【矩阵的特征值和特征向量】

特征值和特征向量

设$A$是$n$阶方阵，若数$\lambda$和$n$维非0列向量$x$使：
$$Ax=\lambda x$$
成立，则：
$\lambda$——$A$的特征值；
$x$——$A$对应于特征值$\lambda$的特征向量。

性质

1、设$n$阶矩阵$A=(a_{ij})$的特征值为${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n$，

1）${\lambda }_1+{\lambda }_2+\cdots +{\lambda }_n=a_{11}+a_{22}+\cdots +a_{nn}$；
2）${\lambda }_1{\lambda }_2\cdots {\lambda }_n=|A|$

$$\{\begin{array}{l}|A|\ne 0\Rightarrow A\text{可逆}\\ \iff {\lambda }_i\ne 0,i=1,2,\cdots ,n\end{array}$$

	$A$	$A^2$	$A^{-1}$	$A^k$	$\phi (A)$
特征值	$\lambda$	${\lambda }^2$	$\frac{1}{\lambda }$	${\lambda }^k$	$\phi (\lambda )$

2、设${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_m$是方阵$A$的$m$个特征值，$p_1,p_2,\cdots ,p_m$依次是与之对应的特征向量，若${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_m$各不相等，则$p_1,p_2,\cdots ,p_m$线性无关。

3、设${\lambda }_1,{\lambda }_2$是方阵$A$的2个不同特征值，${\xi }_1,{\xi }_2,\cdots ,{\xi }_s$和${\eta }_1,{\eta }_2,\cdots ,{\eta }_t$分别是对应于${\lambda }_1,{\lambda }_2$的线性无关的特征向量，则${\xi }_1,{\xi }_2,\cdots ,{\xi }_s,{\eta }_1,{\eta }_2,\cdots ,{\eta }_t$线性无关。

4、对称矩阵的特征值为实数；

5、设${\lambda }_1,{\lambda }_2$是对称矩阵$A$的两个特征值，$p_1,p_2$是对应的特征向量，若${\lambda }_1\ne {\lambda }_2$，则：
$$\Rightarrow p_1\text{与}{p}_2\text{正交}$$

6、设$A$为$n$阶对称矩阵，则必有正交矩阵$P$，使：
$$\begin{gathered} P^{-1}AP=P^{T}AP=\Lambda \\ (\Lambda \text{——以}A\text{的}n\text{个特征值为对角元的对角矩阵}) \end{gathered}$$

7、设$A$为$n$阶对称矩阵，$\lambda$为$A$的特征方程的$k$重根，则矩阵$A-\lambda E$的秩$R(A-\lambda E)=n-k$，
$$\begin{gathered} \Rightarrow \lambda \text{恰有}k\text{个线性无关的特征向量} \\ \iff \text{实对称矩阵一定可对角化} \end{gathered}$$

习题

1、求矩阵$A=\left(\begin{array}{cc}3& -1\\ -1& 3\end{array}\right)$的特征值与特征向量。

解：$Ax=\lambda x(x\ne 0)$
$$\begin{gathered} \Rightarrow (A-\lambda E)x=0 \\ \Rightarrow \text{令}|A-\lambda E|=\left|\begin{array}{cc}3-\lambda & -1\\ -1& 3-\lambda \end{array}\right|=(3-\lambda {)}^2-1=0 \\ \Rightarrow {\lambda }_1=2,{\lambda }_2=4 \end{gathered}$$

1）${\lambda }_1=2$时，
$$\left(\begin{array}{cc}1& -1\\ -1& 1\end{array}\right)\to \left(\begin{array}{cc}1& -1\\ 0& 0\end{array}\right)\Rightarrow \left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=c_1\left(\begin{array}{c}1\\ 1\end{array}\right)\Rightarrow \text{特征向量}{p}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$$

2）${\lambda }_2=4$时，
$$\left(\begin{array}{cc}-1& -1\\ -1& -1\end{array}\right)\to \left(\begin{array}{cc}1& 1\\ 0& 0\end{array}\right)\Rightarrow \left(\begin{array}{c}{x}_1\\ x_2\end{array}\right)=c_2\left(\begin{array}{c}-1\\ 1\end{array}\right)\Rightarrow \text{特征向量}{p}_2=\left(\begin{array}{c}-1\\ 1\end{array}\right)$$

2、求矩阵$A=\left(\begin{array}{ccc}-1& 1& 0\\ -4& 3& 0\\ 1& 0& 2\end{array}\right)$的特征值与特征向量。

解：
$$\begin{gathered} \text{令}|A-\lambda E|=\left|\begin{array}{ccc}-1-\lambda & 1& 0\\ -4& 3-\lambda & 0\\ 1& 0& 2-\lambda \end{array}\right|=(2-\lambda )\left|\begin{array}{cc}-1-\lambda & 1\\ -4& 3-\lambda \end{array}\right|=(2-\lambda )(\lambda -1{)}^2=0 \\ \Rightarrow {\lambda }_1=2,{\lambda }_2={\lambda }_3=1 \end{gathered}$$

1）${\lambda }_1=2$时，
$$\begin{gathered} \left(\begin{array}{ccc}-3& 1& 0\\ -4& 1& 0\\ 1& 0& 0\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 0\\ -4& 1& 0\\ -3& 1& 0\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 1& 0\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 0\\ 0& 1& 0\\ 0& 0& 0\end{array}\right) \\ \Rightarrow \left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=c_1\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\Rightarrow \text{特征向量}{p}_1=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right) \\ \Rightarrow k{p}_1(k\ne 0)\text{是对应于}{\lambda }_1=2\text{的全部特征向量} \end{gathered}$$

2）${\lambda }_2={\lambda }_3=1$时，
$$\begin{gathered} \left(\begin{array}{ccc}-2& 1& 0\\ -4& 2& 0\\ 1& 0& 1\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 1\\ -2& 1& 0\\ -4& 2& 0\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 0\end{array}\right)\Rightarrow \left(\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right)=c_2\left(\begin{array}{c}-1\\ -2\\ 1\end{array}\right)\Rightarrow \text{特征向量}{p}_2=\left(\begin{array}{c}-1\\ -2\\ 1\end{array}\right) \\ \Rightarrow k{p}_2(k\ne 0)\text{是对应于}{\lambda }_2={\lambda }_3=1\text{的全部特征向量} \end{gathered}$$

3、设2阶矩阵$A$的特征值为$1,-1,2$，求$A^{\ast }+3A-2E$的特征值。

解：$A{A}^{\ast }=|A|E\Rightarrow A^{\ast }=A^{-1}|A|,|A|=1\times (-1)\times 2=-2$
$$\begin{gathered} \Rightarrow |A|A^{-1}+3A-2E=-2A^{-1}+3A-2E\text{的特征值} \\ -2\frac{1}{\lambda }+3\lambda -2 \\ {\lambda }_1=-2+3-2=-1 \\ {\lambda }_2=2-3-2=-3 \\ {\lambda }_3=-1+6-2=3 \end{gathered}$$

4、求矩阵的特征值和特征向量：
（1）$\left(\begin{array}{ccc}2& -1& 2\\ 5& -3& 3\\ -1& 0& -2\end{array}\right)$；（2）$\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\\ 3& 3& 6\end{array}\right)$；（3）$\left(\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\\ 1& 0& 0& 0\end{array}\right)$

解：（1）
$$\begin{gathered} \text{令}|A-\lambda E|=0=\left|\begin{array}{ccc}2-\lambda & -1& 2\\ 5& -3-\lambda & 3\\ -1& 0& -2-\lambda \end{array}\right|=\left|\begin{array}{ccc}2-\lambda & -1& {\lambda }^2-2\\ 5& -3-\lambda & -7-5\lambda \\ -1& 0& 0\end{array}\right|=\left|\begin{array}{cc}-1& {\lambda }^2-2\\ 3+\lambda & 7+5\lambda \end{array}\right|=\left|\begin{array}{cc}-1& {\lambda }^2-1\\ 3+\lambda & 4+4\lambda \end{array}\right| \\ =(\lambda +1)\left|\begin{array}{cc}-1& \lambda -1\\ 3+\lambda & 4\end{array}\right|=-(\lambda +1{)}^3 \\ \Rightarrow {\lambda }_1={\lambda }_2={\lambda }_3=-1 \\ \left(\begin{array}{ccc}3& -1& 2\\ 5& -2& 3\\ -1& 0& -1\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 1\\ 0& -2& -2\\ 0& -1& -1\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)\Rightarrow p=\left(\begin{array}{c}-1\\ -1\\ 1\end{array}\right) \end{gathered}$$

（2）
$$\begin{gathered} \text{令}|A-\lambda E|=\left|\begin{array}{ccc}1-\lambda & 2& 3\\ 2& 1-\lambda & 3\\ 3& 3& 6-\lambda \end{array}\right|=\left|\begin{array}{ccc}-1-\lambda & 2& 3\\ \lambda +1& 1-\lambda & 3\\ 0& 3& 6-\lambda \end{array}\right|=(\lambda +1)\left|\begin{array}{ccc}-1& 2& 3\\ 1& 1-\lambda & 3\\ 0& 3& 6-\lambda \end{array}\right| \\ =(\lambda +1)\left|\begin{array}{ccc}0& 3-\lambda & 6\\ 1& 1-\lambda & 3\\ 0& 3& 6-\lambda \end{array}\right|=-(\lambda +1)\left|\begin{array}{cc}3-\lambda & 6\\ 3& 6-\lambda \end{array}\right|=-\lambda (\lambda +1)(\lambda -9)=0 \\ \Rightarrow {\lambda }_1=0,{\lambda }_2=-1,{\lambda }_3=9 \end{gathered}$$

1）${\lambda }_1=0$时，
$$\left(\begin{array}{ccc}1& 2& 3\\ 2& 1& 3\\ 3& 3& 6\end{array}\right)\to \left(\begin{array}{ccc}1& 2& 3\\ 0& -3& -3\\ 0& -3& -3\end{array}\right)\to \left(\begin{array}{ccc}1& 2& 3\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)\Rightarrow p_1=\left(\begin{array}{c}-1\\ -1\\ 1\end{array}\right)$$

2）${\lambda }_2=-1$时，
$$\left(\begin{array}{ccc}2& 2& 3\\ 2& 2& 3\\ 3& 3& 7\end{array}\right)\to \left(\begin{array}{ccc}1& 1& \frac{3}{2}\\ 0& 0& 0\\ 0& 0& \frac{5}{2}\end{array}\right)\to \left(\begin{array}{ccc}1& 1& 0\\ 0& 0& 1\\ 0& 0& 0\end{array}\right)\Rightarrow p_2=\left(\begin{array}{c}-1\\ 1\\ 0\end{array}\right)$$

3）${\lambda }_3=9$时，
$$\left(\begin{array}{ccc}-8& 2& 3\\ 2& -8& 3\\ 3& 3& -3\end{array}\right)\to \left(\begin{array}{ccc}1& 1& -1\\ 2& -8& 3\\ -8& 2& 3\end{array}\right)\to \left(\begin{array}{ccc}1& 1& -1\\ 0& -10& 5\\ 0& 10& -5\end{array}\right)\to \left(\begin{array}{ccc}1& 1& -1\\ 0& 1& -\frac{1}{2}\\ 0& 0& 0\end{array}\right)\to \left(\begin{array}{ccc}1& 0& -\frac{1}{2}\\ 0& 1& -\frac{1}{2}\\ 0& 0& 0\end{array}\right)\Rightarrow p_3=\left(\begin{array}{c}1\\ 1\\ 2\end{array}\right)$$

（3）
$$\begin{gathered} \text{令}|A-\lambda E|=0=\left|\begin{array}{cccc}-\lambda & 0& 0& 1\\ 0& -\lambda & 1& 0\\ 0& 1& -\lambda & 0\\ 1& 0& 0& -\lambda \end{array}\right|=\left|\begin{array}{cccc}-\lambda & 0& 0& 1\\ 0& -\lambda & 1& 0\\ 0& 1& -\lambda & 0\\ 1-{\lambda }^2& 0& 0& 0\end{array}\right|=-\left|\begin{array}{ccc}0& -\lambda & 1\\ 0& 1& -\lambda \\ 1-{\lambda }^2& 0& 0\end{array}\right| \\ =({\lambda }^2-1)\left|\begin{array}{cc}-\lambda & 1\\ 1& -\lambda \end{array}\right|=({\lambda }^2-1{)}^2 \\ \Rightarrow {\lambda }_1={\lambda }_2=-1,{\lambda }_3={\lambda }_4=1 \end{gathered}$$

1）${\lambda }_1={\lambda }_2=-1$时，
$$\left(\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 1& 0\\ 0& 1& 1& 0\\ 1& 0& 0& 1\end{array}\right)\to \left(\begin{array}{cccc}1& 0& 0& 1\\ 0& 1& 1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\Rightarrow p_1=\left(\begin{array}{c}0\\ -1\\ 1\\ 0\end{array}\right),p_2=\left(\begin{array}{c}-1\\ 0\\ 0\\ 1\end{array}\right)$$

2）${\lambda }_3={\lambda }_4=1$时，
$$\left(\begin{array}{cccc}-1& 0& 0& 1\\ 0& -1& 1& 0\\ 0& 1& -1& 0\\ 1& 0& 0& -1\end{array}\right)\to \left(\begin{array}{cccc}1& 0& 0& -1\\ 0& 1& -1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\Rightarrow p_3=\left(\begin{array}{c}0\\ 1\\ 1\\ 0\end{array}\right),p_4=\left(\begin{array}{c}1\\ 0\\ 0\\ 1\end{array}\right)$$

5、已知$p=\left(\begin{array}{c}1\\ 1\\ -1\end{array}\right)$是矩阵$A=\left(\begin{array}{ccc}2& -1& 2\\ 5& a& 3\\ -1& b& -2\end{array}\right)$的一个特征向量。
（1）求参数$a,b$及特征向量$p$所对应的特征值；
（2）问$A$可否对角化？

解：$Ap=\lambda p\Rightarrow (A-\lambda E)p=0$
$$\left(\begin{array}{ccc}2-\lambda & -1& 2\\ 5& a-\lambda & 3\\ -1& b& -2-\lambda \end{array}\right)\left(\begin{array}{c}1\\ 1\\ -1\end{array}\right)=\left(\begin{array}{c}0\\ 0\\ 0\end{array}\right)\Rightarrow \{\begin{array}{l}-\lambda -1=0\\ 2+a-\lambda =0\\ 1+b+\lambda =0\end{array}\Rightarrow \{\begin{array}{l}\lambda =-1\\ a=-3\\ b=0\end{array}$$

（2）
$$\begin{gathered} \text{令}|A-\lambda E|=0=\left|\begin{array}{ccc}2-\lambda & -1& 2\\ 5& -3-\lambda & 3\\ -1& 0& -2-\lambda \end{array}\right|=\left|\begin{array}{ccc}2-\lambda & -1& {\lambda }^2-2\\ 5& -3-\lambda & -5\lambda -7\\ -1& 0& 0\end{array}\right|=\left|\begin{array}{cc}-1& {\lambda }^2-2\\ 3+\lambda & 5\lambda +7\end{array}\right| \\ =\left|\begin{array}{cc}-1& {\lambda }^2-1\\ 3+\lambda & 4\lambda +4\end{array}\right|=(\lambda +1)\left|\begin{array}{cc}-1& \lambda -1\\ 3+\lambda & 4\end{array}\right|=-(\lambda +1{)}^3 \\ \Rightarrow {\lambda }_1={\lambda }_2={\lambda }_3=-1 \\ \left(\begin{array}{ccc}3& -1& 2\\ 5& -2& 3\\ -1& 0& -1\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 1\\ 0& -2& -2\\ 0& -1& -1\end{array}\right)\to \left(\begin{array}{ccc}1& 0& 1\\ 0& 1& 1\\ 0& 0& 0\end{array}\right)\Rightarrow p=\left(\begin{array}{c}-1\\ -1\\ 1\end{array}\right) \\ \text{无3个线性无关的特征向量}\Rightarrow \text{不可对角化} \end{gathered}$$

相似矩阵

设$A,B$均是$n$阶矩阵，若有可逆矩阵$P$，使：
$$P^{-1}AP=B$$
则$B$是$A$的相似矩阵。

性质

1、若$n$阶矩阵$A$与$B$相似，则：
$$\Rightarrow A\text{与}B\text{的特征值相同}$$

2、若$n$阶矩阵$A$与对角阵
$$\Lambda =\left(\begin{array}{ccccc}{\lambda }_1& 0& 0& \cdots & 0\\ 0& {\lambda }_2& 0& \cdots & 0\\ ⋮& ⋮& ⋮& & ⋮\\ 0& 0& 0& \cdots & {\lambda }_n\end{array}\right)$$
相似，则${\lambda }_1,{\lambda }_2,\cdots ,{\lambda }_n$为$A$的$n$个特征值。
$$P^{-1}AP=\Lambda$$

3、$n$阶矩阵$A$与对角阵相似的充要条件：
$$\iff A\text{有}n\text{个线性无关的特征向量}$$

4、若$A$的$n$个特征值互不相等：
$$\Rightarrow A\text{与对角阵相似}$$