ \delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2} \cdots i_{m}}|+ \sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1} } \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2} \cdots i_{m}}|,\quad \forall i \in N_{1}, $$

(4)

then $\mathcal{A}$ is an $\mathcal{H}$-tensor.

Proof

Let

$$\begin{aligned}& M_{i}=\frac{ |a_{ii\cdots i}| s_{i}-r \sum_{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}\\ \delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2} \cdots i_{m}}|- \sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1} }\max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2} \cdots i_{m}}| }{ \sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1} } |a_{ii_{2} \cdots i_{m}}|}, \\& \quad \forall i\in N_{1}. \end{aligned}$$

(5)

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

$$ 0< \varepsilon< \min \Bigl\{ \min_{i\in N_{1}}M_{i}, 1-\max_{i\in N_{2}} t_{i} \Bigr\} . $$

(6)

Let the matrix $X=\operatorname{diag}(x_{1},x_{2},\ldots,x_{n})$, where

$$x_{i}=\left \{ \textstyle\begin{array}{l@{\quad}l} (s_{i} )^{\frac{1}{m-1}}, & i\in N_{1}, \\ (\varepsilon+t_{i} )^{\frac{1}{m-1}}, & i\in N_{2}. \end{array}\displaystyle \right . $$

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

$$\begin{aligned} R_{i}(\mathcal{B}) =&\sum_{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1} \\\delta_{ii_{2}\cdots i_{m}}=0}}|b_{ii_{2}\cdots i_{m}}|+ \sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1}}|b_{ii_{2}\cdots i_{m}}| \\ =&\sum_{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}\\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|x_{i_{2}} \cdots x_{i_{m}}+\sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|x_{i_{2}} \cdots x_{i_{m}} \\ \leq& r \sum_{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}\\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}| < |a_{ii\cdots i}| s_{i}=|b_{ii\cdots i}|. \end{aligned}$$

(7)

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

$$\begin{aligned} R_{i}(\mathcal{B}) =&\sum _{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1} \\\delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|x_{i_{2}}\cdots x_{i_{m}}+\sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|x_{i_{2}} \cdots x_{i_{m}} \\ =&\sum_{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1} \\\delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|x_{i_{2}} \cdots x_{i_{m}} \\ &{}+\sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}| (\varepsilon+t_{i_{2}} )^{\frac{1}{m-1}}\cdots (\varepsilon+t_{i_{m}} )^{\frac{1}{m-1}} \\ \leq&r \sum_{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}\\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|+\sum _{i_{2}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|\Bigl(\varepsilon+ \max _{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \}\Bigr) \\ =&r \sum_{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}\\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|+ \sum _{i_{2}\cdots i_{m}\in N_{2}^{m-1}}\max_{j\in \{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2}\cdots i_{m}}| \\ &{}+\varepsilon\sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}| \\ < & |a_{ii\cdots i}| s_{i} =|b_{ii\cdots i}|. \end{aligned}$$

(8)

Now, we consider $i\in N_{2}$. Since $|a_{ii\cdots i}|>R_{i}(\mathcal {A})$, we have

$$ |a_{ii\cdots i}|-\sum_{\substack{i_{2}i_{3}\cdots i_{m}\in N_{2}^{m-1} \\\delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|>0 $$

(9)

and

$$ r \sum_{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|+ \sum _{\substack{i_{2}\cdots i_{m}\in N_{2}^{m-1}\\ \delta_{ii_{2}\cdots i_{m}}=0}} \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2}\cdots i_{m}}| - R_{i}(\mathcal{A})\leq0. $$

(10)

By inequalities (9), (10), and $\varepsilon>0$, we get

$$ \varepsilon>\frac{r \sum_{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|+ \sum_{\substack{i_{2}\cdots i_{m}\in N_{2}^{m-1} \\ \delta_{ii_{2}\cdots i_{m}}=0}} \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2}\cdots i_{m}}| - R_{i}(\mathcal{A})}{|a_{ii\cdots i}|-\sum_{\substack{i_{2}\cdots i_{m}\in N_{2}^{m-1}\\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|}. $$

(11)

From inequality (11), for any $i\in N_{2}$, we obtain

$$\begin{aligned} |b_{ii\cdots i}|-R_{i}(\mathcal{B}) =&|a_{ii\cdots i}| (\varepsilon+t_{i} )-\sum _{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|x_{i_{2}}\cdots x_{i_{m}} \\ &{}-\sum_{\substack{i_{2}\cdots i_{m}\in N_{2}^{m-1} \\\delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}| ( \varepsilon+t_{i_{2}} )^{\frac{1}{m-1}}\cdots (\varepsilon+t_{i_{m}} )^{\frac{1}{m-1}} \\ \geq& |a_{ii\cdots i}| (\varepsilon+t_{i} )-r \sum _{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}| \\ &{} -\sum_{\substack{i_{2}\cdots i_{m}\in N_{2}^{m-1}\\ \delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}|\Bigl(\varepsilon+ \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} \Bigr) \\ =&\varepsilon \biggl(|a_{ii\cdots i}|-\sum_{\substack{i_{2}\cdots i_{m}\in N_{2}^{m-1} \\\delta_{ii_{2}\cdots i_{m}}=0}}|a_{ii_{2}\cdots i_{m}}| \biggr)+R_{i}(\mathcal{A}) \\ &{} -r \sum_{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}| -\sum _{\substack{i_{2}\cdots i_{m}\in N_{2}^{m-1}\\ \delta_{ii_{2}\cdots i_{m}}=0}} \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2}\cdots i_{m}}| \\ >&0. \end{aligned}$$

(12)

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

$$ |a_{ii\cdots i}| s_{i}\geq r \sum _{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}\\ \delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2} \cdots i_{m}}|+ \sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1} } \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2} \cdots i_{m}}|,\quad \forall i\in N_{1}, $$

(13)

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

$$x_{i}=\left \{ \textstyle\begin{array}{l@{\quad}l} (s_{i} )^{\frac{1}{m-1}}, & i\in N_{1}, \\ (t_{i} )^{\frac{1}{m-1}}, & i\in N_{2}. \end{array}\displaystyle \right . $$

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

$$\begin{aligned}& J_{1}= \biggl\{ i\in N_{1}: |a_{ii\cdots i}| s_{i}> r \sum_{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1} \\\delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2} \cdots i_{m}}| + \sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1} } \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2} \cdots i_{m}}| \biggr\} , \\& J_{2}= \biggl\{ i\in N_{2}: |a_{ii\cdots i}| t_{i}> r \sum_{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1} } |a_{ii_{2} \cdots i_{m}}|+ \sum_{\substack{i_{2}\cdots i_{m}\in N_{2}^{m-1} \\\delta_{ii_{2}\ldots i_{m}}=0}} \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2} \cdots i_{m}}| \biggr\} . \end{aligned}$$

Theorem 5

Let $\mathcal{A}=(a_{i_{1}\cdots i_{m}})$ be an order m dimension n complex tensor. If

$$ |a_{ii\cdots i}| s_{i}\geq r \sum _{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1} \\\delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2} \cdots i_{m}}|+ \sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1} } \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2} \cdots i_{m}}|, $$

(14)

$J_{1}\cup J_{2} \neq\emptyset$, 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

$$x_{i}=\left \{ \textstyle\begin{array}{l@{\quad}l} (s_{i} )^{\frac{1}{m-1}}, & i\in N_{1}, \\ (t_{i} )^{\frac{1}{m-1}}, & i\in N_{2}. \end{array}\displaystyle \right . $$

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

$$ |a_{ii\cdots i}| s_{i}> r \sum _{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}\\ \delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2} \cdots i_{m}}|,\quad \forall i\in N_{1} $$

(15)

and

$$ \sum_{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|=0,\quad \forall i\in N_{2}, $$

(16)

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

$$ |a_{ii\cdots i}| s_{i}> r \sum _{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1} \\\delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2} \cdots i_{m}}| +\frac{1}{K_{i}} \biggl( \sum _{i_{2}\cdots i_{m}\in N_{2}^{m-1}} \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2}\cdots i_{m}}| \biggr). $$

(17)

Let $K\equiv\max_{i\in N_{1}}\{K_{i}\}$. By inequality (17), we obtain

$$\begin{aligned}& |a_{ii\cdots i}| s_{i}> r \sum _{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1} \\\delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2} \cdots i_{m}}| +\frac{1}{K} \biggl( \sum _{i_{2}\cdots i_{m}\in N_{2}^{m-1}} \max_{j\in \{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2}\cdots i_{m}}| \biggr), \\& \quad \forall i\in N_{1}. \end{aligned}$$

(18)

Since $|a_{ii\cdots i}|\leq R_{i}(\mathcal{A})$ ($i\in N_{1}$) and inequality (15), so

$$ \sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|>0, \quad \forall i\in N_{1}. $$

(19)

For any $i\in N_{1}$, denote

$$ T_{i}=\frac{|a_{ii\cdots i}| s_{i}- r \sum_{\substack{i_{2}\cdots i_{m}\in N^{m-1}\setminus N_{2}^{m-1} \\ \delta_{ii_{2}\ldots i_{m}}=0}} |a_{ii_{2} \cdots i_{m}}| -\frac{1}{K} ( \sum_{i_{2}\cdots i_{m}\in N_{2}^{m-1}} \max_{j\in\{i_{2},\ldots, i_{m}\}} \{ t_{j} \} |a_{ii_{2}\cdots i_{m}}| )}{\sum_{i_{2}i_{3}\cdots i_{m}\in N_{2}^{m-1}}|a_{ii_{2}\cdots i_{m}}|}. $$

From inequalities (18) and (19), we have $T_{i}>0$. Therefore there exists a positive number $\varepsilon>0$ such that

$$ 0< \varepsilon< \min \biggl\{ \min_{i\in N_{1}}T_{i}, 1-\max _{i\in N_{2}}\frac{t_{i}}{K} \biggr\} . $$

Let the matrix $X=\operatorname{diag}(x_{1},x_{2},\ldots,x_{n})$, where

$$x_{i}=\left \{ \textstyle\begin{array}{l@{\quad}l} (s_{i} )^{\frac{1}{m-1}}, &i\in N_{1}, \\ (\varepsilon+\frac{t_{i}}{K} )^{\frac{1}{m-1}}, &i\in N_{2}. \end{array}\displaystyle \right . $$

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:

$$\begin{aligned}& \mathcal{A}=\bigl[A(1,:,:),A(2,:,:),A(3,:,:)\bigr], \\& A(1,:,:)=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 15 &1 &0 \\ 1 &10 &0 \\ 1 &1 &10 \end{array}\displaystyle \right ), \qquad A(2,:,:)= \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 1 &1 &0 \\ 0 &12 &0 \\ 1 &0 &1 \end{array}\displaystyle \right ), \\& A(3,:,:)=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 2 &0 &0 \\ 0 &3 &0 \\ 0 &0 &15 \end{array}\displaystyle \right ). \end{aligned}$$

Obviously,

$$\begin{aligned}& |a_{111}|=15, \qquad R_{1}(\mathcal{A})=24,\qquad |a_{222}|=12, \\& R_{2}(\mathcal{A})=4,\qquad |a_{333}|=15, \qquad R_{3}(\mathcal{A})=5, \end{aligned}$$

so $N_{1}=\{1\}$, $N_{2}=\{2, 3\}$. By calculation, we have

$$s_{1}=\frac{|a_{111}|}{R_{1}(\mathcal{A})}=\frac{15}{24}, \qquad t_{2}= \frac {R_{2}(\mathcal{A})}{|a_{222}|}=\frac{1}{3},\qquad t_{3}=\frac{R_{3}(\mathcal {A})}{|a_{333}|}= \frac{1}{3},\qquad r=\frac{15}{24}. $$

Since

$$\begin{aligned} r \sum_{\substack{jk\in N^{2}\setminus N_{2}^{2}\\ \delta_{1jk}=0}}|a_{1jk}|+\sum _{jk\in N_{2}^{2}}\max_{l\in\{j,k\}} \{t_{l} \} |a_{1jk}| =&\frac{15}{24}(1+0+1+1)+\frac{1}{3}(0+1+10+10) \\ =& \frac{213}{24} < \frac{225}{24}=|a_{111}|s_{1}, \end{aligned}$$

we know that $\mathcal{A}$ satisfies the conditions of Theorem 3, then $\mathcal{A}$ is an $\mathcal{H}$-tensor. But

$$ \sum_{jk\in N^{2}\setminus N_{2}^{2}}|a_{2jk}|=3\neq0,\qquad \sum _{jk\in N^{2}\setminus N_{2}^{2}}|a_{3jk}|=2\neq0. $$

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:

$$\begin{aligned}& \mathcal{A}=\bigl[A(1,:,:),A(2,:,:),A(3,:,:)\bigr], \\& A(1,:,:)=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 8 &1 &0 \\ 1 &10 &0 \\ 1 &1 &10 \end{array}\displaystyle \right ), \qquad A(2,:,:)= \left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0 &0 &0 \\ 0 &8 &2 \\ 0 &1 &1 \end{array}\displaystyle \right ), \\& A(3,:,:)=\left ( \textstyle\begin{array}{@{}c@{\quad}c@{\quad}c@{}} 0 &0 &0 \\ 0 &2.5 &1.5 \\ 0 &1 &10 \end{array}\displaystyle \right ). \end{aligned}$$

Obviously,

$$\begin{aligned}& |a_{111}|=8,\qquad R_{1}(\mathcal{A})=24,\qquad |a_{222}|=8, \\& R_{2}(\mathcal{A})=4,\qquad |a_{333}|=10,\qquad R_{3}(\mathcal{A})=5, \end{aligned}$$

so $N_{1}=\{1\}$, $N_{2}=\{2, 3\}$. By calculation, we have

$$s_{1}=\frac{|a_{111}|}{R_{1}(\mathcal{A})}=\frac{1}{3},\qquad t_{2}= \frac {R_{2}(\mathcal{A})}{|a_{222}|}=\frac{1}{2},\qquad t_{3}=\frac{R_{3}(\mathcal {A})}{|a_{333}|}= \frac{1}{2},\qquad r=\frac{1}{2}. $$

Since

$$ r\sum_{\substack{jk\in N^{2}\setminus N_{2}^{2}\\ \delta_{1jk}=0}}|a_{1jk}|= \frac{1}{2}(1+0+1+1)=\frac{3}{2}< \frac {8}{3}=|a_{111}|s_{1} $$

and

$$ \sum_{jk\in N^{2}\setminus N_{2}^{2}}|a_{2jk}|=0,\qquad \sum _{jk\in N^{2}\setminus N_{2}^{2}}|a_{3jk}|=0, $$

we see that $\mathcal{A}$ satisfies the conditions of Theorem 6, then $\mathcal{A}$ is an $\mathcal{H}$-tensor. But

$$\begin{aligned} r \sum_{\substack{jk\in N^{2}\setminus N_{2}^{2}\\ \delta_{1jk}=0}}|a_{1jk}|+\sum _{jk\in N_{2}^{2}}\max_{l\in\{j,k\}} \{t_{l} \}|a_{1jk}| =&\frac{1}{2}(1+0+1+1)+\frac{1}{2}(0+1+10+10) \\ =&12 >\frac{8}{3}=|a_{111}|s_{1}, \end{aligned}$$

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

$$\begin{aligned}& a_{1111}=11,\qquad a_{2222}=18,\qquad a_{3333}=18, \qquad a_{4444}=12, \\& a_{1123}=a_{1132}=a_{1213}=a_{1312}=a_{1231}=a_{1321}=1, \\& a_{2113}=a_{2131}=a_{2311}=a_{3112}=a_{3121}=a_{3211}=1, \\& a_{1234}=a_{1243}=a_{1324}=a_{1342}=a_{1423}=a_{1432}=-1, \\& a_{2134}=a_{2143}=a_{2314}=a_{2341}=a_{2413}=a_{2431}=-1, \\& a_{3124}=a_{3142}=a_{3214}=a_{3241}=a_{3412}=a_{3421}=-1, \\& a_{4123}=a_{4132}=a_{4213}=a_{4231}=a_{4312}=a_{4321}=-1, \end{aligned}$$

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.