本文讨论的广义箭形矩阵具有如下形式

$ \begin{equation} \\{A} = \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}& \cdots &{{b_m}}&{}&{}&{}\\ {{b_1}}&{{a_2}}&{}&{}&{}&{}&{}\\ \vdots &{}& \ddots &{}&{}&{}&{}\\ {{b_m}}&{}&{}&{{a_{m + 1}}}&{{b_{m + 1}}}&{}&{}\\ {}&{}&{}&{{b_{m + 1}}}&{{a_{m + 2}}}& \ddots &{}\\ {}&{}&{}&{}& \ddots & \ddots &{{b_{n{\rm{ - }}1}}}\\ {}&{}&{}&{}&{}&{{b_{n{\rm{ - }}1}}}&{{a_n}} \end{array}} \right], \end{equation} $

(1.1)

其中$ {a_i}\left( {i = 2, \cdots , m + 1} \right) $互不相同, 且$ {b_i} > 0\left( {i{\rm{ = }}m + 1, \cdots , n - 1} \right) $.当$ m = 0 $时, 具有形式(1.1)的矩阵$ {A} $为Jacobi矩阵, 而当$ m = n - 1 $时, 矩阵$ {A} $就变成箭形矩阵. Jacobi矩阵的特征值反问题具有广泛的应用, 关于此类问题研究已取得一些比较好的结果, 详见文献[1-4].箭形矩阵的特征值反问题在现代控制理论中有着广泛的应用, 文献[5, 6]分别讨论了对称三对角矩阵和对称爪形矩阵的特征值反问题.对于一些特殊箭形矩阵的特征值反问题的研究见文献[7-9]. Gladwell从力学角度阐述了振动中的一些特征值反问题^[10], 其中弹簧-质量系统等振动结构参数识别问题, 往往归结为Jacobi矩阵特征值反问题, 星形弹簧质量系统的振动问题则转化为箭形矩阵的特征值反问题.

具有形式(1.1)的广义箭形矩阵的特征值反问题在文献[11-13]中有讨论, 鉴于上述工作以及此类矩阵的重要性, 本文重点研究具有形式(1.1)的广义箭形矩阵的特征值反问题, 我们推广了Jacobi矩阵和箭形矩阵逆特征值问题, 提出了两类逆问题, 给出了问题有唯一解的充分必要条件, 给出了解的表达式及相应数值例子.

本文研究广义箭形矩阵的两类逆问题, 即

问题I   给出三个非零互异实数$ {\lambda _1}, {\lambda _2}, \mu $以及三个非零实向量$ {x_1} = {\left( {{x_1}, \cdots , {x_{m + 1}}} \right)^{\rm T}}, $ $ {x_2} = {\left( {{x_{m + 1}}, \cdots , {x_n}} \right)^{\rm T}}, $ $ {y} = {\left( {{y_1}, \cdots , {y_n}} \right)^{\rm T}}. $求具有形式(1.1)的$ n $阶矩阵$ {A} $使得$ \left( {{\lambda _1}, {x_1}} \right), $ $ \left( {{\lambda _2}, {x_2}} \right), $ $ \left( {\mu , {y}} \right) $分别是$ {{A_{1, m + 1}}} $, $ {A_{m + 1, n}} $和$ {A} $的特征对.

问题II   给出两个非零互异实数$ \lambda , \mu $和两个非零实向量$ {x} = {\left( {{x_1}, \cdots , {x_n}} \right)^{\rm T}}, {y} = {\left( {{y_1}, \cdots , {y_n}} \right)^{\rm T}} $, 求具有形式(1.1)的矩阵$ {A} $和$ {A^*} $, 使得$ \left( {\lambda , {x}} \right), \left( {\mu , {y}} \right) $分别为矩阵$ {A}, {A^*} $的特征对.

矩阵$ {A} $的的主子式$ {A_{1, m + 1}}, {A_{m + 1, n}} $和矩阵$ {A^*} $分别具有如下形式

$ \begin{eqnarray} &&{A_{1, m + 1}} = \left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}& \ldots &{{b_{m - 1}}}&{{b_m}}\\ {{b_1}}&{{a_2}}&{}&{}&{}\\ \vdots &{}& \ddots &{}&{}\\ {{b_{m - 1}}}&{}&{}&{{a_m}}&{}\\ {{b_m}}&{}&{}&{}&{{a_{m + 1}}} \end{array}} \right]{\rm{ }}, \end{eqnarray} $

(1.2)

$ \begin{eqnarray} &&{\rm{ }}{A_{m + 1, n}} = \left[ {\begin{array}{*{20}{c}} {{a_{m + 1}}}&{{b_{m + 1}}}&{}&{}&{}\\ {{b_{m + 1}}}&{{a_{m + 2}}}& \ddots &{}&{}\\ {}& \ddots & \ddots &{}&{}\\ {}&{}&{{b_{n - 2}}}&{{a_{n - 1}}}&{{b_{n - 1}}}\\ {}&{}&{}&{{b_{n - 1}}}&{{a_n}} \end{array}} \right], \end{eqnarray} $

(1.3)

$ \begin{eqnarray} &&{A^{\rm{*}}} = \left[ {\begin{array}{*{20}{c}} {a_1^{\rm{*}}}&{{b_1}}& \cdots &{{b_m}}&{}&{}&{}\\ {{b_1}}&{{a_2}}&{}&{}&{}&{}&{}\\ \vdots &{}& \ddots &{}&{}&{}&{}\\ {{b_m}}&{}&{}&{{a_{m + 1}}}&{{b_{m + 1}}}&{}&{}\\ {}&{}&{}&{{b_{m + 1}}}&{{a_{m + 2}}}& \ddots &{}\\ {}&{}&{}&{}& \ddots & \ddots &{{b_{n{\rm{ - }}1}}}\\ {}&{}&{}&{}&{}&{{b_{n{\rm{ - }}1}}}&{{a_n}} \end{array}} \right], \end{eqnarray} $

(1.4)

其中$ {a_i}\left( {i = 2, \cdots , m + 1} \right) $互不相同, 且$ {b_i} > 0 \left( {i{\rm{ = }}m + 1, \cdots , n - 1} \right) $，可知$ {A_{1, m + 1}}, {A_{m + 1, n}} $分别为箭形矩阵和Jacobi矩阵.式(1.4)中除元素$ a_1^{\rm{*}} $与$ {a_1} $不同外, 其它元素与(1.1)式矩阵$ {A} $中元素相同.

现在作如下约定

$ \begin{eqnarray} &&{D_i} = \left| {\begin{array}{*{20}{c}} {{x_i}}&{{y_i}}\\ {{x_{i + 1}}}&{{y_{i + 1}}} \end{array}} \right|\left( {i = m + 1, \cdots , n - 1} \right), \end{eqnarray} $

(1.5)

$ \begin{eqnarray} &&{E_i} = \left| {\begin{array}{*{20}{c}} {{x_1}}&{{y_1}}\\ {{x_i}}&{{y_i}} \end{array}} \right|\left( {i = 1, 2, \cdots , n - 1} \right), \end{eqnarray} $

(1.6)

$ \begin{eqnarray} &&{d_i} = {x_i}{y_i}{\rm{ }}\left( {i = 1, 2, \cdots , n} \right){\rm{ }}, \end{eqnarray} $

(1.7)

$ \begin{eqnarray} &&\omega {\rm{ = }}{x_1}{x_{m + 1}}{y_{m + 2}} - {x_1}{x_{m + 2}}{y_{m + 1}} + {x_{m + 1}}{x_{m + 2}}{y_1}. \end{eqnarray} $

(1.8)

下面给出本文将要用到的必要引理.

引理2.1^[14]  设$ \lambda $为$ n $阶Jacobi矩阵$ J $的特征值, $ {x} = {\left( {{x_1}, {x_2}, \cdots , {x_n}} \right)^{\rm T}} $为$ J $对应于$ \lambda $的特征向量, 则

(1) $ {x_1}{x_n} \ne 0 $;

(2) $ x $的相邻的两个分量不同时为零;

(3) 若某个$ i\left( {1 < i < n} \right) $使得$ {x_i}{\rm{ = }}0 $, 则$ {x_{i - 1}}{x_{i + 1}} < 0 $.

对于问题Ⅰ给出定理2.1.

定理2.1   问题I有唯一解的充分必要条件为

(i) $ {x_i}\left( {i = m + 1, \cdots , n} \right) $满足引理2.1的条件;

(ii) $ \left\{ \begin{array}{l} {E_i} \ne 0\left( {i = 2, \cdots , m} \right), \omega \ne 0, \\ x_j^2{\rm{ + }}y_j^2 \ne 0{\rm{ }}\left( {j = 1, 2, \cdots , m, m + 2, \cdots n} \right), \\ {D_j} \ne 0\left( {j = m + 1, \cdots , n - 1} \right); \end{array} \right. $

(iii) \(\left({{\lambda _1} - \mu } \right)\sum\limits_{i = 1}^{m + 1} {{x_i}{y_i} + } {b_{m + 1}}{x_{m + 1}}{y_{m + 2}} = 0, \left({{\lambda _2} - \mu } \right)\sum\limits_{j = m + 1}^n {{x_i}} {y_i} + {b_m}{x_{m + 1}}{y_1} = 0\);

(iv) $ \left\{ \begin{aligned} &{b_{m{\rm{ + }}1}} = \frac{{\left[ { - \left( {{\lambda _1} - {\lambda _2}} \right){x_{m + 1}}{y_1} + \left( {\mu - {\lambda _1}} \right){x_1}{y_{m + 1}}} \right]{x_{m + 1}}}}{\omega } > 0, \\ &{b_i} = \frac{{{b_{m{\rm{ + }}1}}{D_{m + 1}} - \left( {{\lambda _2} - \mu } \right)\sum\limits_{j = m + 2}^i {{d_j}} }}{{{D_i}}} > 0 \quad \left ( {i = m + 2, \cdots , n - 1} \right), \end{aligned} \right. $

且当问题有解时, 解由下式给出

$ \begin{eqnarray*} &&{b_i} = \frac{{\left( {{\lambda _1} - \mu } \right){d_{i{\rm{ + }}1}}}}{{{E_{i{\rm{ + }}1}}}}{\rm{ }}\left( {i = 1, \ldots , m{\rm{ - }}1} \right), \\ &&{b_m} = \frac{{\left[ {\left( {{\lambda _1} - {\lambda _2}} \right){x_{m + 1}}{y_{m + 2}} + \left( {\mu - {\lambda _1}} \right){x_{m + 2}}{y_{m + 1}}} \right]{x_{m + 1}}}}{\omega }, \\ &&{b_{m{\rm{ + }}1}} = \frac{{\left[ { - \left( {{\lambda _1} - {\lambda _2}} \right){x_{m + 1}}{y_1} + \left( {\mu - {\lambda _1}} \right){x_1}{y_{m + 1}}} \right]{x_{m + 1}}}}{\omega }, \\ &&{b_i} = \frac{{{b_{m{\rm{ + }}1}}{D_{m + 1}} - \left( {{\lambda _2} - \mu } \right)\sum\limits_{j = m + 2}^i {{d_j}} }}{{{D_i}}} \quad \left( {i = m + 2, \cdots .n - 1} \right), \\ &&{a_1} = \left\{ {\begin{aligned} {\frac{{{\lambda _1}{x_1} - \sum\limits_{i = 1}^m {{b_i}{x_{i + 1}}} }}{{{x_1}}}, {x_1} \ne 0}\\ {\frac{{\mu {y_1} - \sum\limits_{i = 1}^m {{b_i}{y_{i + 1}}} }}{{{y_1}}}, {y_1} \ne 0} \end{aligned}} \right., \\ &&{a_i} = \left\{ {\begin{aligned} {\frac{{{\lambda _1}{x_i} - {b_{i - 1}}{x_1}}}{{{x_i}}}, {x_i} \ne 0}\\ {\frac{{\mu {y_i} - {b_{i - 1}}{y_1}}}{{{y_i}}}, {y_i} \ne 0} \end{aligned}} \right.{\rm{ }}\left( {i = 2, \ldots , m} \right), \\ &&{a_{m{\rm{ + }}1}} = \frac{{{\lambda _1}{x_{m + 1}}{x_{m + 2}}{y_1} + \left( {{\lambda _2}{x_{m + 1}}{y_{m + 2}} - \mu {x_{m + 2}}{y_{m + 1}}} \right){x_1}}}{\omega }, \end{eqnarray*} $

$ \begin{eqnarray*} &&{a_i} = \left\{ {\begin{aligned} {\frac{{{\lambda _2}{x_i} - {b_{i - 1}}{x_{i - 1}} - {b_i}{x_{i + 1}}}}{{{x_i}}}, {x_i} \ne 0, }\\ {\frac{{\mu {y_i} - {b_{i - 1}}{y_{i - 1}} - {b_i}{y_{i + 1}}}}{{{y_i}}}, {y_i} \ne 0, } \end{aligned}} \right.{\rm{ }} \quad {i = m + 2, \cdots , n - 1} , \\ &&{a_n} = \left\{ {\begin{aligned} {\frac{{{\lambda _2}{x_n} - {b_{n - 1}}{x_{n - 1}}}}{{{x_n}}}, {x_n} \ne 0, }\\ {\frac{{\mu {y_n} - {b_{n - 1}}{y_{n - 1}}}}{{{y_n}}}, {y_n} \ne 0.} \end{aligned}} \right.{\rm{ }} \end{eqnarray*} $

证 充分性  因$ ( {{\lambda _1}, {x_1}}), ({\mu , {y}}) $分别是$ {A_{1, m + 1}}, {A} $特征对, 有$ {A_{1, m + 1}}{x_1} = {\lambda _1}{x_1}, $ $ {A}{y} = \mu {y}. $

(1) 当$ 2 \le i \le m $时, 有如下线性方程组

$ \begin{equation} \left\{ \begin{matrix} {{b}_{i-1}}{{x}_{1}}+{{a}_{i}}{{x}_{i}} = {{\lambda }_{1}}{{x}_{i}}, \\ {{b}_{i-1}}{{y}_{1}}+{{a}_{i}}{{y}_{i}} = \mu {{y}_{i}}, \\ \end{matrix} \right.{\rm{ }} \quad {i = 2, \cdots , m} , \end{equation} $

(2.1)

由上式消去$ {a_i} $, 即$ {b_{i{\rm{ - 1}}}}{E_i} = \left( {\lambda - \mu } \right){d_i}. $由条件(ii) $ {E_i} \ne {\rm{0}}\left( {i = 2, \cdots, m}\right), $则$ {a_i}, {b_i} $有唯一解

$ \begin{eqnarray} &&{b_i} = \frac{{\left( {\lambda - \mu } \right){d_{i + 1}}}}{{{E_{i + 1}}}}{\rm{ }}, \quad {i = 1, \cdots , m - 1}, \end{eqnarray} $

(2.2)

$ \begin{eqnarray} &&{a_i} = \left\{ {\begin{aligned} {\frac{{\lambda {x_i} - {b_{i - 1}}{x_1}}}{{{x_i}}}, {x_i} \ne 0}, \\ {\frac{{\mu {y_i} - {b_{i - 1}}{y_1}}}{{{y_i}}}, {y_i} \ne 0, } \end{aligned}} \right.{\rm{ }}\quad {i = 2, \cdots , m}. \end{eqnarray} $

(2.3)

下证(2.3)中$ {a_i} $的两个表达式等价, 由(2.2)式知

$ \begin{equation*} \begin{aligned} &\frac{{\lambda {x_i} - {b_{i - 1}}{x_1}}}{{{x_i}}} - \frac{{\mu {y_i} - {b_{i - 1}}{y_1}}}{{{y_i}}} = \frac{{\left[ {\lambda {x_i} - \left( {\lambda - \mu } \right){d_i}/{E_i}} \right]{x_1}{y_i} - \left[ {\mu {y_i} - \left( {\lambda - \mu } \right){d_i}/{E_i}} \right]{y_1}{x_i}}}{{{x_i}{y_i}}}\\ = & \frac{{\left( {\lambda - \mu } \right){x_i}{y_i} - \left( {\lambda - \mu } \right){d_i}\left( {{x_1}{y_i} - {x_i}{y_1}} \right)/{E_i}}}{{{x_i}{y_i}}} = \frac{{\left( {\lambda - \mu } \right){x_i}{y_i} - \left( {\lambda - \mu } \right){d_i}}}{{{x_i}{y_i}}}{\rm{ = }}0. \end{aligned} \end{equation*} $

所以(2.3)式中$ {a_i} $的两个表达式等价得证.

(2) 当$ i = m + 1 $时, 由$ {A_{1, m + 1}}{x_1} = {\lambda _1}{x_1}, {A}{y} = \mu {y}, {A_{m + 1, n}}{x_2} = {\lambda _2}{x_2} $得下线性方程组

$ \begin{equation} \left\{ {\begin{aligned} &{{b_m}{x_1}{\rm{ + }}{a_{m + 1}}{x_{m + 1}} = {\lambda _1}{x_{m + 1}}, }\\ &{{b_m}{y_1}{\rm{ + }}{a_{m + 1}}{y_{m + 1}} + {b_{m + 1}}{y_{m + 2}} = \mu {y_{m + 1}}, }\\ &{{a_{m + 1}}{x_{m + 1}} + {b_{m + 1}}{x_{m + 2}} = {\lambda _2}{x_{m + 1}}.} \end{aligned}} \right. \end{equation} $

(2.4)

由条件(ii) $ \omega \ne 0 $, 其中$ \omega {\rm{ = }}{x_1}{x_{m + 1}}{y_{m + 2}} - {x_1}{x_{m + 2}}{y_{m + 1}} + {x_{m + 1}}{x_{m + 2}}{y_1} $, 解得

$ \begin{equation} \left\{ {\begin{aligned} &{{b_m} = \frac{{\left[ {\left( {{\lambda _1} - {\lambda _2}} \right){x_{m + 1}}{y_{m + 2}} + \left( {\mu - {\lambda _1}} \right){x_{m + 2}}{y_{m + 1}}} \right]{x_{m + 1}}}}{\omega }, }\\ &{{b_{m{\rm{ + }}1}} = \frac{{\left[ { - \left( {{\lambda _1} - {\lambda _2}} \right){x_{m + 1}}{y_1} + \left( {\mu - {\lambda _1}} \right){x_1}{y_{m + 1}}} \right]{x_{m + 1}}}}{\omega }, }\\ &{{a_{m{\rm{ + }}1}} = \frac{{{\lambda _1}{x_{m + 1}}{x_{m + 2}}{y_1} + {\lambda _2}{x_1}{x_{m + 1}}{y_{m + 2}} - \mu {x_1}{x_{m + 2}}{y_{m + 1}}}}{\omega }.} \end{aligned}} \right. \end{equation} $

(2.5)

(3) 当$ i = {\rm{1}} $时, 由$ {A_{1, m + 1}}{x_1} = {\lambda _1}{x_1}, {A{y_1}} = \mu {y_1} $, 得

$ \begin{equation} \left\{ {\begin{array}{*{20}{c}} &{{a_1}{x_1} + \sum\limits_{i = 1}^m {{b_i}{x_{i + 1}}} = {\lambda _1}{x_1}}, \\ &{{a_1}{y_1} + \sum\limits_{i = 1}^m {{b_i}{y_{i + 1}}} = \mu {y_1}}, \end{array}} \right. \end{equation} $

(2.6)

则有

$ \begin{equation} {a_1} = \left\{ {\begin{aligned} &{\frac{{{\lambda _1}{x_1} - \sum\limits_{i = 1}^m {{b_i}{x_{i + 1}}} }}{{{x_1}}}, {x_1} \ne 0}, \\ &{\frac{{\mu {y_1} - \sum\limits_{i = 1}^m {{b_i}{y_{i + 1}}} }}{{{y_1}}}, {y_1} \ne 0}, \end{aligned}} \right. \end{equation} $

(2.7)

将$ {b_i}\left( {i = 1, \cdots , m} \right) $的表达式带入$ {a_1} $, 利用条件(iii)可证得$ {a_1} $的两个表达式都存在则相等.

下证上述表达式相等.由条件(iii)有

$ \begin{equation*} \left( {{\lambda _1} - \mu } \right)\sum\limits_{i = 1}^{m + 1} {{x_i}{y_i} + } {b_{m + 1}}{x_{m + 1}}{y_{m + 2}} = 0 \Rightarrow \left( {{\lambda _1} - \mu } \right)\sum\limits_{i = 1}^m {{x_i}{y_i} + } {b_m}{E_{m + 1}} = 0, \end{equation*} $

则有

$ \begin{equation*} \left( {{\lambda _1} - \mu } \right){d_1} + \sum\limits_{i = 1}^{m - 1} {{b_i}{E_{i + 1}} + } {b_m}{E_{m + 1}} = 0 \Rightarrow \left( {{\lambda _1} - \mu } \right){d_1} + \sum\limits_{i = 1}^m {{b_i}{E_{i + 1}}} = 0, \end{equation*} $

则可推出

$ \begin{equation} {\lambda _1}{x_1}{y_1} - \sum\limits_{i = 1}^m {{b_i}{x_{i + 1}}{y_1}} {\rm{ = }}\mu {x_1}{y_1} - \sum\limits_{i = 1}^m {{b_i}{x_1}{y_{i + 1}}} . \end{equation} $

(2.8)

在(2.8)式两边同时除以$ {x_1}{y_1}, $得证$ {a_1} $的两个表达式相等.

(4) 当$ m{\rm{ + }}2 \le i \le n{\rm{ - 1}} $时, 由$ {A_{m + 1, n}}{x_2} = {\lambda _2}{x_2}, {A}{y} = \mu {y} $, 得

$ \begin{equation} \left\{ {\begin{array}{*{20}{c}} {{b_{i - 1}}{x_{i - 1}} + {a_i}{x_i} + {b_i}{x_{i + 1}} = {\lambda _2}{x_i}, {\rm{ }}}\\ {{b_{i - 1}}{y_{i - 1}} + {a_i}{y_i} + {b_i}{y_{i + 1}} = \mu {y_i}, {\rm{ }}} \end{array}} \right.\quad {i = m + 2, \cdots , n - 1} , \end{equation} $

(2.9)

由上式消去$ {a_i} $得

$ \begin{equation*} {b_{i - 1}}\left( {{x_{i - 1}}{y_i} - {y_{i - 1}}{x_i}} \right) + {b_i}\left( {{x_{i + 1}}{y_i} - {y_{i + 1}}{x_i}} \right) = \left( {{\lambda _2} - \mu } \right){x_i}{y_i}, \end{equation*} $

即为$ {b_i}{D_i} = {b_{i - 1}}{D_{i - 1}} - \left( {{\lambda _2} - \mu } \right){d_i}. $递推可得

$ \begin{equation} {b_i}{D_i} = {b_{m{\rm{ + }}1}}{D_{m + 1}} - \left( {{\lambda _2} - \mu } \right)\sum\limits_{j = m + 2}^i {{d_j}} \left( {i = m + 2, \cdots .n - 1} \right). \end{equation} $

(2.10)

由条件(ii) $ {D_i} \ne 0, $则$ {a_i}, {b_i}\left( {i = m + 2, \cdots, n - 1} \right) $有唯一解且有

$ \begin{eqnarray} &&{b_i} = \frac{{{b_{m{\rm{ + }}1}}{D_{m + 1}} - \left( {{\lambda _2} - \mu } \right)\sum\limits_{j = m + 2}^i {{d_j}} }}{{{D_i}}}, \quad {i = m + 2, \cdots , n - 1} , \end{eqnarray} $

(2.11)

$ \begin{eqnarray} &&{a_i} = \left\{ {\begin{aligned} {\frac{{{\lambda _2}{x_i} - {b_{i - 1}}{x_{i - 1}} - {b_i}{x_{i + 1}}}}{{{x_i}}}, {x_i} \ne 0}, \\ {\frac{{\mu {y_i} - {b_{i - 1}}{y_{i - 1}} - {b_i}{y_{i + 1}}}}{{{y_i}}}, {y_i} \ne 0}, \end{aligned}} \right.{\rm{ }}\quad {i = m + 2, \cdots, n - 1}. \end{eqnarray} $

(2.12)

下证式(2.12)中$ {a_i} $的两个表达式等价, 由(2.11), (2.12)式知

$ \begin{eqnarray} &&\frac{{{\lambda _2}{x_i} - {b_{i - 1}}{x_{i - 1}} - {b_i}{x_{i + 1}}}}{{{x_i}}} - \frac{{\mu {y_i} - {b_{i - 1}}{y_{i - 1}} - {b_i}{y_{i + 1}}}}{{{y_i}}}\\ & = & \left( {{\lambda _2} - \mu } \right){d_i} - {b_{m + 1}}{D_{m + 1}}+ \left( {{\lambda _2} - \mu } \right)\sum\limits_{j = m + 2}^{i - 1} {{d_j}} + {b_{m + 1}}{D_{m + 1}} - \left( {{\lambda _2} - \mu } \right)\sum\limits_{j = m + 2}^i {{d_j}} = 0, \end{eqnarray} $

(2.13)

所以(2.12)式中$ {a_i} $的两个表达式等价得证.

(5) 当$ i{\rm{ = }}n $时, 可得$ {a_n} $的表达式如下

$ \begin{equation} {a_n} = \left\{ {\begin{aligned} {\frac{{{\lambda _2}{x_n} - {b_{n - 1}}{x_{n - 1}}}}{{{x_n}}}, {x_n} \ne 0}, \\ {\frac{{\mu {y_n} - {b_{n - 1}}{y_{n - 1}}}}{{{y_n}}}, {y_n} \ne 0}. \end{aligned}} \right. \end{equation} $

(2.14)

根据(2.11)式, 利用条件(iii), 可证得$ {a_n} $的两个表达式都存在则相等.

根据条件(i)和(iv)可知所求$ {a_i}, {b_i} $满足问题I要求, 问题I有唯一解, 充分性得证, 且给出解的表达式(2.2), (2.3), (2.5), (2.7), (2.11), (2.12), (2.14).

必要性  若问题Ⅰ有唯一解, 则上述线性方程组(2.1), (2.4), (2.6), (2.9)有唯一解, 则可以推条件(ii)成立, 又因为矩阵$ {A_n} $的顺序主子式$ {A_{m + 1, n}} $为Jacobi矩阵, 若问题Ⅰ有解, 则条件(i)和(iv)成立.下证条件(iii)成立.

若问题Ⅰ有解则要满足$ {A_{1, m + 1}}{x_1} = {\lambda _1}{x_1}, {A}{y} = \mu {y}, $则有

$ \begin{eqnarray} &&\left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}& \ldots &{{b_m}}\\ {{b_1}}&{{a_2}}&{}&{}\\ \vdots &{}& \ddots &{}\\ {{b_m}}&{}&{}&{{a_{m + 1}}} \end{array}} \right]{\rm{ }}\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ \vdots \\ {{x_{m + 1}}} \end{array}} \right]{\rm{ = }}{\lambda _1}\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ \vdots \\ {{x_{m + 1}}} \end{array}} \right], \end{eqnarray} $

(2.15)

$ \begin{eqnarray} &&\left[ {\begin{array}{*{20}{c}} {{a_1}}&{{b_1}}& \ldots &{{b_m}}\\ {{b_1}}&{{a_2}}&{}&{}\\ \vdots &{}& \ddots &{}\\ {{b_m}}&{}&{}&{{a_{m + 1}}} \end{array}} \right]{\rm{ }}\left[ {\begin{array}{*{20}{c}} y\\ {{y_2}}\\ \vdots \\ {{y_{m + 1}}} \end{array}} \right]{\rm{ = }}\mu \left[ {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ \vdots \\ {{y_{m + 1}}} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 0\\ 0\\ \vdots \\ {{b_{m + 1}}{y_{m + 2}}} \end{array}} \right]. \end{eqnarray} $

(2.16)

在(2.15)式两边同时左乘$ {x_1} $得

$ \begin{equation} {\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ \vdots \\ {{x_{m + 1}}} \end{array}} \right]^{\rm T}}\left[ {\begin{array}{*{20}{c}} {{a_1} - \mu }&{{b_1}}& \ldots &{{b_m}}\\ {{b_1}}&{{a_2} - \mu }&{}&{}\\ \vdots &{}& \ddots &{}\\ {{b_m}}&{}&{}&{{a_{m + 1}} - \mu } \end{array}} \right]{\rm{ }}\left[ {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ \vdots \\ {{y_{m + 1}}} \end{array}} \right]{\rm{ = }}{\left[ {\begin{array}{*{20}{c}} {{x_1}}\\ {{x_2}}\\ \vdots \\ {{x_{m + 1}}} \end{array}} \right]^{\rm T}}\left[ {\begin{array}{*{20}{c}} 0\\ 0\\ \vdots \\ {{b_{m + 1}}{y_{m + 2}}} \end{array}} \right], \end{equation} $

(2.17)

结合式(2.14), 则式(2.16)为$ \left( {{\lambda _1} - \mu } \right)\sum\limits_{i = i}^{m + 1} {{x_i}{y_i} + } {b_{m + 1}}{x_{m + 1}}{y_{m + 2}} = 0. $若问题Ⅰ有解同时需要满足$ {A_{m + 1, n}}{x_2} = {\lambda _2}{x_2}, {A}{y} = \mu {y} $, 则有

$ \begin{eqnarray} && {\rm{ }}\left[ {\begin{array}{*{20}{c}} {{a_{m + 1}}}&{{b_{m + 1}}}&{}&{}\\ {{b_{m + 1}}}&{{a_{m + 2}}}& \ddots &{}\\ {}& \ddots & \ddots &{{b_{n - 1}}}\\ {}&{}&{{b_{n - 1}}}&{{a_n}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{x_{m + 1}}}\\ {{x_{m + 2}}}\\ \vdots \\ {{x_n}} \end{array}} \right] = {\lambda _2}\left[ {\begin{array}{*{20}{c}} {{x_{m + 1}}}\\ {{x_{m + 2}}}\\ \vdots \\ {{x_n}} \end{array}} \right], \end{eqnarray} $

(2.18)

$ \begin{eqnarray} && {\rm{ }}\left[ {\begin{array}{*{20}{c}} {{a_{m + 1}}}&{{b_{m + 1}}}&{}&{}\\ {{b_{m + 1}}}&{{a_{m + 2}}}& \ddots &{}\\ {}& \ddots & \ddots &{{b_{n - 1}}}\\ {}&{}&{{b_{n - 1}}}&{{a_n}} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{y_{m + 1}}}\\ {{y_{m + 2}}}\\ \vdots \\ {{y_n}} \end{array}} \right] = \mu \left[ {\begin{array}{*{20}{c}} {{y_{m + 1}}}\\ {{y_{m + 2}}}\\ \vdots \\ {{y_n}} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {{b_m}{y_1}}\\ 0\\ \vdots \\ 0 \end{array}} \right], \end{eqnarray} $

(2.19)

$ \begin{eqnarray} &&{\left[ {\begin{array}{*{20}{c}} {{x_{m + 1}}}\\ {{x_{m + 2}}}\\ \vdots \\ {{x_n}} \end{array}} \right]^{\rm T}}\left[ {\begin{array}{*{20}{c}} {{a_{m + 1}} - \mu }&{{b_{m + 1}}}&{}&{}\\ {{b_{m + 1}}}&{{a_{m + 2}} - \mu }& \ddots &{}\\ {}& \ddots & \ddots &{{b_{n - 1}}}\\ {}&{}&{{b_{n - 1}}}&{{a_n} - \mu } \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{y_{m + 1}}}\\ {{y_{m + 2}}}\\ \vdots \\ {{y_n}} \end{array}} \right]\\ & = & - {\left[ {\begin{array}{*{20}{c}} {{x_{m + 1}}}\\ {{x_{m + 2}}}\\ \vdots \\ {{x_n}} \end{array}} \right]^{\rm T}}\left[ {\begin{array}{*{20}{c}} {{b_m}{y_1}}\\ 0\\ \vdots \\ 0 \end{array}} \right]. \end{eqnarray} $

(2.20)

结合式(2.17), 则式(2.19)为$ \left( {{\lambda _2} - \mu } \right)\sum\limits_{j = m + 1}^n {{x_i}} {y_i} + {b_m}{x_{m + 1}}{y_1} = 0. $条件(iii)得证, 即必要性得证.

下面讨论问题Ⅱ, 给出定理2.2.

定理2.2   问题Ⅱ有唯一解的充分必要条件为

(i) $ {D_i} \ne 0\left( {i = 1, \cdots , m - 1, m + 1, \cdots , n - 1} \right), {E_{m + 1}} \ne 0, $ $ x_i^2 + y_i^2 \ne 0 \, \left( {i = m + 1, \cdots , n} \right), $ $ {x_1} \ne 0, $ $ {y_1} \ne 0 $;

(ii) $ \frac{{\lambda - \mu }}{{{D_i}}}\sum\limits_{j = i + 1}^n {d{}_j} > 0 \, \left( {i = m + 1, \cdots , n - 1} \right) $,

且当问题有解时, 解由下式给出

$ \begin{eqnarray*} &&{b_i} = \frac{{\left( {\lambda - \mu } \right){d_{i{\rm{ + }}1}}}}{{{D_i}}}, \quad {i = 1, \cdots , m - 1}, \\ &&{b_m} = \frac{{\left( {\lambda - \mu } \right){d_{m + 1}} + {b_{m + 1}}{D_{m + 1}}}}{{{E_{m + 1}}}} = \frac{{\lambda - \mu }}{{{E_{m + 1}}}}\sum\limits_{j = m + 1}^n {d{}_j}, \\ &&{b_i} = \frac{{\lambda - \mu }}{{{D_i}}}\sum\limits_{j = i + 1}^n {d{}_j}, \quad {i = m + 1, \cdots , n - 1}, \\ &&\begin{aligned} {a_1} = \frac{{\lambda {x_1} - \sum\limits_{i = 1}^m {{b_i}{x_{i + 1}}} }}{{{x_1}}}, {x_1} \ne 0, a_1^* = \frac{{\mu {y_1} - \sum\limits_{i = 1}^m {{b_i}{y_{i + 1}}} }}{{{y_1}}}, {y_1} \ne 0, \end{aligned}\\ &&{a_{i + 1}} = \frac{{\mu {x_1}{y_{i + 1}} - \lambda {x_{i + 1}}{y_1}}}{{{E_{i + 1}}}}, {\rm{ }}\quad {i = 1, \cdots , m - 1} , \\ &&{a_{m + 1}} = \left\{ {\begin{aligned} {\frac{{\lambda {x_{m + 1}} - {b_m}{x_1} - {b_{m + 1}}{x_{m + 2}}}}{{{x_{m + 1}}}}, {x_{m + 1}} \ne 0}, \\ {\frac{{\mu {y_{m + 1}} - {b_m}{y_1} - {b_{m + 1}}{y_{m + 2}}}}{{{y_{m + 1}}}}, {y_{m + 1}} \ne 0}, \end{aligned}} \right.\\ &&{a_{i{\rm{ + }}1}} = \left\{ {\begin{aligned} {\frac{{\lambda {x_{i{\rm{ + }}1}} - {b_i}{x_i} - {b_{i{\rm{ + }}1}}{x_{i + 2}}}}{{{x_{i{\rm{ + }}1}}}}, {x_{i{\rm{ + }}1}} \ne 0}, \\ {\frac{{\mu {y_{i{\rm{ + }}1}} - {b_i}{y_i} - {b_{i{\rm{ + }}1}}{y_{i + 2}}}}{{{y_{i + 1}}}}, {y_{i{\rm{ + 1}}}} \ne 0}, \end{aligned}} \right.{\rm{ }}\quad {i = m + 1, \cdots , n - 2}, \\ &&{a_n} = \left\{ {\begin{aligned} {\frac{{\lambda {x_n} - {b_{n - 1}}{x_{n - 1}}}}{{{x_n}}}, {x_n} \ne 0}, \\ {\frac{{\mu {y_n} - {b_{n - 1}}{y_{n - 1}}}}{{{y_n}}}, {y_n} \ne 0}. \end{aligned}} \right.{\rm{ }} \end{eqnarray*} $

证 必要性  由于$ \left( {\lambda , {x}} \right), \, \left( {\mu , {y}} \right) $分别为$ {A}, \, {A^*} $的特征对, 所以有

$ \begin{equation} \left\{ {\begin{array}{*{20}{c}} {{A}{x} = \lambda {x}}, \\ {{A^{\rm{*}}}{y} = \mu {y}}. \end{array}} \right. \end{equation} $

(2.21)

则上式可表示为

$ \begin{eqnarray} &&\begin{array}{l} {a_1}{x_1} + \sum\limits_{i = 1}^m {{b_i}{x_{i + 1}}} = \lambda {x_1}, \\ a_1^{\rm{*}}{y_1} + \sum\limits_{i = 1}^m {{b_i}{y_{i + 1}}} = \mu {y_1}, \end{array} \end{eqnarray} $

(2.22)

$ \begin{eqnarray} &&\begin{array}{l} {b_i}{x_1} + {a_{i{\rm{ + }}1}}{x_{i{\rm{ + }}1}} = \lambda {x_{i{\rm{ + }}1}}, \, {i = 1, \cdots , m{\rm{ - }}1}, \\ {b_i}{y_1} + {a_{i{\rm{ + }}1}}{y_{i{\rm{ + }}1}} = \mu {y_{i{\rm{ + }}1}}, {i = 1, \cdots , m{\rm{ - }}1}, \end{array} \end{eqnarray} $

(2.23)

$ \begin{eqnarray} &&\begin{array}{l} {b_m}{x_1} + {a_{m + 1}}{x_{m + 1}} + {b_{m + 1}}{x_{m + 2}} = \lambda {x_{m + 1}}, \\ {b_m}{y_1} + {a_{m + 1}}{y_{m + 1}} + {b_{m + 1}}{y_{m + 2}} = \mu {y_{m + 1}}, \end{array} \end{eqnarray} $

(2.24)

$ \begin{eqnarray} &&\begin{array}{l} {b_i}{x_i} + {a_{i + 1}}{x_{i + 1}} + {b_{i + 1}}{x_{i + 2}} = \lambda {x_{i + 1}}, {i = m + 1, \cdots , n - 2}, \\ {b_i}{y_i} + {a_{i + 1}}{y_{i + 1}} + {b_{i + 1}}{y_{i + 2}} = \mu {y_{i + 1}}, \, {i = m + 1, \cdots , n - 2}, \end{array} \end{eqnarray} $

(2.25)

$ \begin{eqnarray} &&\begin{array}{l} {b_{n - 1}}{x_{n - 1}} + {a_n}{x_n} = \lambda {x_n}, \\ {b_{n - 1}}{y_{n - 1}} + {a_n}{y_n} = \mu {y_n}. \end{array} \end{eqnarray} $

(2.26)

问题Ⅱ的解等价于求解上述线性方程组

由式(2.25)消去$ {a_n} $, 由条件(i) $ {D_{n - 1}} \ne 0, x_n^2 + y_n^2 \ne 0 $知$ {b_{n - 1}}, {a_n} $有唯一解, 所以有

$ \begin{eqnarray} &&{b_{n - 1}} = \frac{{\left( {\lambda - \mu } \right){d_n}}}{{{D_{n - 1}}}}, \end{eqnarray} $

(2.27)

$ \begin{eqnarray} &&{a_n} = \left\{ {\begin{aligned} {\frac{{\lambda {x_n} - {b_{n - 1}}{x_{n - 1}}}}{{{x_n}}}, {x_n} \ne 0}, \\ {\frac{{\mu {y_n} - {b_{n - 1}}{y_{n - 1}}}}{{{y_n}}}, {y_n} \ne 0}. \end{aligned}} \right. \end{eqnarray} $

(2.28)

利用(2.26)式易证明(2.27)式中$ {a_n} $的两个表达式等价.

利用式(2.26)由条件(i)知$ {D_i} \ne 0 $, 则$ {b_i} $有唯一解

$ \begin{equation*} {b_i} = \frac{{\left( {\lambda - \mu } \right){d_{i{\rm{ + }}1}} + {b_{i + 1}}{D_{i + 1}}}}{{{D_i}}}, \quad{i = m + 1, \cdots , n - 2}. \end{equation*} $

通过递推得

$ \begin{equation*} {b_i} = \frac{{\lambda - \mu }}{{{D_i}}}\left( {{d_{i + 1}} + {d_{i + 2}} + \ldots + {d_{n - 1}}} \right) + \frac{{{D_{n - 1}}}}{{{D_i}}}{b_{n - 1}} = \frac{{\lambda - \mu }}{{{D_i}}}\sum\limits_{j = i + 1}^n {d{}_j}, \quad{i = m + 1, \cdots , n - 2}. \end{equation*} $

结合式(2.21), 则$ {b_i} $的表达式为

$ \begin{eqnarray} &&{b_i} = \frac{{\lambda - \mu }}{{{D_i}}}\sum\limits_{j = i + 1}^n {d{}_j}, \quad{i = m + 1, \cdots , n - 1}, \end{eqnarray} $

(2.29)

$ \begin{eqnarray} &&{a_{i{\rm{ + }}1}} = \left\{ {\begin{aligned} {\frac{{\lambda {x_{i{\rm{ + }}1}} - {b_i}{x_i} - {b_{i{\rm{ + }}1}}{x_{i + 2}}}}{{{x_{i{\rm{ + }}1}}}}, {x_{i{\rm{ + }}1}} \ne 0}, \\ {\frac{{\mu {y_{i{\rm{ + }}1}} - {b_i}{y_i} - {b_{i{\rm{ + }}1}}{y_{i + 2}}}}{{{y_{i{\rm{ + }}1}}}}, {y_{i{\rm{ + 1}}}} \ne 0}, \end{aligned}} \right.\quad {i = m + 1, \cdots , n - 2}. \end{eqnarray} $

(2.30)

利用(2.28)式易证(2.29)式中$ {a_{i + 1}} $的两个表达式等价.

由式(2.23)以及条件(i)知$ {E_{m + 1}} \ne 0 $, 则$ {b_m}, {a_{m + 1}} $有唯一解

$ \begin{eqnarray} &&{b_m} = \frac{{\left( {\lambda - \mu } \right){d_{m + 1}} + {b_{m + 1}}{D_{m + 1}}}}{{{E_{m + 1}}}} = \frac{{\lambda - \mu }}{{{E_{m + 1}}}}\sum\limits_{j = m + 1}^n {d{}_j}, \end{eqnarray} $

(2.31)

$ \begin{eqnarray} &&{a_{m + 1}} = \left\{ {\begin{aligned} {\frac{{\lambda {x_{m + 1}} - {b_m}{x_1} - {b_{m + 1}}{x_{m + 2}}}}{{{x_{m + 1}}}}, {x_{m + 1}} \ne 0}, \\ {\frac{{\mu {y_{m + 1}} - {b_m}{y_1} - {b_{m + 1}}{y_{m + 2}}}}{{{y_{m + 1}}}}, {y_{m + 1}} \ne 0}. \end{aligned}} \right. \end{eqnarray} $

(2.32)

利用(2.30)式可证(2.31)式中$ {a_{m + 1}} $的两个表达式等价.

由式(2.21)和条件(i) $ {x_1} \ne 0, {x_2} \ne 0 $, 得

$ \begin{equation} {a_1} = \frac{{\lambda {x_1} - \sum\limits_{i = 1}^m {{b_i}{x_{i + 1}}} }}{{{x_1}}}, {\rm{ }}a_1^* = \frac{{\mu {y_1} - \sum\limits_{i = 1}^m {{b_i}{y_{i + 1}}} }}{{{y_1}}}. \end{equation} $

(2.33)

由条件(i)知$ {D_i} \ne {\rm{0}}\left( {i = 1, \cdots , m - 1} \right) $, 则$ {b_i}, {a_{i + 1}} $有唯一解

$ \begin{eqnarray} &&{b_i} = \frac{{\left( {\lambda - \mu } \right){d_{i{\rm{ + }}1}}}}{{{D_i}}}, \quad{i = 1, \cdots , m - 1}, \end{eqnarray} $

(2.34)

$ \begin{eqnarray} &&{a_{i + 1}} = \frac{{\mu {x_1}{y_{i + 1}} - \lambda {x_{i + 1}}{y_1}}}{{{E_{i + 1}}}}, \quad{i = 1, \cdots , m - 1}. \end{eqnarray} $

(2.35)

充分性得证, 且给出问题Ⅱ解的表达式(2.27)-(2.35).

必要性  若问题Ⅱ有唯一解, 则上述线性方程组(2.21)-(2.25)有唯一解, 则可以推得条件(i)成立.若问题Ⅱ有解, 根据矩阵$ A, {A^*} $的特殊性以及$ {b_i}\left( {i = m + 1, \cdots , n - 1} \right) $的表达式, 则条件(ii)成立.

步骤1  验算所给$ {\lambda _1}, {\lambda _2}, \mu $以及三个非零实向量

$ {x_1} = {\left( {{x_1}, \cdots , {x_{m + 1}}} \right)^{\rm T}}, {x_2} = {\left( {{x_{m + 1}}, \cdots , {x_n}} \right)^{\rm T}}, {y} = {\left( {{y_1}, \cdots , {y_n}} \right)^{\rm T}} $

是否满足定理2.1的条件(i)-(iv).是, 则进行下一步; 否则, 停止.

步骤2  根据定理2.1中的公式(2.2), (2.3), (2.5), (2.7), (2.11), (2.12), (2.14), 求解$ {a_i}, {b_i} $, 形成广义箭形矩阵$ {A} $.

例1   给实数$ {\lambda _1}{\rm{ = }}1.0879, {\lambda _2}{\rm{ = }}0.5689, \mu {\rm{ = }}0.9730, m = 3, n = 7 $给定实向量$ {x_1}, {x_2}, {y} $如下

$ \begin{eqnarray*} &&{x_1} = {\left( { - 0.5667, 0.0961, - 0.3903, 0.7193} \right)^{\rm T}}, {x_2} = {\left( {0.7193, 0.1702, - 0.4556, - 0.2676} \right)^{\rm T}}, \\ &&{y} = {\left( {0.4109, - 0.0752, 0.3080, - 0.4426, 0.1225, 0.6568, 0.2972} \right)^{\rm T}}. \end{eqnarray*} $

根据算法1中的步骤将$ {\lambda _1}, {\lambda _2}, \mu, {x_1}, {x_2}, {y} $带入定理2.1的条件(i)-(iv), 验算可知所给数据满足有唯一解的条件, 利用公式(2.2), (2.3), (2.5), (2.7), (2.11), (2.12), (2.14), 通过MATLAB编程计算$ {a_i}, {b_i} $, 形成广义箭形矩阵$ {A} $, 得到$ {A} $如下

$ {A} = \left[ {\begin{array}{*{20}{c}} { - 0.8026}&{ - 0.2651}&{0.9760}&{ - 0.9245}&{}&{}&{}\\ { - 0.2651}&{ - 0.4763}&{}&{}&{}&{}&{}\\ {0.9760}&{}&{ - 0.3293}&{}&{}&{}&{}\\ { - 0.9245}&{}&{}&{0.3595}&{0.8852}&{}&{}\\ {}&{}&{}&{0.8852}&{ - 0.7269}&{0.9133}&{}\\ {}&{}&{}&{}&{0.9133}&{0.4425}&{0.7962}\\ {}&{}&{}&{}&{}&{0.7962}&{ - 0.7865} \end{array}} \right]. $

容易验证$ \left( {{\lambda _1}, {x_1}} \right) $是$ {A_{1, 6}} $的一个特征对, $ \left( {{\lambda _2}, {x_2}} \right) $是$ {A_{5, 10}} $的一个特征对, $ \left( {\mu , y} \right) $是$ A $的一个特征对, 所以$ {A} $是所要求的矩阵.

步骤1  验算所给数据$ \lambda, $ $ \mu $和两个非零实向量$ {x} = {\left( {{x_1}, \cdots , {x_n}} \right)^{\rm T}}, $ $ {y} = {\left( {{y_1}, \cdots , {y_n}} \right)^{\rm T}} $是否满足定理2.2的要求.是, 则进行下一步; 否则, 停止.

步骤2  根据定理2.2中公式(2.27)-(2.35), 求解

$ {a_1}, a_1^*, {a_i}\left( {i = 2, \cdots , n} \right), {b_i}\left( {i = 1, \cdots , n - 1} \right) $

分别形成广义箭形矩阵$ {A}, {A^*} $, 使得$ \left( {\lambda , {x}} \right) $和$ \left( {\mu , {y}} \right) $分别为矩阵$ {A}, {A^*} $的特征对.

例2   给定实数$ \lambda {\rm{ = }} - 0.5435, \mu = - 0.0032, m = 3, n = 7 $, 给定实向量$ {x}, {y} $如下

$ \begin{eqnarray*} &&{x} = \left( { - 0.0047, - 0.0076, 0.0041, - 0.0472, - 0.2254, - 0.8072, - 0.5435} \right), \\ &&{y} = \left( { - 0.2201, 0.8056, 0.5498, 0.0149, 0.0056, - 0.0016, - 0.0032} \right). \end{eqnarray*} $

根据算法2中的步骤将$ \lambda , \mu, {x}, {y} $带入定理2.2中的(i)和(ii), 验算可知所给数据满足有唯一解的条件, 根据式(2.27)-(2.35), 通过Matlab编程计算得到$ {a_1}, a_1^*, {a_i}\, \left( {i = 2, \cdots , n} \right), $ $ {b_i}\, \left( {i = 1, \cdots , n - 1} \right) $, 并形成广义箭形矩阵$ {A} $和$ {A^*} $, 如下所示

$ \begin{eqnarray*} &&{A} = \left[ {\begin{array}{*{20}{c}} { - 0.3551}&{0.8320}&{ - 0.9977}&{ - 0.0751}&{}&{}&{}\\ {0.8320}&{0.5695}&{}&{}&{}&{}&{}\\ { - 0.9977}&{}&{ - 0.0573}&{}&{}&{}&{}\\ { - 0.0751}&{}&{}&{ - 0.9285}&{0.4243}&{}&{}\\ {}&{}&{}&{0.4243}&{ - 0.6483}&{0.4609}&{}\\ {}&{}&{}&{}&{0.4609}&{0.4435}&{0.7702}\\ {}&{}&{}&{}&{}&{0.7702}&{ - 0.0530} \end{array}} \right], \\&&{A^*} = \left[ {\begin{array}{*{20}{c}} {0.8903}&{0.8320}&{ - 0.9977}&{ - 0.0751}&{}&{}&{}\\ {0.8320}&{0.5695}&{}&{}&{}&{}&{}\\ { - 0.9977}&{}&{ - 0.0573}&{}&{}&{}&{}\\ { - 0.0751}&{}&{}&{ - 0.9285}&{0.4243}&{}&{}\\ {}&{}&{}&{0.4243}&{ - 0.6483}&{0.4609}&{}\\ {}&{}&{}&{}&{0.4609}&{0.4435}&{0.7702}\\ {}&{}&{}&{}&{}&{0.7702}&{ - 0.0530}. \end{array}} \right]. \end{eqnarray*} $

易知$ \left( {\lambda , {x}} \right) $是矩阵$ {A} $的一个特征对, $ \left( {\mu , {y}} \right) $为矩阵$ {A^*} $的一个特征对, 所以$ {A}, {A^*} $是所要求的矩阵.