# 关于特殊矩阵求特征值的新方法A New Method to Find Eigenvalues of Special Matrices

DOI: 10.12677/AAM.2018.74037, PDF, HTML, XML, 下载: 1,101  浏览: 2,090

Abstract: The eigenvalues of the matrix have an important role in many fields, but they are very complicated in practice. In this paper, we use the correlation property of determinant calculation and the multiplication of partitioned matrix to provide two general and simple methods for calculating eigenvalues of special matrices, simplifying calculation steps to improve computational efficiency.

1. 引言

2. 特殊矩阵的分块求特征值

$|{A}_{2n}-\lambda E|=|\left(A-\lambda E\right)\left(D-\lambda E\right)-CB|$

$|\begin{array}{cc}A& B\\ C& D\end{array}|=|AD-CB|$

$|{A}_{2n}-\lambda E|=|\begin{array}{cc}A-\lambda E& B\\ C& D-\lambda E\end{array}|$

$\left(A-\lambda E\right)C=AC-\lambda EC=AC-\lambda C=CA-C\lambda E$

$\left(A-\lambda E\right)C=C\left(A-\lambda E\right)$ 且矩阵 $A-\lambda E$ 可逆；

$|{A}_{2n}-\lambda E|=|\left(A-\lambda E\right)\left(D-\lambda E\right)-CB|$

$\begin{array}{c}|{A}_{4}-\lambda E|=|\begin{array}{cccc}1-\lambda & -1& -1& 1\\ -1& 1-\lambda & 1& -1\\ -1& 1& 1-\lambda & -1\\ 1& -1& -1& 1-\lambda \end{array}|=|\begin{array}{cccc}1-\lambda & -1& -1& 1\\ -1& 1-\lambda & 1& -1\\ 0& \lambda & -\lambda & 0\\ 0& 0& -\lambda & -\lambda \end{array}|\\ =\left(1-\lambda \right)|\begin{array}{ccc}1-\lambda & 1& -1\\ \lambda & -\lambda & 0\\ 0& -\lambda & -\lambda \end{array}|+|\begin{array}{ccc}-1& -1& 1\\ \lambda & -\lambda & 0\\ 0& -\lambda & -\lambda \end{array}|\\ =\left(1-\lambda \right)\left[\left(1-\lambda \right)|\begin{array}{cc}-\lambda & 0\\ -\lambda & -\lambda \end{array}|-\lambda |\begin{array}{cc}1& -1\\ -\lambda & -\lambda \end{array}|\right]+|\begin{array}{ccc}-1& -1& 1\\ 0& -2\lambda & \lambda \\ 0& -\lambda & -\lambda \end{array}|\\ ={\lambda }^{3}\left(\lambda -4\right)\end{array}$

2) 对于简便求解法：

$|\begin{array}{cc}A& B\\ C& D\end{array}|=|A-B{D}^{-1}C||D|=|AD-B{D}^{-1}CD|=|AD-BC|$

$|{A}_{2n}-\lambda E|=|\left(A-\lambda E\right)\left(D-\lambda E\right)-BC|$

$\begin{array}{c}|{A}_{4}-\lambda E|=|{A}_{2n}-\lambda E|=|\left(A-\lambda E\right)\left(D-\lambda E\right)-BC|\\ =|\left[\begin{array}{cc}{\left(2-\lambda \right)}^{2}& 0\\ 3-2\lambda & {\left(2-\lambda \right)}^{2}\end{array}\right]+\left[\begin{array}{cc}1& -0.75\\ 0& 0.25\end{array}\right]|\\ =\frac{1}{4}\lambda \left(2\lambda -3\right)\left(2\lambda -5\right)\left(\lambda -2\right)\end{array}$

3. 特殊矩阵的分解求特征值

${M}_{n}$ 的特征向量为：

$\lambda =0$ 时：

${\eta }_{1}=\left(\begin{array}{c}{a}_{2}\\ \begin{array}{c}-{a}_{1}\\ 0\\ ⋮\\ 0\end{array}\end{array}\right)$${\eta }_{2}=\left(\begin{array}{c}{a}_{3}\\ \begin{array}{c}0\\ -{a}_{1}\\ ⋮\\ 0\end{array}\end{array}\right)$${\eta }_{n-1}=\left(\begin{array}{c}{a}_{n}\\ \begin{array}{c}0\\ 0\\ ⋮\\ -{a}_{1}\end{array}\end{array}\right)$

$\lambda ={a}_{1}^{2}+{a}_{2}^{2}+\cdots +{a}_{n-1}^{2}+{a}_{n}^{2}$ 时：

${\eta }_{n}=\left(\begin{array}{c}{a}_{1}\\ \begin{array}{c}{a}_{2}\\ {a}_{3}\\ ⋮\\ {a}_{n}\end{array}\end{array}\right)$

$\begin{array}{c}{A}_{n}=|\lambda E-{A}^{\text{T}}A|=|\begin{array}{cccc}\lambda -{a}_{1}^{2}& -{a}_{1}{a}_{2}& \cdots & -{a}_{1}{a}_{n}\\ -{a}_{2}{a}_{1}& \lambda -{a}_{2}^{2}& \cdots & -{a}_{2}{a}_{n}\\ ⋮& ⋮& \ddots & ⋮\\ -{a}_{n}{a}_{1}& -{a}_{n}{a}_{2}& \cdots & \lambda -{a}_{n}^{2}\end{array}|\\ =|\begin{array}{cccc}\lambda & 0& \cdots & 0\\ -{a}_{2}{a}_{1}& \lambda -{a}_{2}^{2}& \cdots & -{a}_{2}{a}_{n}\\ ⋮& ⋮& \ddots & ⋮\\ -{a}_{n}{a}_{1}& -{a}_{n}{a}_{2}& \cdots & \lambda -{a}_{n}^{2}\end{array}|+|\begin{array}{cccc}-{a}_{1}^{2}& -{a}_{1}{a}_{2}& \cdots & -{a}_{1}{a}_{n}\\ -{a}_{2}{a}_{1}& \lambda -{a}_{2}^{2}& \cdots & -{a}_{2}{a}_{n}\\ ⋮& ⋮& \ddots & ⋮\\ -{a}_{n}{a}_{1}& -{a}_{n}{a}_{2}& \cdots & \lambda -{a}_{n}^{2}\end{array}|\\ =\lambda |\begin{array}{cccc}\lambda -{a}_{2}^{2}& 0& \cdots & 0\\ -{a}_{3}{a}_{2}& \lambda -{a}_{3}^{2}& \cdots & -{a}_{3}{a}_{n}\\ ⋮& ⋮& \ddots & ⋮\\ -{a}_{n}{a}_{2}& -{a}_{n}{a}_{3}& \cdots & \lambda -{a}_{n}^{2}\end{array}|+{a}_{1}|\begin{array}{cccc}{a}_{1}& {a}_{2}& \cdots & {a}_{n}\\ 0& \lambda & \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & \lambda \end{array}|\\ =\lambda {A}_{n-1}-{a}_{1}^{2}{\lambda }^{n-1}\end{array}$

$\left\{\begin{array}{l}{A}_{n}=\lambda {A}_{n-1}-{a}_{1}^{2}{\lambda }^{n-1}\\ {A}_{n-1}=\lambda {A}_{n-2}-{a}_{2}^{2}{\lambda }^{n-2}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}⋮\\ {A}_{2}=\lambda {A}_{1}-{a}_{n-1}^{2}\lambda \end{array}$

${A}_{n}={\lambda }^{n-1}{A}_{1}-\left({a}_{1}^{2}+{a}_{2}^{2}+\cdots +{a}_{n-1}^{2}\right){\lambda }^{n-1}={\lambda }^{n}-\left({a}_{1}^{2}+{a}_{2}^{2}+\cdots +{a}_{n-1}^{2}+{a}_{n}^{2}\right){\lambda }^{n-1}$

${\lambda }_{1}=0$ ( $n-1$ 重)， ${\lambda }_{2}={a}_{1}^{2}+{a}_{2}^{2}+\cdots +{a}_{n-1}^{2}+{a}_{n}^{2}$

$\lambda =0$ 时：

${\xi }_{1}=\left(\begin{array}{c}{a}_{2}\\ \begin{array}{c}-{a}_{1}\\ 0\\ ⋮\\ 0\end{array}\end{array}\right)$ , ${\xi }_{2}=\left(\begin{array}{c}{a}_{3}\\ \begin{array}{c}0\\ -{a}_{1}\\ ⋮\\ 0\end{array}\end{array}\right)$ , ${\xi }_{n-1}=\left(\begin{array}{c}{a}_{n}\\ \begin{array}{c}0\\ 0\\ ⋮\\ -{a}_{1}\end{array}\end{array}\right)$

$\lambda ={a}_{1}^{2}+{a}_{2}^{2}+\cdots +{a}_{n-1}^{2}+{a}_{n}^{2}$ 时：

${\xi }_{n}=\left(\begin{array}{c}{a}_{1}\\ \begin{array}{c}{a}_{2}\\ {a}_{3}\\ ⋮\\ {a}_{n}\end{array}\end{array}\right)$

2) 对于简便解答方法：

1) 若 $\left({a}_{1}\text{\hspace{0.17em}}{a}_{2}\text{\hspace{0.17em}}{a}_{3}\text{\hspace{0.17em}}\cdots \text{\hspace{0.17em}}{a}_{n}\right)\left(\begin{array}{c}{a}_{1}\\ \begin{array}{c}{a}_{2}\\ {a}_{3}\\ ⋮\\ {a}_{n}\end{array}\end{array}\right)=0$ ，则 $\lambda =0$${\xi }_{1}=\left(\begin{array}{c}{a}_{2}\\ \begin{array}{c}-{a}_{1}\\ 0\\ ⋮\\ 0\end{array}\end{array}\right)$${\xi }_{2}=\left(\begin{array}{c}{a}_{3}\\ \begin{array}{c}0\\ -{a}_{1}\\ ⋮\\ 0\end{array}\end{array}\right)$${\xi }_{n-1}=\left(\begin{array}{c}{a}_{n}\\ \begin{array}{c}0\\ 0\\ ⋮\\ -{a}_{1}\end{array}\end{array}\right)$

2) 若 $\left({a}_{1}\text{\hspace{0.17em}}{a}_{2}\text{\hspace{0.17em}}{a}_{3}\text{\hspace{0.17em}}\cdots \text{\hspace{0.17em}}{a}_{n}\right)\left(\begin{array}{c}{a}_{1}\\ \begin{array}{c}{a}_{2}\\ {a}_{3}\\ ⋮\\ {a}_{n}\end{array}\end{array}\right)\ne 0$ ，则 $\lambda ={a}_{1}^{2}+{a}_{2}^{2}+\cdots +{a}_{n-1}^{2}+{a}_{n}^{2}$

${\xi }_{n}=\left(\begin{array}{c}{a}_{1}\\ \begin{array}{c}{a}_{2}\\ {a}_{3}\\ ⋮\\ {a}_{n}\end{array}\end{array}\right)$

${\xi }_{1}=\left(\begin{array}{c}3\\ -1\\ 0\\ 0\end{array}\right)$${\xi }_{2}=\left(\begin{array}{c}5\\ 0\\ -1\\ 0\end{array}\right)$${\xi }_{3}=\left(\begin{array}{c}7\\ 0\\ 0\\ -1\end{array}\right)$

${\xi }_{4}=\left(\begin{array}{c}1\\ 3\\ 5\\ 7\end{array}\right)$

4. 结论

 [1] 陈志杰. 高等代数与解析几何(上) [M]. 北京: 高等教育出版社, 2008: 330-340.