0 引言

张量是矩阵的高阶推广, 在医学成像、期权定价和多人非合作博弈等实际应用问题中有着广泛的应用^[1-5]。作为H-矩阵的推广, H-张量具有特殊结构, 其在张量理论及应用研究上被广泛使用, 尤其是H-张量的数值性质、判别方法等受到越来越多学者的关注^[6-10]。同时, 齐次多项式因其在众多领域的重要应用，使其成为一个重要的研究课题^[11-13]。由于主对角元为正的H-张量与张量正定性具有一致性, 因此可基于此解决多项式正定性的判定。本文得出了判定H-张量的新条件, 所得结果拓展了文献[11-13]的结论。最后, 举例说明了新条件的有效性。

1 预备知识

记C(R)为复(实)数域, [n]: ={1, 2, …, n}。一个复(实)m阶n维张量A=(a_i1i2…im)由n^m个复(实)数元素构成^[1-5], 其中a_i1i2…im∈ C(R), i_j∈[n], j∈[m]。显然, 2阶张量即为矩阵。此外, 张量A=(a_i1i2…im)被称为对称的^[6-7], 若a_i1i2…im=a_{π(i1i2…im)}, ∀π∈Π_m, 其中Π_m为m个指标的置换群。

定义 1^[5]  张量I=(δ_i1i2…im)称为单位张量, 其中

$ \delta_{i_{1} i_{2} \cdots i_{m}}=\left\{\begin{array}{ll} 1, & i_{1}=i_{2}=\cdots=i_{m}, \\ 0, & \text {其他 }。\end{array}\right. $

定义 2^[4]  给定一个m阶n维张量A=(a_i1i2…im), 若存在一个复数λ和一个非零复向量x=(x₁, x₂, …, x_n)^T∈Cⁿ, 使得Ax^m-1=λx^[m-1], 那么称λ为A的特征值, x为A的关于特征值λ的特征向量, 向量Ax^m-1和x^[m-1]的第i个分量分别为${(\mathit{\boldsymbol{A}}{\mathit{\boldsymbol{x}}^{m - 1}})_i} = \mathop \sum \limits_{{i_2}, \ldots , {i_m} \in [n]} {a_{i{i_2}}}_ \ldots {i_m}{x_{{i_2}}} \ldots {x_{{i_m}}}, {({\mathit{\boldsymbol{x}}^{[m - 1]}})_i} = x_i^{m - 1}$。

记m阶n次齐次多项式f(x)为

$ f(\boldsymbol{x})=\sum\limits_{i_{1}, \cdots, i_{m} \in[n]} a_{i_{1} i_{2} \cdots i_{m}} x_{i_{1}} \cdots x_{i_{m}}, $

(1)

其中x=(x₁, x₂, …, x_n)^T∈Rⁿ。当m为偶数时, f(x)是正定的, 若f(x)>0, ∀x∈Rⁿ, x≠0。式(1)中的齐次多项式f(x)可以表示为m阶n维对称张量A与x^m的乘积, $f(\mathit{\boldsymbol{x}}) = \mathit{\boldsymbol{A}}{\mathit{\boldsymbol{x}}^m} = \mathop \sum \limits_{{i_1}, \ldots ,{i_m} \in [n]} {a_{{i_1}{i_2}_ \ldots {i_m}}}{x_{{i_1}}} \ldots {x_{{i_m}}}$, 当f(x)是正定时, 对称张量A也是正定的^[9]。

定义 3^[7]  设A=(a_i1i2…im)为m阶n维张量, 若存在正向量x=(x₁, x₂, …, x_n)^T∈Rⁿ, 使得对∀i∈[n], $\left| {{a_{ii \ldots i}}} \right|x_i^{m - 1} > \mathop \sum \limits_{\begin{array}{*{20}{c}} {{i_2},{i_3}, \ldots ,{i_m} \in [n]}\\ {{\delta _{i{i_2} \ldots {i_m}}} = 0} \end{array}} \left| {{a_{i{i_2} \ldots {i_m}}}} \right|{x_{{i_2}}} \ldots {x_{{i_m}}}$, 则称A是H-张量。

定义 4^[7]  m阶n维张量A=(a_i1i2…im)与矩阵X=diag(x₁, x₂, …, x_n)的乘积表示为

$ \boldsymbol{B}=\left(b_{i_{1} \cdots i_{m}}\right)=\boldsymbol{A} \boldsymbol{X}^{m-1}, 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}}, i_{j} \in[n], j \in[m]。$

假设Q表示[n]的任意非空子集, 令Q^m-1={i₂i₃…i_m: i_j∈Q, j=2, 3, …, m}, [n]^m-1\Q^m-1={i₂i₃…i_m: i₂i₃…i_m∈[n]^m-1且i₂i₃…i_m∉Q^m-1}。给定一个m阶n维张量A=(a_i1i2…im), 令${R_i}(\mathit{\boldsymbol{A}}) = \mathop \sum \limits_{\begin{array}{*{20}{c}} {{i_2}, \cdots ,{i_m} \in [n]}\\ {{\delta _{i{i_2} \cdots {i_m}}} = 0} \end{array}} = \left| {{a_{i{i_2} \ldots {i_m}}}} \right| = \mathop \sum \limits_{{i_2},{i_3}, \cdots ,{i_m} \in [n]} \left| {{a_{i{i_2} \cdots {i_m}}}} \right| - \left| {{a_{ii \cdots i}}} \right|$, N₁={i∈[n]: 0＜|a_ii…i|=R_i(A)}, N₂={i∈[n]: 0＜|a_ii…i|＜R_i(A)}, N₃={i∈[n]: |a_ii…i|>R_i(A)}, N₀^m-1=[n]^m-1\(N₂^m-1∪N₃^m-1)。

引理 1^[8]  若A为严格对角占优张量，则A为H-张量。

引理 2^[9]  设A=(a_i1i2…im)为m阶n维张量。如果存在一个正对角矩阵X, 使得AX^m-1是H-张量, 则A为H-张量。

引理 3^[9]  设A=(a_i1i2…im)为m阶n维张量。若A是不可约的, 且|a_ii…i|≥R_i(A), ∀i∈[n], 又至少存在一个i, 使得严格不等式成立, 则A为H-张量。

引理 4^[10]  设A=(a_i1i2…im)为m阶n维张量。若1)|a_ii…i ≥R_i(A), ∀i∈[n]，2) N₃={i∈[n]: |a_ii…i|>R_i(A)}≠Ø，3) ∀i∉N₃, 从i到j存在一个非零元素链, 使得j∈N₃，则A为H-张量。

文献[11]中给出了如下结果。

定理 1  设A=(a_i1i2…im)为m阶n维张量。若对∀i∈N₁∪N₂, 满足

$ \left| {{a}_{ii\cdots i}} \right|>\sum\limits_{\begin{smallmatrix} {{i}_{2}}{{i}_{3}}\cdots {{i}_{m}}\in {{[n]}^{m-1}}\backslash N_{3}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+\sum\limits_{{{i}_{2}}{{i}_{3}}\cdots {{i}_{m}}\in N_{3}^{m-j}}{\mathop {\max }\limits_{j \in \{ {i_2},{i_3}, \cdots ,{i_m}\} } }\left\{ {{R}_{j}}(A)/\left| {{a}_{jj\cdots j}} \right| \right\}\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|, $

则A为H-张量。

2 主要结果

为了叙述方便，引入以下符号，记：

$ r=\max \limits_{i \in N_{3}}\,\left\{ \sum\limits_{{{i}_{2}}{{i}_{3}}\cdots {{i}_{m}}\in N_{0}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+\sum\limits_{{{i}_{2}}{{i}_{3}}\cdots {{i}_{m}}\in N_{2}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}/\left| {{a}_{ii\cdots i}} \right|-\sum\limits_{\begin{smallmatrix} {{i}_{2}}{{i}_{3}}\cdots {{i}_{m}}\in N_{3}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|} \right\}; $

$ m_{i}=R_{i}(\boldsymbol{A})-\left|a_{i i \cdots i}\right| / R_{i}(\boldsymbol{A}), m=\max \limits_{i \in N_{2}}\left\{m_{i}\right\}, \delta=\max \{r, m\} ; $

$ {{P}_{i,r}}(\boldsymbol{A})=\sum\limits_{{{i}_{2}}{{i}_{3}}\cdots {{i}_{m}}\in N_{0}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+\sum\limits_{{{i}_{2}}{{i}_{3}}\cdots {{i}_{m}}\in N_{2}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+r\sum\limits_{\begin{smallmatrix} {{i}_{2}}{{i}_{3}}\cdots {{i}_{m}}\in N_{3}^{m-1} \\ {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|},\forall i\in {{N}_{3}}; $

$ h = \mathop {\max }\limits_{i \in {N_3}} \left\{ {\delta \left( {\sum\limits_{{i_2}{i_3} \cdots {i_m} \in N_0^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} + \sum\limits_{{i_2}{i_3} \cdots {i_m} \in N_2^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} } \right)/{P_{i,r}}(\boldsymbol{A})} \right. - \begin{array}{*{20}{l}} {\sum\limits_{\begin{array}{*{20}{c}} {{i_2}{i_3} \cdots {i_m} \in N_3^{m - 1}}\\ {{\delta _{i{i_2} \cdots {i_m}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} \left. {\frac{{{P_{j,r}}(\boldsymbol{A})}}{{\left| {{a_{jj \cdots j}}} \right|}}\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} \right\}。} } \end{array} $

定理 2  设A=(a_i1i2…_im)为m阶n维张量。若对任意的i∈N₂, A满足

$ \begin{array}{l} \left| {{a_{ii \cdots i}}} \right|{m_i} > \delta \sum\limits_{{i_2}{i_3} \cdots {i_m} \in N_0^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} + \sum\limits_{\begin{array}{*{20}{c}} {{i_2}{i_3} \cdots {i_m} \in N_2^{m - 1}}\\ {{\delta _{i{i_2} \cdots {i_m}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} }\left\{ {{m_j}} \right\}\left| {{a_{i{i_2} \cdots {i_m}}}} \right| + \\ h\sum\limits_{{i_2} \cdots {i_m} \in N_3^{m - j}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} \left\{ {{P_{j,r}}(\mathit{\boldsymbol{A}})/\left| {{a_{jj \cdot \cdots j}}} \right|} \right\}} \left| {{a_{i{i_2} \cdots {i_m}}}} \right|, \end{array} $

(2)

且

$ \left| {{a}_{ii\cdots i}} \right|\ne \sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|},\forall i\in {{N}_{1}}, $

(3)

则A是H-张量。

证明  由r、m、δ的表达式知0≤r＜1, 0＜m＜1, 0＜δ＜1, 因此对∀i∈N₃,

$ {{P}_{i,r}}(\boldsymbol{A})=\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{2}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+r\sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}\le r\left| {{a}_{ii\cdots i}} \right|, $

则0≤P_i(A)/|a_ii…i ≤r≤δ＜1, ∀i∈N₃。又由P_{i, r}(A)的定义, 对∀i∈N₁, 可得

$ \begin{array}{l} \left( {\delta \left( {\sum\limits_{{i_2} \cdots {i_m} \in N_0^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} + \sum\limits_{{i_2} \cdots {i_m} \in N_2^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} } \right)} \right)/\left( {{P_{i,r}}(\boldsymbol{A}) - \begin{array}{*{20}{l}} {\sum\limits_{\begin{array}{*{20}{c}} {{i_2}{i_3} \cdots {i_m} \in N_3^{m - 1}}\\ {{\delta _{i{i_2} \cdots {i_m}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} {P_{j,r}}(\boldsymbol{A})/\left| {{a_{jj \cdots j}}} \right|\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} } \end{array}} \right)\\ \left( {{P_{i,r}}(\boldsymbol{A}) - \sum\limits_{\begin{array}{*{20}{c}} {{i_2} \cdots {i_m} \in N_3^{m - 1}}\\ {{\delta _{i{i_2} \cdots {i_m}}} = 0} \end{array}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} } \right)/\left( {{P_{i,r}}(\boldsymbol{A}) - \sum\limits_{\begin{array}{*{20}{c}} {{i_2} \cdots {i_m} \in N_3^{m - 1}}\\ {{\delta _{i{i_2} \cdots {i_m}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} \left\{ {{P_{i,r}}(\boldsymbol{A})/\left| {{a_{jj \cdots j}}} \right|} \right\}\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} } \right) \le 1。\end{array} $

因此, 对∀i∈N₃, 0≤h＜1, 且

$ h{P_{i,r}}({\boldsymbol{A}}) \ge \delta \sum\limits_{{i_2} \cdots {i_m} \in N_0^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} + \delta \sum\limits_{{i_2} \cdots {i_m} \in N_2^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} + \sum\limits_{\begin{array}{*{20}{c}} {{i_2} \cdots {i_m} \in N_3^{m - 1}}\\ {{\delta _{i{i_2} \cdots {i_m}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} \left\{ {{P_{i,r}}(\boldsymbol{A})/\left| {{a_{jj \cdots j}}} \right|} \right\}\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} 。$

(4)

令

$ \begin{array}{*{20}{l}} {{M_i} = \left( {\left| {{a_{ii \cdots i}}} \right|{m_i} - \delta \sum\limits_{{i_2} \cdots {i_m} \in N_0^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} - \sum\limits_{\begin{array}{*{20}{c}} {{i_2} \cdots {i_m} \in N_2^{m - 1}}\\ {{\delta _{i{i_2} \cdots {i_m}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} } \{ {m_j}\} \left| {{a_{i{i_2} \cdots {i_m}}}} \right| - } \right.}\\ {\left. {h\sum\limits_{{i_2} \cdots {i_m} \in N_3^{m - 1}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} \left\{ {{P_j}(\mathit{\boldsymbol{A}})/\left| {{a_{jj \cdots j}}} \right|} \right\}} \left| {{a_{i{i_2} \cdots {i_m}}}} \right|} \right)/\left( {\sum\limits_{{i_2} \cdots {i_m} \in N_3^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} } \right),} \end{array} $

(5)

若$\sum\limits_{{i_2} \cdots {i_{\rm{m}}} \in N_3^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} = 0$, 记M_i=+∞。由式(2)、式(3)及式(5)知, M_i>0(∀i∈N₂), 且0≤hP_{i, r}(A)/|a_ii…i|＜δ＜1, ∀i∈N₃, 从而一定存在充分小的正数ε, 使得

$ 0 < \varepsilon < \mathop {\min }\limits_{i \in {N_2}} \left\{ {{M_i}} \right\} \le + \infty ,\mathop {\max }\limits_{i \in {N_3}} \left\{ {h{P_{i,r}}(\mathit{\boldsymbol{A}})/\left| {{a_{ii \cdots i}}} \right| + \varepsilon } \right\} < \delta < 1。$

(6)

构造正对角矩阵D=diag(d₁, d₂, …, d_n), 记B=AD^m-1=(b_i1i2…im)，其中：

$ d_{i}=(\delta)^{\frac{1}{m-1}}, i \in N_{1} ; d_{i}=\left(m_{i}\right)^{\frac{1}{m-1}}, i \in N_{2} ; d_{i}=\left(h P_{i, r}(\boldsymbol{A}) /\left|a_{i i \cdots i}\right|+\varepsilon\right)^{\frac{1}{m-1}}, i \in N_{3}。$

1) 对∀i∈N₁, 由(4)式及(6)式知

$ \begin{align} & {{R}_{i}}(\boldsymbol{B})=\sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}{{x}_{{{i}_{2}}}}\cdots {{x}_{{{i}_{m}}}}+\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{2}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}{{\left( {{m}_{{{i}_{2}}}} \right)}^{\frac{1}{m-1}}}\cdots {{\left( {{m}_{{{i}_{m}}}} \right)}^{\frac{1}{m-1}}}+ \\ & \sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}{{\left( h{{P}_{{{i}_{2}},r}}(\boldsymbol{A})/\left| {{a}_{{{i}_{2}}{{i}_{2}}\cdots {{i}_{2}}}} \right|+\varepsilon \right)}^{\frac{1}{m-1}}}\cdots {{\left( h{{P}_{{{i}_{m}},r}}(\boldsymbol{A})/\left| {{a}_{{{i}_{m}}{{i}_{m}}\cdots {{i}_{m}}}} \right|+\varepsilon \right)}^{\frac{1}{m-1}}}\le \\ & \delta \sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{2}^{m-1}}\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}}\left\{ {{m}_{i}} \right\}\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|+ \\ & \sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1}}{\left( \underset{j\in \left\{ {{i}_{2}},{{i}_{3}},\cdots ,{{i}_{m}} \right\}}{\mathop{\max }}\,\left\{ h{{P}_{j,r}}(A)/\left| {{a}_{jj\cdots j}} \right| \right\}+\varepsilon \right)}\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|<\left| {{a}_{ii\cdots i}} \right|\delta =\left| {{b}_{ii\cdots i}} \right| 。\\ \end{align} $

(7)

2) 对∀i∈N₂, 由式(2)、式(5)及式(6)知

$ \begin{align} & {{R}_{i}}(\boldsymbol{B})=\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}{{x}_{{{i}_{2}}}}\cdots {{x}_{{{i}_{m}}}}+\sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{2}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}{{\left( {{m}_{{{i}_{2}}}} \right)}^{\frac{1}{m-1}}}\cdots {{\left( {{m}_{{{i}_{m}}}} \right)}^{\frac{1}{m-1}}}+ \\ & \sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}{{\left( h{{P}_{{{i}_{2}},r}}(\boldsymbol{A})/\left| {{a}_{{{i}_{2}}{{i}_{2}}\cdots {{i}_{2}}}} \right|+\varepsilon \right)}^{\frac{1}{m-1}}}\cdots {{\left( h{{P}_{{{i}_{m}},r}}(\boldsymbol{A})/\left| {{a}_{{{i}_{m}}{{i}_{m}}\cdots {{i}_{m}}}} \right|+\varepsilon \right)}^{\frac{1}{m-1}}}\le \delta \sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+ \\ & \sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{2}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}}\left\{ {{m}_{i}} \right\}\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|+\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1}}{\left( \underset{j\in \left\{ {{i}_{2}},{{i}_{3}},\cdots ,{{i}_{m}} \right\}}{\mathop{\max }}\,\left\{ h{{P}_{j,r}}(\boldsymbol{A})/\left| {{a}_{jj\cdots j}} \right| \right\}+\varepsilon \right)}\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|< \\ & \delta \sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+\sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{2}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}}\left\{ {{m}_{i}} \right\}\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|+\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1}}\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}}\left\{ h{{P}_{j,r}}(\boldsymbol{A})/\left| {{a}_{jj\cdots j}} \right| \right\}\cdot \\ & \left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|+{{M}_{i}}\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}=\left| {{a}_{ii\cdots i}} \right|{{m}_{i}}=\left| {{b}_{ii\cdots i}} \right| 。\\ \end{align} $

(8)

3) 对∀i∈N₃, 由式(4)及式(6)知

$ \begin{align} & {{R}_{i}}(\boldsymbol{B})=\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}{{x}_{{{i}_{2}}}}\cdots {{x}_{{{i}_{m}}}}+\sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{2}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}{{\left( {{m}_{{{i}_{2}}}} \right)}^{\frac{1}{m-1}}}\cdots {{\left( {{m}_{{{i}_{m}}}} \right)}^{\frac{1}{m-1}}}+ \\ & \sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}{{\left( h{{P}_{{{i}_{2}},r}}(\boldsymbol{A})/\left| {{a}_{{{i}_{2}}{{i}_{2}}\cdots {{i}_{2}}}} \right|+\varepsilon \right)}^{\frac{1}{m-1}}}\cdots {{\left( h{{P}_{{{i}_{m}},r}}(\boldsymbol{A})/\left| {{a}_{{{i}_{m}}{{i}_{m}}\cdots {{i}_{m}}}} \right|+\varepsilon \right)}^{\frac{1}{m-1}}}\le \delta \sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+ \\ & \sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{2}^{m-1}}\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}}\left\{ {{m}_{i}} \right\}\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|+\sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left( \underset{j\in \left\{ {{i}_{2}},{{i}_{3}},\cdots ,{{i}_{m}} \right\}}{\mathop{\max }}\,\left\{ h{{P}_{j,r}}(\boldsymbol{A})/\left| {{a}_{jj\cdots j}} \right| \right\}+\varepsilon \right)}\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|= \\ & h{{P}_{j,r}}(\boldsymbol{A})\varepsilon \sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{3}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}<h{{P}_{{{i}_{2}},r}}(\boldsymbol{A})+\varepsilon \left| {{a}_{ii\cdots i}} \right|=\left| {{b}_{ii\cdots i}} \right| 。\\ \end{align} $

(9)

综上所述，由式(7)、式(8)及式(9)知b_ii…i >R_i(B)(∀i∈[n]), 即B是严格对角占优的, 故由引理1知B是H-张量。进而, 由引理2知A是H-张量。

类似于定理2的证明, 借助引理3和引理4, 可得如下结论。

定理 3  设A=(a_i1i2…im)为m阶n维张量, A不可约, 若对任意的i∈N₂，

$ \begin{array}{*{35}{l}} \left| {{a}_{ii\cdots i}} \right|{{m}_{i}}\ge \delta \sum\limits_{{{i}_{2}}\cdots {{i}_{m}}\in N_{0}^{m-1}}{\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|}+\sum\limits_{\begin{smallmatrix} {{i}_{2}}\cdots {{i}_{m}}\in N_{2}^{m-1} \\ {{\delta }_{i{{i}_{2}}\cdots {{i}_{m}}}}=0 \end{smallmatrix}}{\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}}\,\left\{ {{m}_{i}} \right\}\left| {{a}_{i{{i}_{2}}\cdots {{i}_{m}}}} \right|+} \\ h\sum\limits_{{i_2} \cdots {i_m} \in N_3^{m - 1}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} \left\{ {{P_{j,r}}({\boldsymbol{A}})/\left| {{a_{jj \cdot \cdots j}}} \right|} \right\}\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} , \end{array} $

(10)

且(10)式中至少有一个严格不等式成立, 则A是H-张量，记

$ \begin{array}{*{20}{l}} {K(\mathit{\boldsymbol{A}}) = \left\{ {i \in {N_2}:\left| {{a_{ii \cdots i}}} \right|{m_i} > \delta \sum\limits_{{i_2} \cdots {i_m} \in N_0^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} + \sum\limits_{\begin{array}{*{20}{c}} {{i_2} \cdots {i_m} \in N_2^{m - 1}}\\ {{\delta _{i{i_2} \cdots {i_m}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} } \left\{ {{m_i}} \right\}\left| {{a_{i{i_2} \cdots {i_m}}}} \right| + } \right.}\\ {\left. {h\sum\limits_{{i_2} \cdots {i_m} \in N_3^{m - 1}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} } \left\{ {h{P_{j,r}}(\mathit{\boldsymbol{A}})/\left| {{a_{jj \cdot \cdot j}}} \right|} \right\}\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} \right\}}。\end{array} $

定理 4  设A=(a_i1i2…im)为m阶n维张量。若对任意的i∈N₂, 有

$ \left| {{a_{ii \cdots i}}} \right|{m_i} \ge \delta \sum\limits_{{i_2} \cdots {i_m} \in N_0^{m - 1}} {\left| {{a_{i{i_2} \cdots {i_m}}}} \right|} + \sum\limits_{\begin{array}{*{20}{c}} {{i_2} \cdots {i_m} \in N_2^{m - 1}}\\ {{\delta _{i{i_2} \ldots {i_m}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \{ {i_2},{i_3}, \ldots ,{i_m}\} } } \left\{ {{m_j}} \right\}\left| {{a_{i{i_2} \cdots {i_m}}}} \right| + \\h\mathop \sum \limits_{{i_2} \cdots {i_m} \in N_3^{m - 1}} \mathop {\max }\limits_{j \in \{ {i_2},{i_3}, \cdots ,{i_m}\} } \left\{ {{P_{j,r}}(\mathit{\boldsymbol{A}})/\left| {{a_{jj \cdots j}}} \right|} \right\}\left| {{a_{i{i_2} \cdots {i_m}}}} \right|, $

且对∀i∈[n]\K(A), 存在从i到j的非零元素链，使得j∈K(A)≠Ø, 则A是H-张量。

例 1  给定A=[A(1, : , : ), A(2, : , : ), A(3, : , : )], 其中

$ \boldsymbol{A}(1,:,:)=\left(\begin{array}{ccc} 12 & 1 & 0 \\ 1 & 6 & 0 \\ 1 & 0 & 16 \end{array}\right), \boldsymbol{A}(2,:,:)=\left(\begin{array}{lll} 1 & 1 & 0 \\ 0 & 6 & 0 \\ 0 & 0 & 1 \end{array}\right), \boldsymbol{A}(3,:,:)=\left(\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 16 \end{array}\right) \text { 。} $

由张量A的元素得到

$ \left| {{a_{111}}} \right| = 12,{R_1}(\mathit{\boldsymbol{A}}) = 25,\left| {{a_{222}}} \right| = 6,{R_2}(\mathit{\boldsymbol{A}}) = 3,\left| {{a_{333}}} \right| = 16,{R_3}(\mathit{\boldsymbol{A}}) = 2, $

所以N₁=Ø, N₂={1}, N₃={2, 3}。当i=1时, 有

$ \begin{array}{l} \delta \sum\limits_{{i_2}{i_3} \in N_0^2} {\left| {{a_{1{i_2}{i_3}}}} \right|} + \sum\limits_{\begin{array}{*{20}{c}} {{i_2}{i_3} \in N_2^2}\\ {{\delta _{1{i_2}{i_3}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} } \left\{ {{m_j}} \right\}\left| {{a_{1{i_2}{i_3}}}} \right| + h\sum\limits_{{i_2}{i_3} \in N_3^2} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}} \right\}} } \left\{ {{P_{j,r}}(\mathit{\boldsymbol{A}})/\left| {{a_{ij \cdot \cdot j}}} \right|} \right\}\left| {{a_{1{i_2}{i_3}}}} \right| = \\ 767/125 < 156/25 = \left| {{a_{111}}} \right|{m_1}。\end{array} $

所以张量A满足定理2的条件, 可知张量A是H-张量。但

$ \sum\limits_{\begin{smallmatrix} {{i}_{2}}{{i}_{3}}\in {{[n]}^{2}}\backslash N_{3}^{2} \\ {{\delta }_{1{{i}_{2}}{{i}_{3}}}}=0 \end{smallmatrix}}{\left| {{a}_{1{{i}_{2}}{{i}_{3}}}} \right|}+\sum\limits_{{{i}_{2}}{{i}_{3}}\in N_{3}^{2}}{\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}}\,\left\{ {{R}_{j}}(\boldsymbol{A})/\left| {{a}_{jj\cdots j}} \right| \right\}}\left| {{a}_{1{{i}_{2}}{{i}_{3}}}} \right|=14>12=\left| {{a}_{111}} \right|, $

因此, A不满足定理1的条件。

3 应用

基于H-张量的判定条件, 下面给出了偶数阶实对称张量正定性判定的新条件(多元形式的正定)。

引理 5^[9]  设m阶n维张量A=(a_i1i2…im)为偶数阶实对称张量, 且对∀i∈[n], 满足a_ii…i>0。如果A是H-张量，则A是正定的。

根据引理5、定理2、定理3及定理4, 可得到以下结论。

定理 5  设张量A=(a_i1i2…im)是m阶n维的偶数阶实对称张量, 且对∀i∈[n], 满足a_ii…i>0。如果A满足下列条件之一：

1) 定理2的所有条件；2) 定理3的所有条件；3) 定理4的所有条件；则A是正定的。

例 2  设四次齐次多项式f(x)=Ax⁴=16x₁⁴+84x₂⁴+87x₃⁴+75x₄⁴-8x₁³x₄-12x₁²x₂x₃-12x₂x₃²x₄+24x₁x₂x₃x₄, 其中A=(a_i1i2…im)是一个4阶4维的实对称张量, 且

$ {{a_{1111}} = 16,{a_{2222}} = 84,{a_{3333}} = 87,{a_{4444}} = 75,{a_{1114}} = {a_{1141}} = {a_{1411}} = {a_{4111}} = - 2,} $

$ {{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_{2334}} = {a_{2343}} = {a_{2433}} = {a_{4233}} = {a_{4323}} = {a_{4332}} = - 1,{a_{3234}} = {a_{3243}} = {a_{3324}} = {a_{3342}} = {a_{3423}} = {a_{3432}} = - 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,} $

其余的a_i1i2i3i4=0。计算得

$ {a_{1111}} = 16 < 18 = {R_1}(\mathit{\boldsymbol{A}}),{a_{4444}}\left( {{a_{1111}} - {R_1}(\mathit{\boldsymbol{A}}) + \left| {{a_{1444}}} \right|} \right) = - 150 < 0 = {R_4}(\mathit{\boldsymbol{A}})\left| {{a_{1444}}} \right| $

因此, A既不是严格对角占优张量, 也不是拟双严格对角占优张量, 所以不能用文献[12-13]中的定理来判定A的正定性。但可以证明A满足定理2的所有条件。因为

$ \left|a_{1111}\right|=16, R_{1}(\boldsymbol{A})=18,\left|a_{2222}\right|=84, R_{2}(\boldsymbol{A})=12,\left|a_{3333}\right|=87, R_{3}(\boldsymbol{A})=15,\left|a_{4444}\right|=75, R_{4}(\boldsymbol{A})=11, $

所以N₁=Ø, N₂={1}, N₃={2, 3, 4}。当i=1时，有

$ \delta \sum\limits_{{i_2}{i_3} \in N_0^2} {\left| {{a_{1{i_2}{i_3}}}} \right|} + \sum\limits_{\begin{array}{*{20}{c}} {{i_2}{i_3} \in N_2^2}\\ {{\delta _{1{i_2}{i_3}}} = 0} \end{array}} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}, \cdots ,{i_m}} \right\}} } \left\{ {{m_j}} \right\}\left| {{a_{1{i_2}{i_3}}}} \right| + h\sum\limits_{{i_2}{i_3} \in N_3^2} {\mathop {\max }\limits_{j \in \left\{ {{i_2},{i_3}} \right\}} } \left\{ {{P_{j,r}}(\mathit{\boldsymbol{A}})/\left| {{a_{jj \cdot \cdots j}}} \right|} \right\}\left| {{a_{1{i_2}{i_3}}}} \right| = 38/27 < 16/9 = \left| {{a_{111}}} \right|{m_1}。$

根据定理5, A是正定的, 即f(x)是正定的。

4 结论

本文给出了判别张量是否是H-张量新的充分条件, 该条件可用于判别偶数次齐次多项式f(x)=Ax^m的正定性。数值算例表明了本文所得条件的有效性。