线代强化NO20|矩阵的相似与相似对角化|综合运用

原创已于 2025-11-25 10:55:51 修改 · 507 阅读

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#矩阵 #线性代数 #机器学习

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$\begin{aligned} &分析：求方阵的n次幂 \\ &① \text{稀疏矩阵（零比较多）：找规律} \\ &② \text{秩为1的矩阵：} A^n = (\text{tr}A)^{n-1}A \\ &③ \text{对角线元素为相同值的上（下）三角矩阵：二项展开} \\ &④ \text{可相似对角化：} A^n = P\Lambda^nP^{-1} \end{aligned}$
$\begin{aligned} &解：(1)\ |\lambda E - A| = \begin{vmatrix} \lambda & 1 & -1 \\ -2 & \lambda + 3 & 0 \\ 0 & 0 & \lambda \end{vmatrix} = \lambda(\lambda + 1)(\lambda + 2) = 0 \\ &故A的特征值为-2, -1, 0. \\ \\ &A + 2E = \begin{pmatrix} 2 & -1 & 1 \\ 2 & -1 & 0 \\ 0 & 0 & 2 \end{pmatrix} \to \begin{pmatrix} 2 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & -\frac{1}{2} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}，\begin{pmatrix} \frac{1}{2} \\ 1 \\ 0 \end{pmatrix}为\lambda = -2的一个特征向量. \\ \\ &A + E = \begin{pmatrix} 1 & -1 & 1 \\ 2 & -2 & 0 \\ 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & -1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix}，\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}为\lambda = -1的一个特征向量. \\ \\ &A = \begin{pmatrix} 0 & -1 & 1 \\ 2 & -3 & 0 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & -\frac{3}{2} & 0 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & -\frac{3}{2} \\ 0 & 1 & -1 \\ 0 & 0 & 0 \end{pmatrix}，\begin{pmatrix} \frac{3}{2} \\ 1 \\ 1 \end{pmatrix}为\lambda = 0的一个特征向量. \\ \\ &A^{99} = \begin{pmatrix} \frac{1}{2} & 1 & \frac{3}{2} \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} -2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}^{99} \begin{pmatrix} \frac{1}{2} & 1 & \frac{3}{2} \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^{-1} \end{aligned}$
$\begin{aligned} &\begin{pmatrix} \frac{1}{2} & 1 & \frac{3}{2} & \bigg| & 1 & 0 & 0 \\ 1 & 1 & 1 & \bigg| & 0 & 1 & 0 \\ 0 & 0 & 1 & \bigg| & 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 2 & 3 & \bigg| & 2 & 0 & 0 \\ 0 & -1 & -2 & \bigg| & -2 & 1 & 0 \\ 0 & 0 & 1 & \bigg| & 0 & 0 & 1 \end{pmatrix} \\ &\to \begin{pmatrix} 1 & 2 & 0 & \bigg| & 2 & 0 & -3 \\ 0 & -1 & 0 & \bigg| & -2 & 1 & 2 \\ 0 & 0 & 1 & \bigg| & 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 & \bigg| & -2 & 2 & 1 \\ 0 & 1 & 0 & \bigg| & 2 & -1 & -2 \\ 0 & 0 & 1 & \bigg| & 0 & 0 & 1 \end{pmatrix} \\ \\ &A^{99} = \begin{pmatrix} \frac{1}{2} & 1 & \frac{3}{2} \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} -2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix}^{99} \begin{pmatrix} \frac{1}{2} & 1 & \frac{3}{2} \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix}^{-1} \\ &= \begin{pmatrix} \frac{1}{2} & 1 & \frac{3}{2} \\ 1 & 1 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} (-2)^{99} & 0 & 0 \\ 0 & (-1)^{99} & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} -2 & 2 & 1 \\ 2 & -1 & -2 \\ 0 & 0 & 1 \end{pmatrix} \\ &= \begin{pmatrix} -2^{98} & -1 & 0 \\ -2^{99} & -1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} -2 & 2 & 1 \\ 2 & -1 & -2 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -2+2^{99} & 1-2^{99} & 2-2^{98} \\ 2^{100} - 2 & -2^{100} + 1 & 2-2^{99} \\ 0 & 0 & 0 \end{pmatrix} \end{aligned}$
（2）
$\begin{aligned} &分析: \\ &思路1: \ B^{100} = (B^2)^{50} = (BA)^{50} = BABABA\cdots BA\qquad无法继续 \\ &思路2: \ B^{100} = B^{98}B^2 = B^{98} \cdot BA = B^{99}A = B^{97}B^2A \\ & \quad \quad = B^{97}BAA = B^{98}A^2 \\ & \quad \quad \vdots \\ & \quad \quad B^{100} = BA^{99} \\& 思路3: \ B^3 = B^2A = BAA = BA^2\end{aligned}$
$\begin{aligned} &(2)\ B^{100} = B^{98}B^2 = \cdots = BA^{99} = (\alpha_1, \alpha_2, \alpha_3) \begin{pmatrix} -2 + 2^{99} & 1 - 2^{99} & 2 - 2^{98} \\ 2^{100} - 2 & -2^{100} + 1 & 2 - 2^{99} \\ 0 & 0 & 0 \end{pmatrix} \\ &= (\beta_1, \beta_2, \beta_3) \\ \\ &\beta_1 = (2^{99} - 2)\alpha_1 + (2^{100} - 2)\alpha_2 \\ &\beta_2 = (1 - 2^{99})\alpha_1 + (-2^{100} + 1)\alpha_2 \\ &\beta_3 = (2 - 2^{98})\alpha_1 + (2 - 2^{99})\alpha_2 \end{aligned}$
【小结】
如果矩阵 $A$ 可以相似对角化，则可以按照如下方法计算 $A^n$ ：

先求出可逆矩阵 $P$ 及对角矩阵 $Λ\Lambda$ 使得 $P−1AP=ΛP^{-1}AP=\Lambda$ ；
由矩阵的运算法则可得 $A=PΛP−1A=P\Lambda P^{-1}$ ，则 $An=PΛnP−1A^n=P\Lambda^n P^{-1}$ 。

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$(1)\begin{cases} x_{n+1} = \dfrac{5}{6}x_n + \dfrac{2}{5}\left( \dfrac{1}{6}x_n + y_n \right) \\ y_{n+1} = \dfrac{3}{5}\left( \dfrac{1}{6}x_n + y_n \right) \end{cases}$ $\begin{cases} x_{n+1} = \dfrac{9}{10}x_n + \dfrac{2}{5}y_n \\ y_{n+1} = \dfrac{1}{10}x_n + \dfrac{3}{5}y_n \end{cases}$ $\begin{pmatrix} \dfrac{9}{10} & \dfrac{2}{5} \\ \dfrac{1}{10} & \dfrac{3}{5} \end{pmatrix}$
$(2)\ A\eta_1 = \begin{pmatrix} \dfrac{9}{10} & \dfrac{2}{5} \\ \dfrac{1}{10} & \dfrac{3}{5} \end{pmatrix} \begin{pmatrix} 4 \\ 1 \end{pmatrix} = \begin{pmatrix} 4 \\ 1 \end{pmatrix},\ 1为特征值$ $A\eta_2 = \begin{pmatrix} \dfrac{9}{10} & \dfrac{2}{5} \\ \dfrac{1}{10} & \dfrac{3}{5} \end{pmatrix} \begin{pmatrix} -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -\dfrac{1}{2} \\ \dfrac{1}{2} \end{pmatrix},\ \dfrac{1}{2}\ 为特征值$
$(3)\ \begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} = A\begin{pmatrix} x_n \\ y_n \end{pmatrix} = AA\begin{pmatrix} x_{n-1} \\ y_{n-1} \end{pmatrix} = \cdots = A^n\begin{pmatrix} x_1 \\ y_1 \end{pmatrix} = A^n\begin{pmatrix} \dfrac{1}{2} \\ \dfrac{1}{2} \end{pmatrix}$ $计算：A^n = \begin{pmatrix} 4 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \dfrac{1}{2} \end{pmatrix}^n \begin{pmatrix} 4 & -1 \\ 1 & 1 \end{pmatrix}^{-1} = \begin{pmatrix} 4 & -1 \\ 1 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 0 & \left( \dfrac{1}{2} \right)^n \end{pmatrix} \dfrac{1}{5} \begin{pmatrix} 1 & 1 \\ -1 & 4 \end{pmatrix}$ $\dfrac{1}{5} \begin{pmatrix} 4 + \dfrac{1}{2^n} & 4 - \dfrac{4}{2^n} \\ 1 - \dfrac{1}{2^n} & 1 + \dfrac{4}{2^n} \end{pmatrix},\ \begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} = \dfrac{1}{10} \begin{pmatrix} 8 - \dfrac{3}{2^n} \\ 2 + \dfrac{3}{2^n} \end{pmatrix}$
$设\begin{pmatrix} 1 \\ 1 \end{pmatrix} = x_1\begin{pmatrix} 4 \\ 1 \end{pmatrix} + x_2\begin{pmatrix} -1 \\ 1 \end{pmatrix},\ 解得x_1 = \dfrac{2}{5},\ x_2 = \dfrac{3}{5}$
$\begin{pmatrix} 4 & -1 & \bigg| & 1 \\ 1 & 1 & \bigg| & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 1 & \bigg| & 1 \\ 0 & -5 & \bigg| & -3 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & \bigg| & \dfrac{2}{5} \\ 0 & 1 & \bigg| & \dfrac{3}{5} \end{pmatrix}$
$\begin{pmatrix} x_{n+1} \\ y_{n+1} \end{pmatrix} = \dfrac{1}{2}A^n\begin{pmatrix} 1 \\ 1 \end{pmatrix} = \dfrac{1}{2}A^n\left( \dfrac{2}{5}\begin{pmatrix} 4 \\ 1 \end{pmatrix} + \dfrac{3}{5}\begin{pmatrix} -1 \\ 1 \end{pmatrix} \right) = \dfrac{1}{5} \cdot 1^n\begin{pmatrix} 4 \\ 1 \end{pmatrix} + \dfrac{3}{10} \cdot \dfrac{1}{2^n}\begin{pmatrix} -1 \\ 1 \end{pmatrix}$

【小结】无论矩阵 $A\boldsymbol{A}$ 是否可以相似对角化，都可以按照如下方法计算 $Anβ\boldsymbol{A}^n\boldsymbol{\beta}$ ：

先求出 $A\boldsymbol{A}$ 的特征值和特征向量，其中 $Aα1=λ1α1\boldsymbol{A}\boldsymbol{\alpha}_1 = \lambda_1\boldsymbol{\alpha}_1$ ， $Aα2=λ2α2\boldsymbol{A}\boldsymbol{\alpha}_2 = \lambda_2\boldsymbol{\alpha}_2$ ， $⋯\cdots$ ， $Aαn=λnαn\boldsymbol{A}\boldsymbol{\alpha}_n = \lambda_n\boldsymbol{\alpha}_n$ ；
将 $β\boldsymbol{\beta}$ 用特征向量 $α1\boldsymbol{\alpha}_1$ ， $α2\boldsymbol{\alpha}_2$ ， $⋯\cdots$ ， $αn\boldsymbol{\alpha}_n$ 线性表示为 $β=k1α1+k2α2+⋯+knαn\boldsymbol{\beta} = k_1\boldsymbol{\alpha}_1 + k_2\boldsymbol{\alpha}_2 + \cdots + k_n\boldsymbol{\alpha}_n$ ；
$Anβ=An(k1α1+k2α2+⋯+knαn)=k1λ1nα1+k2λ2nα2+⋯+knλnnαn\boldsymbol{A}^n\boldsymbol{\beta} = \boldsymbol{A}^n(k_1\boldsymbol{\alpha}_1 + k_2\boldsymbol{\alpha}_2 + \cdots + k_n\boldsymbol{\alpha}_n) = k_1\lambda_1^n\boldsymbol{\alpha}_1 + k_2\lambda_2^n\boldsymbol{\alpha}_2 + \cdots + k_n\lambda_n^n\boldsymbol{\alpha}_n$ 。

$\begin{aligned} &?\quad A = P^{-1}\Lambda P\quad ① \\ &= P\Lambda P^{-1}\quad ②\quad \checkmark \\ &\Lambda = P^{-1}AP\quad ③\quad \checkmark \\ &\Lambda = PAP^{-1}\quad ④ \\ \end{aligned}$ $\begin{aligned} &A(\alpha_1\ \alpha_2\ \alpha_3) = (A\alpha_1\ A\alpha_2\ A\alpha_3) \\ &= (\lambda_1\alpha_1\ \lambda_2\alpha_2\ \lambda_3\alpha_3) = (\alpha_1\ \alpha_2\ \alpha_3)\begin{pmatrix} \lambda_1 \\ & \lambda_2 \\ & & \lambda_3 \end{pmatrix} \\ &A = (\alpha_1\ \alpha_2\ \alpha_3)\begin{pmatrix} \lambda_1 \\ & \lambda_2 \\ & & \lambda_3 \end{pmatrix}(\alpha_1\ \alpha_2\ \alpha_3)^{-1} \\ \end{aligned}$ $\begin{aligned} &①\ 求矩阵P,使得A = P\Lambda P^{-1} \\ &\qquad \text{P为特征向量}\quad (P^{-1}AP = \Lambda) \\ &②\ 求矩阵P,使得A = P^{-1}\Lambda P \\ &\qquad \text{P为特征向量的逆}\quad (PAP = \Lambda) \\ \end{aligned}$
$\begin{cases} x - 4 = y + 1 \\ -2y = 4x-8 \end{cases} \Rightarrow \begin{cases} x = 3 \\ y = -2 \end{cases}$
$\begin{aligned} &\boldsymbol{A} = \boldsymbol{P}_1\boldsymbol{\Lambda}\boldsymbol{P}_1^{-1},\ \boldsymbol{B} = \boldsymbol{P}_2\boldsymbol{\Lambda}\boldsymbol{P}_2^{-1} \\ &\boldsymbol{P}_1^{-1}\boldsymbol{A}\boldsymbol{P}_1 = \boldsymbol{\Lambda} = \boldsymbol{P}_2^{-1}\boldsymbol{B}\boldsymbol{P}_2 \\ &\boldsymbol{P}_2\boldsymbol{P}_1^{-1}\boldsymbol{A}\boldsymbol{P}_1\boldsymbol{P}_2^{-1} = \boldsymbol{B} \\ &(\boldsymbol{P}_1\boldsymbol{P}_2^{-1})^{-1}\boldsymbol{A}\boldsymbol{P}_1\boldsymbol{P}_2^{-1} = \boldsymbol{B} \\ &\text{故取}\ \boldsymbol{P} = \boldsymbol{P}_1\boldsymbol{P}_2^{-1} \end{aligned}$
$\begin{aligned} &解: (2)\ \boldsymbol{B}的特征值为2,-1,-2,\boldsymbol{A}与\boldsymbol{B}相似,故\boldsymbol{A}的特征值为2,-1,-2. \\& \boldsymbol{A}-2\boldsymbol{E}=\begin{pmatrix} -4 & -2 & 1 \\ 2 & 1 & -2 \\ 0 & 0 & -4 \end{pmatrix}\to\begin{pmatrix} 1 & \dfrac{1}{2} & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix},\ \begin{pmatrix} -1 \\ 2 \\ 0 \end{pmatrix}为2的一个特征向量. \\& \boldsymbol{A}+\boldsymbol{E}=\begin{pmatrix} -1 & -2 & 1 \\ 2 & 4 & -2 \\ 0 & 0 & 1 \end{pmatrix}\to\begin{pmatrix} 1 & 2 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix},\ \begin{pmatrix} -2 \\ 1 \\ 0 \end{pmatrix}为-1的一个特征向量. \\& \boldsymbol{A}+2\boldsymbol{E}=\begin{pmatrix} 0 & -2 & 1 \\ 2 & 5 & -2 \\ 0 & 0 & 0 \end{pmatrix}\to\begin{pmatrix} 1 & \dfrac{5}{2} & -1 \\ 0 & 1 & -\dfrac{1}{2} \\ 0 & 0 & 0 \end{pmatrix}\to\begin{pmatrix} 1 & 0 & \dfrac{1}{4} \\ 0 & 1 & -\dfrac{1}{2} \\ 0 & 0 & 0 \end{pmatrix},\ \begin{pmatrix} 1 \\ -2 \\ -4 \end{pmatrix}为-2的一个特征向量. \end{aligned}$
$令\boldsymbol{P_{1}} = \begin{pmatrix} -1 & -2 & -1 \\ 2 & 1 & 2 \\ 0 & 0 & 4 \end{pmatrix} , \boldsymbol{P}_1^{-1}\boldsymbol{A}\boldsymbol{P}_1 = \begin{pmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{pmatrix}$ $\begin{aligned} &\boldsymbol{B}-2\boldsymbol{E} = \begin{pmatrix} 0 & 1 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & -4 \end{pmatrix} \to \begin{pmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix},\ \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}为2的一个特征向量 \\ &\boldsymbol{B}+\boldsymbol{E} = \begin{pmatrix} 3 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix} \to \begin{pmatrix} 3 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{pmatrix},\ \begin{pmatrix} 1 \\ -3 \\ 0 \end{pmatrix}为-1的一个特征向量 \\ &\boldsymbol{B}+2\boldsymbol{E} = \begin{pmatrix} 4 & 1 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix},\ \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}为-2的一个特征向量 \\ &令\boldsymbol{P}_2 = \begin{pmatrix} 1 & 1 & 0 \\ 0 & -3 & 0 \\ 0 & 0 & 1 \end{pmatrix} ，\boldsymbol{P}_2^{-1}\boldsymbol{A}\boldsymbol{P}_2= \begin{pmatrix} 2 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{pmatrix}\end{aligned}$
$\begin{aligned} &\boldsymbol{P}_1^{-1}\boldsymbol{A}\boldsymbol{P}_1 =\ \boldsymbol{P}_2^{-1}\boldsymbol{B}\boldsymbol{P}_2 \\ &\boldsymbol{P}_2\boldsymbol{P}_1^{-1}\boldsymbol{A}\boldsymbol{P}_1\boldsymbol{P}_2^{-1} = \boldsymbol{B},\ (\boldsymbol{P}_1\boldsymbol{P}^{-1}_2)^{-1}\boldsymbol{A}(\boldsymbol{P}_1\boldsymbol{P}^{-1}_2) = \boldsymbol{B} \\ &令\boldsymbol{P} = \boldsymbol{P}_1\boldsymbol{P}_2^{-1} = \begin{pmatrix} -1 & -2 & 1 \\ 2 & 1 & -2 \\ 0 & 0 & -4 \end{pmatrix} \begin{pmatrix} 1 & \dfrac{1}{3} & 0 \\ 0 & -\dfrac{1}{3} & 0 \\ 0 & 0 & 1 \end{pmatrix} = \begin{pmatrix} -1 & \dfrac{1}{3} & -1 \\ 2 & \dfrac{1}{3} & -2 \\ 0 & 0 & -4 \end{pmatrix} \end{aligned}$
验证
$\begin{aligned} \boldsymbol{A}\boldsymbol{P} &= \begin{pmatrix} -2 & -2 & 1 \\ 2 & 3 & -2 \\ 0 & 0 & -2 \end{pmatrix} \begin{pmatrix} -1 & \dfrac{1}{3} & 1 \\ 2 & \dfrac{1}{3} & 2 \\ 0 & 0 & 4 \end{pmatrix} = \begin{pmatrix} -2 & -\dfrac{4}{3} & 2 \\ 4 & \dfrac{5}{3} & -4 \\ 0 & 0 & -8 \end{pmatrix} \\ \boldsymbol{P}\boldsymbol{B} &= \begin{pmatrix} -1 & \dfrac{1}{3} & 1 \\ 2 & \dfrac{1}{3} & 2 \\ 0 & 0 & 4 \end{pmatrix} \begin{pmatrix} 2 & 1 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -2 \end{pmatrix} = \begin{pmatrix} -2 & -\dfrac{4}{3} & 2 \\ 4 & \dfrac{5}{3} & -4 \\ 0 & 0 & -8 \end{pmatrix} \end{aligned}$
$\begin{aligned} &\text{【小结】如果矩阵 } \boldsymbol{A},\ \boldsymbol{B} \text{ 相似且都可以相似对角化，则可以按照如下方法计算可逆矩阵 } \boldsymbol{P},\\&\ \text{使得 } \boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P} = \boldsymbol{B}: \\ &① \text{ 分别求出可逆矩阵 } \boldsymbol{P}_1,\ \boldsymbol{P}_2 \text{ 使得 } \boldsymbol{P}_1^{-1}\boldsymbol{A}\boldsymbol{P}_1 = \boldsymbol{P}_2^{-1}\boldsymbol{B}\boldsymbol{P}_2 = \boldsymbol{\varLambda}; \\ &② \text{ 由 } \boldsymbol{P}_1^{-1}\boldsymbol{A}\boldsymbol{P}_1 = \boldsymbol{P}_2^{-1}\boldsymbol{B}\boldsymbol{P}_2 \text{ 可得 } \boldsymbol{P}_2\boldsymbol{P}_1^{-1}\boldsymbol{A}\boldsymbol{P}_1\boldsymbol{P}_2^{-1} = (\boldsymbol{P}_1\boldsymbol{P}_2^{-1})^{-1}\boldsymbol{A}(\boldsymbol{P}_1\boldsymbol{P}_2^{-1}) = \boldsymbol{B},\\&\ \text{故令 } \boldsymbol{P} = \boldsymbol{P}_1\boldsymbol{P}_2^{-1} \text{ 则有 } \boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P} = \boldsymbol{B}。 \end{aligned}$
$\begin{aligned} &\boldsymbol{A} \sim \boldsymbol{B} \ (\boldsymbol{A}与\boldsymbol{B}相似) \\ &①\ 若\boldsymbol{A},\boldsymbol{B}均可以相似对角化 \\ &②\ 一个可,一个不可.\ \times \\ &③\ 若两个均不可.\ (\text{没考点, 只能用定义}) \\ &\quad \text{参考87题.} \end{aligned}$
$\begin{aligned} &\text{eg: } \boldsymbol{A} = \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix},\ \boldsymbol{B} = \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \\ &\text{问：求所有可逆矩阵} \boldsymbol{P},\ \text{使得} \boldsymbol{P}^{-1}\boldsymbol{A}\boldsymbol{P} = \boldsymbol{B}？ \\ &\text{设} \boldsymbol{P} = \begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix},\ \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix} = \begin{pmatrix} x_1 & x_2 \\ x_3 & x_4 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} \\ &\begin{pmatrix} x_1 + 2x_3 & x_2 + 2x_4 \\ x_3 & x_4 \end{pmatrix} = \begin{pmatrix} x_1 + 2x_2 & x_2 \\ x_3 + 2x_4 & x_4 \end{pmatrix} \\ &\begin{cases} x_1 + 2x_3 = x_1 + 2x_2 \\ x_2 + 2x_4 = x_2 \\ x_3 = x_3 + 2x_4 \\ x_4 = x_4 \end{cases} \implies \begin{cases} x_2 - x_3 = 0 \\ x_4 = 0 \\ x_4 = 0 \\ 0 = 0 \end{cases} \\ &\begin{pmatrix} 0 & 1 & -1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix},\ k_1 \begin{pmatrix} 1 \\ 0 \\ 0 \\ 0 \end{pmatrix} + k_2 \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} k_1 \\ k_2 \\ k_2 \\ 0 \end{pmatrix} \end{aligned}$
验证
$\begin{aligned} \begin{pmatrix} 1 & 2 \\ 0 & 1 \end{pmatrix} \begin{pmatrix} k_1 & k_2 \\ k_2 & 0 \end{pmatrix} &= \begin{pmatrix} k_1 + 2k_2 & k_2 \\ k_2 & 0 \end{pmatrix} \\ \begin{pmatrix} k_1 & k_2 \\ k_2 & 0 \end{pmatrix} \begin{pmatrix} 1 & 0 \\ 2 & 1 \end{pmatrix} &= \begin{pmatrix} k_1 + 2k_2 & k_2 \\ k_2 & 0 \end{pmatrix} \\ \begin{vmatrix} k_1 & k_2 \\ k_2 & 0 \end{vmatrix} &= -k_2^2 \neq 0 \\ k_2 &\neq 0 \end{aligned}$
得
$\boldsymbol{P} = \begin{pmatrix} k_1 & k_2 \\ k_2 & 0 \end{pmatrix}, \quad k_2 \neq 0$