Section02_特征值与特征向量的性质

一般性质

1. (重点) $\boldsymbol{A}_{n\times n}, \lambda_{1}\ne \lambda_{2}$ $$\begin{array}{cc} & (\lambda_{1}\boldsymbol{E}-\boldsymbol{A}) \boldsymbol{X} = \boldsymbol{0} \Rightarrow \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{s} \quad \lambda_{1}\text{对应的线性无关组} \\ & (\lambda_{2}\boldsymbol{E}-\boldsymbol{A}) \boldsymbol{X} = \boldsymbol{0} \Rightarrow \boldsymbol{\beta}_{1},\cdots, \boldsymbol{\beta}_{t} \quad \lambda_{2}\text{对应的线性无关组} \\ \Rightarrow & \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{s},\boldsymbol{\beta}_{1},\cdots,\boldsymbol{\beta}_{t}\text{线性无关} \end{array}$$

   Proof $$\begin{array}{ll} & \boldsymbol{A\alpha}_{1} = \lambda_{1}\boldsymbol{\alpha}_{1},\cdots,\boldsymbol{A\alpha}_{s} = \lambda_{1}\boldsymbol{\alpha}_{s} \\ & \boldsymbol{A\beta}_{1} = \lambda_{2}\boldsymbol{\beta}_{1},\cdots,\boldsymbol{A\beta}_{s} = \lambda_{2}\boldsymbol{\beta}_{s} \\ & \text{令}k_{1}\boldsymbol{\alpha}_{1} +\cdots +k_{s}\boldsymbol{\alpha}_{s} + \ell_{1}\boldsymbol{\beta}_{1}+\cdots + \ell_{t}\boldsymbol{\beta}_{t} = \boldsymbol{0} \quad(*) \\ \therefore & \boldsymbol{A}\times (*): \\ & \lambda_{1}(k_{1}\boldsymbol{\alpha}_{1} + \cdots + k_{s}\boldsymbol{\alpha}_{s}) + \lambda_{2}(\ell_{1}\boldsymbol{\beta}_{1} + \cdots + \ell_{t}\boldsymbol{\beta}_{t}) = \boldsymbol{0} \quad (**) \\ & \lambda_{2}(*) - (**): \\ & (\lambda_{2}-\lambda_{1})(k_{1}\boldsymbol{\alpha}_{1} + \cdots + k_{s}\boldsymbol{\alpha}_{s}) = \boldsymbol{0} \\ \because & \lambda_{2} \ne \lambda_{1} \\ \therefore & k_{1} = \cdots = k_{s} = 0 \\ \therefore & \ell_{1} = \cdots = \ell_{t} = 0 \\ \therefore & \boldsymbol{\alpha}_{1},\cdots,\boldsymbol{\alpha}_{s},\boldsymbol{\beta}_{1},\cdots,\boldsymbol{\beta}_{t}\text{线性无关} \end{array}$$

2. (==重点==) $\boldsymbol{A\alpha} = \lambda_{0}\boldsymbol{\alpha}$
   1. $f(\boldsymbol{A})\boldsymbol{\alpha} = f(\lambda_{0})\boldsymbol{\alpha}$，如 $\boldsymbol{A\alpha} = \lambda_{0}\boldsymbol{\alpha} \Rightarrow \boldsymbol{A}^{2}\boldsymbol{\alpha} = \lambda_{0}^{2}\boldsymbol{\alpha}$
   2. 若 $\boldsymbol{A}$ 可逆，有 $\begin{cases} \boldsymbol{A}^{-1}\boldsymbol{\alpha} = \frac{1}{\lambda_{0}}\boldsymbol{\alpha} \\ \boldsymbol{A}^{*}\boldsymbol{\alpha} = \frac{\det(\boldsymbol{A})}{\lambda_{0}}\boldsymbol{\alpha} \end{cases}$

      Notes

      1. $\boldsymbol{A}$ 可逆，$\boldsymbol{A}, \boldsymbol{A}^{-1},\boldsymbol{A}^{*}$ 的特征向量相同
      2. $f(\boldsymbol{A})\begin{cases} \text{像多项式一样处理，因式分解} \\ \boldsymbol{AX} = \lambda_{0}\boldsymbol{X}\Rightarrow f(\boldsymbol{A})\boldsymbol{X} = f(\lambda)\boldsymbol{X} \end{cases}$
      3. $\lambda_{0}$ 为 $\boldsymbol{A}$ 的 $r$ 重特征值: $(\lambda_{0}\boldsymbol{E}-\boldsymbol{A})\boldsymbol{X} = \boldsymbol{0}\Rightarrow \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{s}$，则 $s\le r$

3. $\boldsymbol{A}_{n\times n}$ 可对角化 $\Leftrightarrow$ $\boldsymbol{A}$ 存在 $n$ 个无关特征向量

例题

1. $\boldsymbol{A},\boldsymbol{B}$ 为 $4$ 阶矩阵，$\boldsymbol{A}\sim \boldsymbol{B}$，$\boldsymbol{A}$ 的特征值为 $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5}$，求 $\det(\boldsymbol{B}^{-1} - \boldsymbol{E})$ $$\begin{array}{ll} \because & \boldsymbol{A}\sim \boldsymbol{B} \\ \therefore & \boldsymbol{B}\text{的特征值: } \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{5} \\ \therefore & \boldsymbol{B}^{-1}\text{的特征值: } 2,3,4,5 \\ \therefore & \det(\boldsymbol{B}^{-1} - \boldsymbol{E}) = (2-1)(3-1)(4-1)(5-1) = 24 \\ \end{array}$$
2. $\boldsymbol{A}_{3\times 3}, \boldsymbol{A}^{2} + \boldsymbol{A} - 2 \boldsymbol{E} = \boldsymbol{0}, \det(\boldsymbol{A}) = -2$，求特征值 $$\begin{array}{ll} \because & \boldsymbol{A}^{2} + \boldsymbol{A}- 2 \boldsymbol{E} = \boldsymbol{0} \\ & \text{令} \boldsymbol{AX}= \lambda \boldsymbol{X}\quad \boldsymbol{X} \ne \boldsymbol{0} \\ \therefore & (\boldsymbol{A}^{2} + \boldsymbol{A} - 2 \boldsymbol{E})\boldsymbol{X} = (\lambda^{2} + \lambda -2) \boldsymbol{X} = 0 \\ \therefore & \lambda^{2} + \lambda - 2 = 0 \\ \therefore & \lambda_{1} = -2, \lambda_{2} = 1 \\ \because & \det(\boldsymbol{A}) = \lambda_{1}\lambda_{2}\lambda_{3} \\ \therefore & \lambda_{3} = 1 \\ \end{array}$$

实对称矩阵 $\boldsymbol{A}^{\intercal} = \boldsymbol{A}$

1. $\boldsymbol{A}^{\intercal}\boldsymbol{A}\Rightarrow \lambda_{1}\in \mathbb{R}, \cdots, \lambda_{n}\in \mathbb{R}$
2. $\boldsymbol{A}^{\intercal} = \boldsymbol{A}, \lambda_{1}\ne \lambda_{2}$
   • $(\lambda_{1}\boldsymbol{E} - \boldsymbol{A})\boldsymbol{X} = \boldsymbol{0} \Rightarrow \boldsymbol{\alpha}_{1},\cdots, \boldsymbol{\alpha}_{s}$
   • $(\lambda_{2}\boldsymbol{E} - \boldsymbol{A})\boldsymbol{X} = \boldsymbol{0} \Rightarrow \boldsymbol{\beta}_{1},\cdots, \boldsymbol{\beta}_{t}$
   • 则 $\boldsymbol{\alpha}_{i}\perp \boldsymbol{\beta}_{j}\quad (i=1,\cdots, s, j = 1,\cdots, t)$

     Proof $$\begin{array}{ll} & \boldsymbol{A\alpha}_{i} = \lambda_{1} \boldsymbol{\alpha}_{i} \Rightarrow \boldsymbol{\alpha}_{i}^{\intercal}\boldsymbol{A}^{\intercal} = \lambda_{1} \boldsymbol{\alpha}_{i}^{\intercal} \\ \Rightarrow & \boldsymbol{\alpha}_{i}^{\intercal}\boldsymbol{A\beta}_{j} = \lambda_{1}\boldsymbol{\alpha}_{i}^{\intercal}\boldsymbol{\beta}_{j} \Rightarrow \lambda_{2} \boldsymbol{\alpha}^{\intercal}_{i}\boldsymbol{\beta}_{j} = \lambda_{1}\boldsymbol{\alpha}^{\intercal}_{i}\boldsymbol{\beta}_{j}\\ \Rightarrow & (\lambda_{2} - \lambda_{1}) \boldsymbol{\alpha}_{i}^{\intercal} \boldsymbol{\beta}_{j} = 0 \\ \because & \lambda_{1} \ne \lambda_{2} \\ \therefore & \boldsymbol{\alpha}_{i}^{\intercal} \boldsymbol{\beta}_{j} = 0 \\ \therefore & \boldsymbol{\alpha}_{i}\perp \boldsymbol{\beta}_{j} \\ \end{array}$$

3. $\boldsymbol{A}^{\intercal} = \boldsymbol{A} \Rightarrow \boldsymbol{A}$ 一定可对角化