5 , 设a₁ , a₂ , a₃为两两正交的单位向量组

b₁=$\frac{- 1}{3}$a₁+$\frac{2}{3}$a₂+$\frac{2}{3}$a₃ , b₂=$\frac{2}{3}$a₁+$\frac{2}{3}$a₂+$\frac{- 1}{3}$a₃ , b₃=$\frac{- 2}{3}$a₁+$\frac{1}{3}$a₂+$\frac{- 2}{3}$a₃

试证明b₁ , b₂ , b₃也是两两正交的单位向量组

证：

要证明$b_{1},b_{2},b_{3}$是两两正交的单位向量组，

需分别证明每个向量是单位向量（模长为1）

和任意两个向量的内积为0（两两正交）。

已知$a_{1},a_{2},a_{3}$是两两正交的单位向量组，即满足内积性质：$a_{i} \cdot a_{j} = \left\{ \begin{matrix} 1, & i = j \\ 0, & i \neq j \end{matrix} \right.\$

一、证明$b_{1},b_{2},b_{3}$是单位向量

单位向量的定义是向量自身的内积等于1，即$b_{i} \cdot b_{i} = 1$（$i = 1,2,3$）。

1. 证明$b_{1} \cdot b_{1} = 1$ 已知$b_{1} = - \frac{1}{3}a_{1} + \frac{2}{3}a_{2} + \frac{2}{3}a_{3}$，则：

$$b_{1} \cdot b_{1} = \left( - \frac{1}{3}a_{1} + \frac{2}{3}a_{2} + \frac{2}{3}a_{3} \right) \cdot \left( - \frac{1}{3}a_{1} + \frac{2}{3}a_{2} + \frac{2}{3}a_{3} \right)$$

$$= \left( - \frac{1}{3} \right)\left( - \frac{1}{3} \right)(a_{1} \cdot a_{1}) + \left( - \frac{1}{3} \right)\left( \frac{2}{3} \right)(a_{1} \cdot a_{2}) + \left( - \frac{1}{3} \right)\left( \frac{2}{3} \right)(a_{1} \cdot a_{3})$$

$$+ \left( \frac{2}{3} \right)\left( - \frac{1}{3} \right)(a_{2} \cdot a_{1}) + \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{2} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{2} \cdot a_{3})$$

$$+ \left( \frac{2}{3} \right)\left( - \frac{1}{3} \right)(a_{3} \cdot a_{1}) + \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{3} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{3} \cdot a_{3})$$

根据$a_{i} \cdot a_{j}$的正交性，当$i \neq j$时，$a_{i} \cdot a_{j} = 0$，

因此所有交叉项（$i \neq j$的项）都为0，仅保留$i = j$的项：

$$b_{1} \cdot b_{1} = \left( \frac{1}{9} \right)(a_{1} \cdot a_{1}) + \left( \frac{4}{9} \right)(a_{2} \cdot a_{2}) + \left( \frac{4}{9} \right)(a_{3} \cdot a_{3}) = \frac{1}{9} \times 1 + \frac{4}{9} \times 1 + \frac{4}{9} \times 1 = 1$$

2. 证明$b_{2} \cdot b_{2} = 1$ 已知$b_{2} = \frac{2}{3}a_{1} + \frac{2}{3}a_{2} - \frac{1}{3}a_{3}$，则：

$$b_{2} \cdot b_{2} = \left( \frac{2}{3}a_{1} + \frac{2}{3}a_{2} - \frac{1}{3}a_{3} \right) \cdot \left( \frac{2}{3}a_{1} + \frac{2}{3}a_{2} - \frac{1}{3}a_{3} \right)$$

$$= \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{1} \cdot a_{1}) + \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{1} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( - \frac{1}{3} \right)(a_{1} \cdot a_{3})$$

$$+ \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{2} \cdot a_{1}) + \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{2} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( - \frac{1}{3} \right)(a_{2} \cdot a_{3})$$

$$+ \left( - \frac{1}{3} \right)\left( \frac{2}{3} \right)(a_{3} \cdot a_{1}) + \left( - \frac{1}{3} \right)\left( \frac{2}{3} \right)(a_{3} \cdot a_{2}) + \left( - \frac{1}{3} \right)\left( - \frac{1}{3} \right)(a_{3} \cdot a_{3})$$

同理，交叉项（$i \neq j$）的内积为0，仅保留$i = j$的项：

$$b_{2} \cdot b_{2} = \left( \frac{4}{9} \right)(a_{1} \cdot a_{1}) + \left( \frac{4}{9} \right)(a_{2} \cdot a_{2}) + \left( \frac{1}{9} \right)(a_{3} \cdot a_{3}) = \frac{4}{9} \times 1 + \frac{4}{9} \times 1 + \frac{1}{9} \times 1 = 1$$

3. 证明$b_{3} \cdot b_{3} = 1$

已知$b_{3} = - \frac{2}{3}a_{1} + \frac{1}{3}a_{2} - \frac{2}{3}a_{3}$，则：

$$b_{3} \cdot b_{3} = \left( - \frac{2}{3}a_{1} + \frac{1}{3}a_{2} - \frac{2}{3}a_{3} \right) \cdot \left( - \frac{2}{3}a_{1} + \frac{1}{3}a_{2} - \frac{2}{3}a_{3} \right)$$

$$= \left( - \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{1} \cdot a_{1}) + \left( - \frac{2}{3} \right)\left( \frac{1}{3} \right)(a_{1} \cdot a_{2}) + \left( - \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{1} \cdot a_{3})$$

$$+ \left( \frac{1}{3} \right)\left( - \frac{2}{3} \right)(a_{2} \cdot a_{1}) + \left( \frac{1}{3} \right)\left( \frac{1}{3} \right)(a_{2} \cdot a_{2}) + \left( \frac{1}{3} \right)\left( - \frac{2}{3} \right)(a_{2} \cdot a_{3})$$

$$+ \left( - \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{3} \cdot a_{1}) + \left( - \frac{2}{3} \right)\left( \frac{1}{3} \right)(a_{3} \cdot a_{2}) + \left( - \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{3} \cdot a_{3})$$

交叉项（$i \neq j$）的内积为0，仅保留$i = j$的项：

$$b_{3} \cdot b_{3} = \left( \frac{4}{9} \right)(a_{1} \cdot a_{1}) + \left( \frac{1}{9} \right)(a_{2} \cdot a_{2}) + \left( \frac{4}{9} \right)(a_{3} \cdot a_{3}) = \frac{4}{9} \times 1 + \frac{1}{9} \times 1 + \frac{4}{9} \times 1 = 1$$

综上，$b_{1},b_{2},b_{3}$的自身内积均为1，因此都是单位向量。

二、证明$b_{1},b_{2},b_{3}$两两正交

两两正交的定义是任意两个不同向量的内积为0，即$b_{i} \cdot b_{j} = 0$（$i \neq j$），

需证明$b_{1} \cdot b_{2} = 0$、$b_{1} \cdot b_{3} = 0$、$b_{2} \cdot b_{3} = 0$。

1. 证明$b_{1} \cdot b_{2} = 0$

$$\begin{aligned} b_{1} \cdot b_{2} & = \left( - \frac{1}{3}a_{1} + \frac{2}{3}a_{2} + \frac{2}{3}a_{3} \right) \cdot \left( \frac{2}{3}a_{1} + \frac{2}{3}a_{2} - \frac{1}{3}a_{3} \right) \\ & = \left( - \frac{1}{3} \right)\left( \frac{2}{3} \right)(a_{1} \cdot a_{1}) + \left( - \frac{1}{3} \right)\left( \frac{2}{3} \right)(a_{1} \cdot a_{2}) + \left( - \frac{1}{3} \right)\left( - \frac{1}{3} \right)(a_{1} \cdot a_{3}) \\ & \quad + \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{2} \cdot a_{1}) + \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{2} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( - \frac{1}{3} \right)(a_{2} \cdot a_{3}) \\ & \quad + \left( \frac{2}{3} \right)\left( \frac{2}{3} \right)(a_{3} \cdot a_{1}) + \left( \frac{2}{3} \right)(\frac{2}{3})(a_{3} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( - \frac{1}{3} \right)(a_{3} \cdot a_{3}) \end{aligned}$$

交叉项（$i \neq j$）的内积为0，仅保留$i = j$的项：

$$b_{1} \cdot b_{2} = - \frac{2}{9}(a_{1} \cdot a_{1}) + \frac{4}{9}(a_{2} \cdot a_{2}) - \frac{2}{9}(a_{3} \cdot a_{3}) = - \frac{2}{9} \times 1 + \frac{4}{9} \times 1 - \frac{2}{9} \times 1 = 0$$

2. 证明$b_{1} \cdot b_{3} = 0$

$$\begin{aligned} b_{1} \cdot b_{3} & = \left( - \frac{1}{3}a_{1} + \frac{2}{3}a_{2} + \frac{2}{3}a_{3} \right) \cdot \left( - \frac{2}{3}a_{1} + \frac{1}{3}a_{2} - \frac{2}{3}a_{3} \right) \\ & = \left( - \frac{1}{3} \right)\left( - \frac{2}{3} \right)(a_{1} \cdot a_{1}) + \left( - \frac{1}{3} \right)\left( \frac{1}{3} \right)(a_{1} \cdot a_{2}) + \left( - \frac{1}{3} \right)\left( - \frac{2}{3} \right)(a_{1} \cdot a_{3}) \\ & \quad + \left( \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{2} \cdot a_{1}) + \left( \frac{2}{3} \right)\left( \frac{1}{3} \right)(a_{2} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{2} \cdot a_{3}) \\ & \quad + \left( \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{3} \cdot a_{1}) + \left( \frac{2}{3} \right)\left( \frac{1}{3} \right)(a_{3} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{3} \cdot a_{3}) \end{aligned}$$

交叉项（$i \neq j$）的内积为0，仅保留$i = j$的项：

$$b_{1} \cdot b_{3} = \frac{2}{9}(a_{1} \cdot a_{1}) + \frac{2}{9}(a_{2} \cdot a_{2}) - \frac{4}{9}(a_{3} \cdot a_{3}) = \frac{2}{9} \times 1 + \frac{2}{9} \times 1 - \frac{4}{9} \times 1 = 0$$

3. 证明$b_{2} \cdot b_{3} = 0$

$$\begin{aligned} b_{2} \cdot b_{3} & = \left( \frac{2}{3}a_{1} + \frac{2}{3}a_{2} - \frac{1}{3}a_{3} \right) \cdot \left( - \frac{2}{3}a_{1} + \frac{1}{3}a_{2} - \frac{2}{3}a_{3} \right) \\ & = \left( \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{1} \cdot a_{1}) + \left( \frac{2}{3} \right)\left( \frac{1}{3} \right)(a_{1} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{1} \cdot a_{3}) \\ & \quad + \left( \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{2} \cdot a_{1}) + \left( \frac{2}{3} \right)\left( \frac{1}{3} \right)(a_{2} \cdot a_{2}) + \left( \frac{2}{3} \right)\left( - \frac{2}{3} \right)(a_{2} \cdot a_{3}) \\ & \quad + \left( - \frac{1}{3} \right)\left( - \frac{2}{3} \right)(a_{3} \cdot a_{1}) + \left( - \frac{1}{3} \right)\left( \frac{1}{3} \right)(a_{3} \cdot a_{2}) + \left( - \frac{1}{3} \right)\left( - \frac{2}{3} \right)(a_{3} \cdot a_{3}) \end{aligned}$$

交叉项（$i \neq j$）的内积为0，仅保留$i = j$的项：

$$b_{2} \cdot b_{3} = - \frac{4}{9}(a_{1} \cdot a_{1}) + \frac{2}{9}(a_{2} \cdot a_{2}) + \frac{2}{9}(a_{3} \cdot a_{3}) = - \frac{4}{9} \times 1 + \frac{2}{9} \times 1 + \frac{2}{9} \times 1 = 0$$

三、结论 $b_{1},b_{2},b_{3}$既满足自身内积为1（是单位向量），

又满足任意两个不同向量的内积为0（两两正交），

因此$b_{1},b_{2},b_{3}$是两两正交的单位向量组。

6 , 求下列矩阵的特征值和特征向量:

(1) $\begin{bmatrix} 2 & - 1 & 2 \\ 5 & - 3 & 3 \\ - 1 & 0 & - 2 \end{bmatrix}$ (2) $\begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 3 & 6 \end{bmatrix}$ (3) $\left\lbrack \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 0 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack$

解:(1)|A -λE|=$\left| \begin{matrix} 2 - \lambda & - 1 & 2 \\ 5 & - 3 - \lambda & 3 \\ - 1 & 0 & - 2 - \lambda \end{matrix} \right|\overset{c_{3} - (\lambda + 2)c_{1}}{\Rightarrow}\left| \begin{matrix} 2 - \lambda & - 1 & \lambda^{2} - 2 \\ 5 & - 3 - \lambda & - 7 - 5\lambda \\ - 1 & 0 & 0 \end{matrix} \right|$

$\overset{按r_{3}展开}{\Rightarrow}\left| \begin{matrix} - 1 & \lambda^{2} - 2 \\ 3 + \lambda & 7 + 5\lambda \end{matrix} \right|\overset{\begin{matrix} c_{2} - c_{1} \\ c_{2} \div (1 + \lambda) \end{matrix}}{\Rightarrow}(1 + \lambda)\left| \begin{matrix} - 1 & \lambda - 1 \\ 3 + \lambda & 4 \end{matrix} \right|$=-(1+λ)³

所以A的特征值为λ₁=λ₂=λ₃=-1 (三重根)

对于特征值-1 , 解方程(A-(-1)E)x=0

由A+E=$\begin{bmatrix} 3 & - 1 & 2 \\ 5 & - 2 & 3 \\ - 1 & 0 & - 1 \end{bmatrix}\overset{r}{\Rightarrow}\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$

得同解方程组：$\left\{ \begin{array}{r} x_{1} + x_{3} = 0 \\ x_{2} + x_{3} = 0 \end{array} \right.\$

得参数形式：$\left\{ \begin{array}{r} x_{1} = - x_{3} \\ x_{2} = - x_{3} \end{array} \right.\$ (x₃ 可任意取值)

得参数形式：$\left\{ \begin{matrix} x_{1} = - c \\ x_{2} = - c \\ x_{3} = c\ \end{matrix} \right.\$ (其中x₃=c )

得齐次通解：$\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} - c \\ - c \\ c \end{bmatrix} = \begin{bmatrix} - 1 \\ - 1 \\ 1 \end{bmatrix}$(c₃为任意实数)

得特征向量p=$\begin{bmatrix} - 1 \\ - 1 \\ 1 \end{bmatrix}$

(2)|A -λE|=$\left| \begin{matrix} 1 - \lambda & 2 & 3 \\ 2 & 1 - \lambda & 3 \\ 3 & 3 & 6 - \lambda \end{matrix} \right|\overset{c_{1} - c_{2}}{\Rightarrow}\left| \begin{matrix} - (1 + \lambda) & 2 & 3 \\ 1 + \lambda & 1 - \lambda & 3 \\ 0 & 3 & 6 - \lambda \end{matrix} \right|$

$\overset{\begin{matrix} c_{1} \div (1 + \lambda) \\ r_{2} + r_{1} \end{matrix}}{\Rightarrow}(1 + \lambda)\left| \begin{matrix} - 1 & 2 & 3 \\ 0 & 3 - \lambda & 6 \\ 0 & 3 & 6 - \lambda \end{matrix} \right|$=-(1+λ)$\left| \begin{matrix} 3 - \lambda & 6 \\ 3 & 6 - \lambda \end{matrix} \right|$=-λ(λ+1)(λ-9)

所以A的特征值为λ₁=-1 , λ₂=0 , λ₃=9

当λ₁=-1时 , 解方程(A-(-1)E)x=0

由A+E=$\begin{bmatrix} 2 & 2 & 3 \\ 2 & 2 & 3 \\ 3 & 3 & 7 \end{bmatrix}\overset{r}{\Rightarrow}\begin{bmatrix} 1 & 1 & 4 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}\overset{r}{\Rightarrow}\begin{bmatrix} 1 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$

得同解方程组：$\left\{ \begin{array}{r} x_{1} + x_{2} = 0 \\ x_{3} = 0 \end{array} \right.\$

得参数形式：$\left\{ \begin{array}{r} x_{1} = - x_{2} \\ x_{3} = 0 \end{array} \right.\$ (x₂ 可任意取值)

得参数形式：$\left\{ \begin{matrix} x_{1} = - c \\ x_{2} = c \\ x_{3} = 0\ \end{matrix} \right.\$ (其中x₂=c )

得齐次通解：$\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} - c \\ c \\ 0 \end{bmatrix} = c\begin{bmatrix} - 1 \\ 1 \\ 0 \end{bmatrix}$

得对应的特征向量p₁=$\begin{bmatrix} - 1 \\ 1 \\ 0 \end{bmatrix}$

当λ₂=0时 , 解方程(A-0E)x=Ax=0

由A=$\begin{bmatrix} 1 & 2 & 3 \\ 2 & 1 & 3 \\ 3 & 3 & 6 \end{bmatrix}\overset{r}{\Rightarrow}\begin{bmatrix} 1 & 2 & 3 \\ 0 & - 3 & - 3 \\ 0 & - 3 & - 3 \end{bmatrix}\overset{r}{\Rightarrow}\begin{bmatrix} 1 & 2 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}\overset{r}{\Rightarrow}\begin{bmatrix} 1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 0 \end{bmatrix}$

得同解方程组：$\left\{ \begin{array}{r} x_{1} + x_{3} = 0 \\ x_{2} + x_{3} = 0 \end{array} \right.\$

得参数形式：$\left\{ \begin{array}{r} x_{1} = - x_{3} \\ x_{2} = - x_{3} \end{array} \right.\$ (x₃ 可任意取值)

得参数形式：$\left\{ \begin{matrix} x_{1} = - c \\ x_{2} = - c \\ x_{3} = c\ \end{matrix} \right.\$ (其中x₃=c )

得齐次通解：$\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} - c \\ - c \\ c \end{bmatrix} = c\begin{bmatrix} - 1 \\ - 1 \\ 1 \end{bmatrix}$

得对应的特征向量p₂=$\begin{bmatrix} - 1 \\ - 1 \\ 1 \end{bmatrix}$

当λ₃=9时 , 解方程(A-9E)x=0

由A-9E=$\begin{bmatrix} - 8 & 2 & 3 \\ 2 & - 8 & 3 \\ 3 & 3 & - 3 \end{bmatrix}\overset{r}{\Rightarrow}\begin{bmatrix} - 5 & 5 & 0 \\ 5 & - 5 & 0 \\ - 1 & - 1 & 1 \end{bmatrix}\overset{r}{\Rightarrow}\begin{bmatrix} 0 & 0 & 0 \\ 1 & - 1 & 0 \\ 0 & - 2 & 1 \end{bmatrix}$

$$\overset{r}{\Rightarrow}\begin{bmatrix} 1 & - 1 & 0 \\ 0 & - 2 & 1 \\ 0 & 0 & 0 \end{bmatrix}\overset{r}{\Rightarrow}\begin{bmatrix} 1 & - 1 & 0 \\ 0 & 1 & \frac{- 1}{2} \\ 0 & 0 & 0 \end{bmatrix}$$

得同解方程组：$\left\{ \begin{array}{r} x_{1} - x_{2} = 0 \\ x_{2} - \frac{1}{2}x_{3} = 0 \end{array} \right.\$

得参数形式：$\left\{ \begin{array}{r} x_{1} = x_{2} \\ x_{2} = \frac{1}{2}x_{3} \end{array} \right.\$ (x₃ 可任意取值)

得参数形式：$\left\{ \begin{matrix} x_{1} = \frac{1}{2}c \\ x_{2} = \frac{1}{2}c \\ x_{3} = c\ \end{matrix} \right.\$ (其中x₃=c )

得齐次通解：$\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} \frac{1}{2}c \\ \frac{1}{2}c \\ c \end{bmatrix} = \frac{1}{2}c\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$

得对应的特征向量p₃=$\begin{bmatrix} 1 \\ 1 \\ 2 \end{bmatrix}$

(3)特征多项式为

|A -λE|=$\left| \begin{array}{r} \begin{matrix} - \lambda \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ - \lambda \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} - \lambda \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ - \lambda \end{matrix} \end{array} \right|\overset{\begin{matrix} r_{4} \leftrightarrow r_{2} \\ c_{4} \leftrightarrow c_{2} \end{matrix}}{\Rightarrow}\left| \begin{array}{r} \begin{matrix} - \lambda \\ 1 \end{matrix} \\ \begin{matrix} \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ - \lambda \end{matrix} \\ \begin{matrix} \\ \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ \end{matrix} \\ \begin{matrix} - \lambda \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} \\ \end{matrix} \\ \begin{matrix} 1 \\ - \lambda \end{matrix} \end{array} \right|$

所以A的特征值为λ₁=λ₂=-1 , λ₃=λ₄=1

当λ₁=λ₂=-1时 , 解方程(A+E)x=0

由A +E=$\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack\overset{r}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack$

得同解方程组：$\left\{ \begin{array}{r} x_{1} + x_{4} = 0 \\ x_{2} + x_{3} = 0 \end{array} \right.\$

得参数形式：$\left\{ \begin{array}{r} x_{1} = - x_{4} \\ x_{2} = - x_{3} \end{array} \right.\$ (x₃ , x₄ 可任意取值)

得参数形式：$\left\{ \begin{array}{r} \begin{matrix} x_{1} = - c_{2} \\ x_{2} = - c_{1} \end{matrix} \\ \begin{matrix} x_{3} = c_{1} \\ x_{4} = c_{2} \end{matrix} \end{array} \right.\$ (其中x₃=c₁ , x₄=c₂ )

得齐次通解：$\left\lbrack \begin{array}{r} \begin{matrix} x_{1} \\ x_{2} \end{matrix} \\ \begin{matrix} x_{3} \\ x_{4} \end{matrix} \end{array} \right\rbrack$=$\left\lbrack \begin{array}{r} \begin{matrix} - c_{2} \\ - c_{1} \end{matrix} \\ \begin{matrix} c_{1} \\ \ c_{2} \end{matrix} \end{array} \right\rbrack$=$c_{1}\left\lbrack \begin{array}{r} \begin{matrix} 0 \\ - 1 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack$+$c_{2}\left\lbrack \begin{array}{r} \begin{matrix} - 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack$

得对应的线性无关特征向量为p₁=$\left\lbrack \begin{array}{r} \begin{matrix} 0 \\ - 1 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack$ , p₂=$\left\lbrack \begin{array}{r} \begin{matrix} - 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack$

当λ₃=λ₄=1时,解方程(A-E)x=0

由A -E=$\left\lbrack \begin{array}{r} \begin{matrix} - 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ - 1 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} - 1 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ - 1 \end{matrix} \end{array} \right\rbrack\overset{r}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} - 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ - 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack\overset{r}{\Rightarrow}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} 0 \\ - 1 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array}\ \ \ \ \begin{array}{r} \begin{matrix} - 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 0 \end{matrix} \end{array} \right\rbrack$

得同解方程组：$\left\{ \begin{array}{r} x_{1} - x_{4} = 0 \\ x_{2} - x_{3} = 0 \end{array} \right.\$

得参数形式：$\left\{ \begin{array}{r} x_{1} = x_{4} \\ x_{2} = x_{3} \end{array} \right.\$ (x₃ , x₄ 可任意取值)

得参数形式：$\left\{ \begin{array}{r} \begin{matrix} x_{1} = c_{2} \\ x_{2} = c_{1} \end{matrix} \\ \begin{matrix} x_{3} = c_{1} \\ x_{4} = c_{2} \end{matrix} \end{array} \right.\$ (其中x₃=c₁ , x₄=c₂ )

得齐次通解：$\left\lbrack \begin{array}{r} \begin{matrix} x_{1} \\ x_{2} \end{matrix} \\ \begin{matrix} x_{3} \\ x_{4} \end{matrix} \end{array} \right\rbrack$=$\left\lbrack \begin{array}{r} \begin{matrix} c_{2} \\ c_{1} \end{matrix} \\ \begin{matrix} c_{1} \\ \ c_{2} \end{matrix} \end{array} \right\rbrack$=$c_{1}\left\lbrack \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack$+$c_{2}\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack$

得对应的线性无关特征向量为p₃=$\left\lbrack \begin{array}{r} \begin{matrix} 0 \\ 1 \end{matrix} \\ \begin{matrix} 1 \\ 0 \end{matrix} \end{array} \right\rbrack$ , p₄=$\left\lbrack \begin{array}{r} \begin{matrix} 1 \\ 0 \end{matrix} \\ \begin{matrix} 0 \\ 1 \end{matrix} \end{array} \right\rbrack$