Appendix_常见题型

型一 性质

1. $\boldsymbol{A}_{3\times 3}, \boldsymbol{A}^{2}- \boldsymbol{A} - 2 \boldsymbol{E} = \boldsymbol{0}, \det(\boldsymbol{A}) = 2$，求 $A_{11} + A_{22} + A_{33}$ $$\begin{array}{ll} & \text{令} \boldsymbol{AX} = \lambda \boldsymbol{X}\quad \boldsymbol{X} \ne \boldsymbol{0} \\ \therefore & (\boldsymbol{A}^{2} - \boldsymbol{A} - 2 \boldsymbol{E})\boldsymbol{X} = (\lambda^{2} - \lambda - 2)\boldsymbol{X} = \boldsymbol{0} \\ \therefore & \lambda^{2} - \lambda - 2 = 0 \Rightarrow \lambda_{1} = 2, \lambda_{2} = -1 \\ \because & \det(\boldsymbol{A}) = \lambda_{1}\lambda_{2}\lambda_{3} \\ \therefore & \lambda_{3} = \lambda_{2} = -1 \\ \therefore & \boldsymbol{A}^{*}\text{的特征值为} 1, -2 ,-2 \\ \therefore & \text{tr}(\boldsymbol{A}^{*}) = A_{11} + A_{22} + A_{33} = -3 \end{array}$$
2. 设 $\boldsymbol{\alpha} = \left( \begin{smallmatrix} 1 \\ 1 \\ 2 \end{smallmatrix} \right)$ 为矩阵 $\boldsymbol{A} = \left( \begin{smallmatrix} 1 & -3 & 3 \\ 6 & x & -6 \\ y & -9 &13 \end{smallmatrix} \right)$ 的逆矩阵 $\boldsymbol{A}^{-1}$ 的特征向量，求 $x,y$ 及 $\boldsymbol{A}^{-1}$ 的特征值 $\mu$ $$\begin{array}{ll} & \text{由题意得}\boldsymbol{\alpha}\text{为}\boldsymbol{A}\text{的特征向量} \\ \therefore & \boldsymbol{A\alpha} = \lambda \boldsymbol{\alpha}\\ & \boldsymbol{A\alpha}= \left( \begin{smallmatrix} 4 \\ x-6 \\ y+17 \end{smallmatrix} \right) = \lambda \alpha \\ \therefore & \frac{4}{1} = \frac{x-6}{1} = \frac{y+17}{2} \\ \therefore & x = 10, y = -9,\lambda = 4\\ \therefore & \mu = \frac{1}{\lambda} = \frac{1}{4} \end{array}$$

型二 求 $\lambda$

方法

1. 公式法 $\det(\lambda \boldsymbol{E} - \boldsymbol{A}) = 0$
2. 定义法 $\boldsymbol{AX} = \lambda X \quad \boldsymbol{X} \ne \boldsymbol{0}$
   • 特征 $f(\boldsymbol{A}), \boldsymbol{AB} = \boldsymbol{C}$
3. 关联法 $\begin{cases} \boldsymbol{A}, \boldsymbol{A}^{-1}, \boldsymbol{A}^{*} \\ \boldsymbol{P}^{-1}\boldsymbol{AP} = \boldsymbol{B} \text{, 即} \boldsymbol{A}\sim \boldsymbol{B} \end{cases}$

例题

1. $\boldsymbol{A}_{4\times 4}, \boldsymbol{A}^{2} = \boldsymbol{E}, \text{tr}(\boldsymbol{A}) = 0$，求 $\boldsymbol{A}$ 的特征值 $$\begin{array}{ll} \because & \boldsymbol{A}^{2} = \boldsymbol{E} \\ \therefore & \boldsymbol{A}^{2} - \boldsymbol{E} = \boldsymbol{0} \\ & \text{令} \boldsymbol{AX} = \lambda \boldsymbol{X} \quad \boldsymbol{X} \ne \boldsymbol{0} \\ \therefore & (\boldsymbol{A}^{2} - \boldsymbol{E}) \boldsymbol{X} = (\lambda^{2} - 1) \boldsymbol{X} = \boldsymbol{0} \\ \therefore & \lambda^{2} - 1 = 0\Rightarrow \lambda = \pm 1 \\ \because & \text{tr}(\boldsymbol{A}) = 0 \\ \therefore & \lambda_{1} = \lambda_{2} = -1, \lambda_{3} = \lambda_{4} = 1 \\ \end{array}$$
2. $\boldsymbol{\alpha}= \left( \begin{smallmatrix} a_{1} \\ a_{2} \\ a_{3} \\ a_{4} \\ \end{smallmatrix} \right), \boldsymbol{\beta} = \left( \begin{smallmatrix} b_{1} \\ b_{2} \\ b_{3} \\ b_{4} \\ \end{smallmatrix} \right), (\boldsymbol{\alpha},\boldsymbol{\beta}) = 3, \boldsymbol{A} = \boldsymbol{\alpha\beta}^{\intercal}$，求特征值 $$\begin{array}{ll} \because & \boldsymbol{A}^{2} = \boldsymbol{\alpha\beta}^{\intercal}\boldsymbol{\alpha\beta}^{\intercal} \\ \because & (\boldsymbol{\alpha},\boldsymbol{\beta}) = \boldsymbol{\beta}^{\intercal}\boldsymbol{\alpha} = 3 \\ \therefore & \boldsymbol{A}^{2} = 3 \boldsymbol{A} \Rightarrow \boldsymbol{A}^{2} - 3 \boldsymbol{A} = \boldsymbol{0} \\ & \text{令} \boldsymbol{AX} = \lambda \boldsymbol{X} \quad \boldsymbol{X} \ne \boldsymbol{0} \\ \therefore & \lambda^{2} - 3\lambda = 0 \Rightarrow \lambda_{1} = 3,\lambda_{2} = 0 \\ \because & \text{tr}(\boldsymbol{A}) = \boldsymbol{\alpha}^{\intercal}\boldsymbol{\beta} = 3 \\ \therefore & \lambda_{3} = \lambda_{4} = \lambda_{2} = 0 \\ \end{array}$$
3. $\boldsymbol{A}_{3\times 3}, \boldsymbol{A} \left( \begin{smallmatrix} -1 & 1 \\ 1 & 0 \\ 0 & 1 \end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 & 0 \\ -1 & 0 \\ 0 & 0 \end{smallmatrix} \right)$，$\boldsymbol{A}$ 每行之和为 $2$，求 $\boldsymbol{A}$ $$\begin{array}{ll} \because & \boldsymbol{A} \left( \begin{smallmatrix} -1 & 1 \\ 1 & 0 \\ 0 & 1 \end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 & 0 \\ -1 & 0 \\ 0 & 0 \end{smallmatrix} \right) \\ \therefore & \lambda_{1} = -1, \lambda_{2} = 0 \\ \because & \boldsymbol{A} \text{每行之和为}2 \\ \therefore & \boldsymbol{A} \left( \begin{smallmatrix} 1 \\ 1 \\ 1 \end{smallmatrix} \right) = \left( \begin{smallmatrix} 2 \\ 2 \\ 2 \end{smallmatrix} \right)\\ \therefore & \lambda_{3} = 2 \\ \therefore & \text{令} \boldsymbol{P} = \left( \begin{smallmatrix} -1 & 1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{smallmatrix} \right)\\ \because & \boldsymbol{P}^{-1}\boldsymbol{AP} = \left( \begin{smallmatrix} -1 \\ & 0 \\ & & 2 \end{smallmatrix} \right)\\ \therefore & \boldsymbol{A} = \boldsymbol{P} \left( \begin{smallmatrix} -1 \\ & 0 \\ & & 2 \end{smallmatrix} \right)\boldsymbol{P}^{-1}\\ \therefore & \boldsymbol{A} = \left( \begin{smallmatrix} 1 & 0 & 2 \\ -1 & 0 & 2 \\ 0 & 0 & 2 \\ \end{smallmatrix} \right)\left( \begin{smallmatrix} -1 & 0 & 1 \\ -1 & -1 & 2 \\ 1 & 1 & -1 \end{smallmatrix} \right) = \left( \begin{smallmatrix} 1 & 2 & -1 \\ 3 & 2 & -3 \\ 2 & 2 & -2 \end{smallmatrix} \right)\\ \end{array}$$
4. $\boldsymbol{A}_{3\times 3}$，$\boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2},\boldsymbol{\alpha}_{3}$ 线性无关，$\boldsymbol{A\alpha}_{1} = \boldsymbol{\alpha}_{2} + \boldsymbol{\alpha}_{3},\boldsymbol{A\alpha}_{2} = \boldsymbol{\alpha}_{1} +\boldsymbol{\alpha}_{3}, \boldsymbol{A\alpha}_{3} = \boldsymbol{\alpha}_{1} + \boldsymbol{\alpha}_{2}$，求 $\boldsymbol{A}$ 的特征值 $$\begin{array}{ll} & \text{令} \boldsymbol{P} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} & \boldsymbol{\alpha}_{3} \end{matrix} \right) \\ & \text{由题意得} \boldsymbol{AP} = \boldsymbol{P} \left( \begin{smallmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{smallmatrix} \right) \\ \therefore & \boldsymbol{A} \sim \left( \begin{smallmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \\ \end{smallmatrix} \right) \\ \therefore & \det(\lambda \boldsymbol{E} - \boldsymbol{A}) = \left\vert \begin{smallmatrix} \lambda & -1 & -1 \\ -1 & \lambda & -1 \\ -1 & -1 & \lambda \\ \end{smallmatrix} \right\vert = (\lambda-2) \left\vert \begin{smallmatrix} 1 & 1 & 1 \\ 0 & \lambda +1 & 0 \\ 0 & 0 & \lambda +1 \\ \end{smallmatrix} \right\vert = (\lambda-2)(\lambda+1)^{2}\\ \therefore & \lambda_{1} = 2, \lambda_{2} = \lambda_{3} = -1 \\ \end{array}$$

型三 矩阵对角化的判断

方法

1. 若 $\boldsymbol{A}^{\intercal} = \boldsymbol{A}$，则 $\boldsymbol{A}$ 可对角化
2. 若 $\boldsymbol{A}^{\intercal}\ne \boldsymbol{A}$:
   • Case1 $\boldsymbol{A}\sim \boldsymbol{B} \Rightarrow \boldsymbol{A}, \boldsymbol{B}$ 可对角化特性相同
   • Case2 求 $\lambda_{1},\cdots, \lambda_{n}$
     1. 单值 $\Rightarrow$ $\boldsymbol{A}$ 可对角化
     2. 每个特征值重数和无关特征向量个数一致 $\Rightarrow$ $\boldsymbol{A}$ 可对角化
     3. 存在 $n$ 个线性无关特征向量 $\Rightarrow$ $\boldsymbol{A}$ 可对角化

例题

1. $\boldsymbol{A} = \left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$，且 $ad - bc < 0$，证: $\boldsymbol{A}$ 可相似对角化 $$\begin{array}{ll} & \det(\lambda \boldsymbol{E} - \boldsymbol{A}) = \left\vert \begin{smallmatrix} \lambda - a & -b \\ -c & \lambda - d \end{smallmatrix} \right\vert = \lambda^{2} - (a+d)\lambda + ad - bc \\ \because & ad - bc < 0 \\ \therefore & [-(a+d)]^{2} - 4 (ad-bc) > 0\\ \therefore & \boldsymbol{A} \text{必有两个不同特征值} \\ \therefore & \boldsymbol{A}\text{可相似对角化} \\ \end{array}$$
2. $\boldsymbol{\alpha} = \left( \begin{smallmatrix} a_{1} \\ a_{2} \\ a_{3} \end{smallmatrix} \right), \boldsymbol{\beta} = \left( \begin{smallmatrix} b_{1} \\ b_{2} \\ b_{3} \end{smallmatrix} \right)$ 单位且正交，$\boldsymbol{A} = \boldsymbol{\alpha\beta}^{\intercal}+ \boldsymbol{\beta\alpha}^{\intercal}$
   1. 证明: $\boldsymbol{\alpha} + \boldsymbol{\beta}, \boldsymbol{\alpha}- \boldsymbol{\beta}$ 为 $\boldsymbol{A}$ 特征向量
   2. 证明: $\boldsymbol{A}$ 可对角化 $$\begin{array}{ll} \because & \boldsymbol{A}(\boldsymbol{\alpha} + \boldsymbol{\beta}) = \boldsymbol{\alpha\beta}^{\intercal}\boldsymbol{\alpha}+ \boldsymbol{\beta\alpha}^{\intercal}\boldsymbol{\alpha} + \boldsymbol{\alpha\beta}^{\intercal}\boldsymbol{\beta}+ \boldsymbol{\beta\alpha}^{\intercal}\boldsymbol{\beta} \\ & \boldsymbol{A}(\boldsymbol{\alpha} - \boldsymbol{\beta}) = \boldsymbol{\alpha\beta}^{\intercal}\boldsymbol{\alpha}+ \boldsymbol{\beta\alpha}^{\intercal}\boldsymbol{\alpha} - \boldsymbol{\alpha\beta}^{\intercal}\boldsymbol{\beta}- \boldsymbol{\beta\alpha}^{\intercal}\boldsymbol{\beta} \\ \because & \boldsymbol{\alpha}, \boldsymbol{\beta}\text{单位且正交} \\ \therefore & \boldsymbol{A}(\boldsymbol{\alpha}+ \boldsymbol{\beta}) = 1\times(\boldsymbol{\alpha} + \boldsymbol{\beta})\\ & \boldsymbol{A}(\boldsymbol{\alpha}- \boldsymbol{\beta}) = -1\times(\boldsymbol{\alpha} - \boldsymbol{\beta})\\ \therefore & \boldsymbol{\alpha},\boldsymbol{\beta}\text{为}\boldsymbol{A}\text{的特征向量} \\\\ \because & \boldsymbol{A} = \boldsymbol{\alpha\beta}^{\intercal} + \boldsymbol{\beta\alpha}^{\intercal} \\ \therefore & \text{tr}(\boldsymbol{A}) = \boldsymbol{\alpha}^{\intercal}\boldsymbol{\beta} + \boldsymbol{\beta}^{\intercal}\boldsymbol{\alpha} = 0\\ \therefore & \lambda_{3} = 0\ne \lambda_{1} \ne \lambda_{2} \\ \therefore & \boldsymbol{A}\text{可对角化} \\ \end{array}$$
3. $\boldsymbol{A} = \left( \begin{smallmatrix} -1 & 2 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{smallmatrix} \right)$，$\boldsymbol{A}$ 可否对角化 $$\begin{array}{ll} &\det(\lambda\boldsymbol{E} - \boldsymbol{A}) = \left\vert \begin{smallmatrix} \lambda + 1 & -2 & 0 \\ -1 & \lambda & 0 \\ 0 & 0 & \lambda -1 \end{smallmatrix} \right\vert = (\lambda - 1) \left\vert \begin{smallmatrix} \lambda+1 & -2 \\ -1 & \lambda \end{smallmatrix} \right\vert = (\lambda-1)^{2}(\lambda+2) = 0 \\ \therefore & \lambda_{1} = \lambda_{2} = 1,\lambda_{3} = -2 \\ & \lambda = 1 \text{对应的线性无关特征向量} \boldsymbol{\alpha}_{1} = \left( \begin{smallmatrix} 1 \\ 1 \\ 0 \end{smallmatrix} \right),\boldsymbol{\alpha}_{2} = \left( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \right) \\ \therefore & \boldsymbol{A}\text{可对角化} \\ \end{array}$$
4. $\boldsymbol{A}_{3\times 3}, \boldsymbol{\alpha}_{1}, \boldsymbol{\alpha}_{2}, \boldsymbol{\alpha}_{3}$ 线性无关，$\boldsymbol{A\alpha}_{1} = \boldsymbol{\alpha}_{1}, \boldsymbol{A\alpha}_{2} = 2\boldsymbol{\alpha}_{1}+\boldsymbol{\alpha}_{2}, \boldsymbol{A\alpha}_{3} = -\boldsymbol{\alpha}_{1} + \boldsymbol{\alpha}_{3}$，$\boldsymbol{A}$ 可否对角化? $$\begin{array}{ll} & \text{令} \boldsymbol{P} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} & \boldsymbol{\alpha}_{3} \end{matrix} \right) \\ & \text{由题意得} \boldsymbol{AP} = \boldsymbol{P} \left( \begin{smallmatrix} 1 & 2 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix} \right) \\ \therefore & \boldsymbol{A} \sim \left( \begin{smallmatrix} 1 & 2 & -1 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{smallmatrix} \right)\triangleq \boldsymbol{B}\\ \therefore & \det(\lambda \boldsymbol{E} - \boldsymbol{A}) = \det(\lambda \boldsymbol{E} - \boldsymbol{B}) = \left\vert \begin{smallmatrix} \lambda-1 & -2 & 1 \\ 0 & \lambda-1 & 0 \\ 0 & 0 & \lambda-1 \end{smallmatrix} \right\vert = (\lambda-1)^{3} = 0 \\ \therefore & \lambda =1 \\ \because & r(\boldsymbol{E} - \boldsymbol{B}) = 1 \\ \therefore & (\boldsymbol{E} - \boldsymbol{B})\boldsymbol{X} = \boldsymbol{0}\text{仅有两个线性无关解} \\ \therefore & \boldsymbol{A}\text{不可对角化} \\ \end{array}$$
5. 设 $\boldsymbol{A} = \left( \begin{smallmatrix} 2 & 2 & 0 \\ 8 & 2 & a \\ 0 & 0 & 6 \end{smallmatrix} \right)$ 相似于对角阵，求常数 $a$，并求可逆矩阵 $\boldsymbol{P}$ 使得 $\boldsymbol{P}^{-1}\boldsymbol{AP}$ 为对角矩阵 $$\begin{array}{ll} & \det(\lambda\boldsymbol{E} - \boldsymbol{A}) = \left\vert \begin{smallmatrix} \lambda -2 & -2 & 0 \\ -8 & \lambda -2 & -a \\ 0 & 0 & \lambda-6 \end{smallmatrix} \right\vert = (\lambda-6) \left\vert \begin{smallmatrix} \lambda - 2 & -2 \\ -8 & \lambda-2 \end{smallmatrix} \right\vert = (\lambda-6)^{2}(\lambda+2) = 0 \\ \therefore & \lambda_{1} =\lambda_{2} = 6, \lambda_{3} = -2\\ \because & \boldsymbol{A}\text{可对角化} \\ \therefore & r(6 \boldsymbol{E} - \boldsymbol{A}) = 1 \\ & 6 \boldsymbol{E} - \boldsymbol{A} = \left( \begin{smallmatrix} 4 & -2 & 0 \\ -8 & 4 & -a \\ 0 & 0 & 0 \end{smallmatrix} \right) \\ \therefore & a = 0 \\\\ & 6 \boldsymbol{E} - \boldsymbol{A} = \left( \begin{smallmatrix} 1 & -\frac{1}{2} & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \\ \end{smallmatrix} \right)\\ \therefore & \lambda = 6 \text{对应的线性无关特征向量} \boldsymbol{\alpha}_{1} = \left( \begin{smallmatrix} \frac{1}{2} \\ 1 \\0 \end{smallmatrix} \right), \boldsymbol{\alpha}_{2} = \left( \begin{smallmatrix} 0 \\ 0 \\ 1 \end{smallmatrix} \right)\\ & 2 \boldsymbol{E} + \boldsymbol{A} = \left( \begin{smallmatrix} 1 & \frac{1}{2} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{smallmatrix} \right) \\ \therefore & \lambda = -2 \text{对应的线性无关特征向量} \boldsymbol{\alpha}_{3} = \left( \begin{smallmatrix} -\frac{1}{2} \\ 1 \\ 0 \end{smallmatrix} \right)\\ & \text{令} \boldsymbol{P} = \left( \begin{matrix} \boldsymbol{\alpha}_{1} & \boldsymbol{\alpha}_{2} & \boldsymbol{\alpha}_{3} \end{matrix} \right) = \left( \begin{smallmatrix} \frac{1}{2} & 0 & -\frac{1}{2} \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{smallmatrix} \right) \\ \therefore & \boldsymbol{P}^{-1}\boldsymbol{AP} = \left( \begin{smallmatrix} 6 \\ & 6 \\ &&-2 \end{smallmatrix} \right)\\ \end{array}$$
6. 设 $\boldsymbol{A} = \left( \begin{smallmatrix} 0 & 0 & 1\\ x & 1 & y \\ 1 & 0 & 0 \end{smallmatrix} \right)$ 有三个线性无关特征向量，求 $x,y$ 所满足的条件 $$\begin{array}{ll} & \det(\lambda\boldsymbol{E} - \boldsymbol{A}) = \left\vert \begin{smallmatrix} \lambda & 0 & -1 \\ -x & \lambda - 1 & -y \\ -1 & 0 & \lambda \end{smallmatrix} \right\vert = \lambda^{2}(\lambda -1) - (\lambda - 1) = (\lambda -1)^{2}(\lambda +1) = 0 \\ \therefore & \lambda_{1} = \lambda_{2} = 1,\lambda_{3} = -1 \\ \because & \boldsymbol{A} \text{有三个线性无关特征向量} \\ & r(\boldsymbol{E} - \boldsymbol{A}) = 1 \\ \therefore & r \left( \begin{smallmatrix} 1 & 0 & -1 \\ -x & 0 & -y \\ -1 & 0 & 1 \end{smallmatrix} \right) = 1\\ \therefore & x = -y \\ \end{array}$$