线代强化NO22|实对称矩阵

实对称矩阵与正定矩阵分析

最新推荐文章于 2025-11-27 05:38:45 发布

原创最新推荐文章于 2025-11-27 05:38:45 发布 · 299 阅读

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#矩阵 #线性代数 #机器学习

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$\begin{aligned} &\text{解：(I) 设}\boldsymbol{B} = \begin{pmatrix} 1 & 1 & -1 \\ 1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix},\ \text{则}\boldsymbol{A} = \boldsymbol{B} + (\alpha - 1)\boldsymbol{E},\ \boldsymbol{B}\text{的特征值为}0,0,3. \\ &\boldsymbol{A}\text{的特征值为}\alpha - 1,\ \alpha - 1,\ \alpha + 2. \\ &\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\ \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}\text{为}\alpha - 1\text{的两个特征向量；}\begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix}\text{为}\alpha + 2\text{的特征向量.} \\ & \\ &\text{注：用施密特正交化法} \\ &\text{令}\boldsymbol{\xi}_1 = \begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\ \boldsymbol{\xi}_2 = \boldsymbol{\xi}_2 - \frac{(\boldsymbol{\xi}_1,\boldsymbol{\xi}_2)}{(\boldsymbol{\xi}_1,\boldsymbol{\xi}_1)}\boldsymbol{\xi}_1 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} - \frac{-1}{2}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{1}{2} \\ \frac{1}{2} \\ 1 \end{pmatrix}. \end{aligned}$
$\begin{aligned} &(\boldsymbol{\xi}_1, \boldsymbol{\xi}_2) = (-1) \times \frac{1}{2} + 1 \times \frac{1}{2} + 0 \times 1 = -\frac{1}{2} + \frac{1}{2} = 0, \\ &\text{说明} \boldsymbol{\xi}_1 \text{和} \boldsymbol{\xi}_2 \text{已正交，施密特正交化成功。} \end{aligned}$
$\begin{aligned} &\text{单位化：}\ \frac{1}{\sqrt{2}}\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix},\ \frac{1}{\sqrt{6}}\begin{pmatrix} 1 \\ 1 \\ 2 \end{pmatrix}, \\ &\text{令}\ \boldsymbol{P} = \begin{pmatrix} -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} \\ 0 & \frac{2}{\sqrt{6}} & -\frac{1}{\sqrt{3}} \end{pmatrix}, \\ &\boldsymbol{P}^T\boldsymbol{A}\boldsymbol{P}\ \text{为对角阵}. \end{aligned}$
$\begin{aligned} &\text{解：(Ⅱ) 令}\boldsymbol{D}=(\alpha+3)\boldsymbol{E}-\boldsymbol{A}=\begin{pmatrix} 3 & -1 & 1 \\ -1 & 3 & 1 \\ 1 & 1 & 3 \end{pmatrix},\ \boldsymbol{D}\text{的特征值为}1,4,4. \\ &\text{法1：}\boldsymbol{P}^T\boldsymbol{D}\boldsymbol{P}=\begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{pmatrix},\ \boldsymbol{D}=\boldsymbol{P}\begin{pmatrix} 4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1 \end{pmatrix}\boldsymbol{P}^T=\boldsymbol{C}^2. \\ &\text{令}\boldsymbol{C}=\boldsymbol{P}\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}\boldsymbol{P}^T, \\ &\boldsymbol{C}^2=\boldsymbol{P}\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}\boldsymbol{P}^T\boldsymbol{P}\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}\boldsymbol{P}^T \\ &=\boldsymbol{P}\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}^2\boldsymbol{P}^T=\boldsymbol{D}. \\ &\text{又}\boldsymbol{C}\text{的特征值全为正，且}\boldsymbol{C}^T=\boldsymbol{C},\ \text{故}\boldsymbol{C}\text{为正定矩阵.} \\ &\boldsymbol{C}=\boldsymbol{P}\begin{pmatrix} 2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 1 \end{pmatrix}\boldsymbol{P}^T=\boldsymbol{P}\left(\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & -1 \end{pmatrix}+2\boldsymbol{E}\right)\boldsymbol{P}^T=\frac{1}{3}\begin{pmatrix} 5 & -1 & 1 \\ -1 & 5 & 1 \\ 1 & 1 & 5 \end{pmatrix}. \end{aligned}$
$\begin{aligned} &\text{法2：设}\ \boldsymbol{B} = \begin{pmatrix} 1 & 1 & -1 \\ 1 & 1 & -1 \\ -1 & -1 & 1 \end{pmatrix}\ , \\ &\boldsymbol{B}^n = (\text{tr}(\boldsymbol{B}))^{n-1}\boldsymbol{B},\boldsymbol{B}^2 = 3\boldsymbol{B} \\ &\boldsymbol{A} = \boldsymbol{B} + (\alpha - 1)\boldsymbol{E}. \end{aligned}$
$\begin{aligned} \boldsymbol{C}^2 &= (\alpha + 3)\boldsymbol{E} - (\boldsymbol{B} + (\alpha - 1)\boldsymbol{E}) = 4\boldsymbol{E} - \boldsymbol{B} \\ &= 4\boldsymbol{E} - \frac{4}{3}\boldsymbol{B} + \frac{1}{3}\boldsymbol{B} = 4\boldsymbol{E} - \frac{4}{3}\boldsymbol{B} + \frac{\boldsymbol{B}^2}{9} = \left(\frac{\boldsymbol{B}}{3} - 2\boldsymbol{E}\right)^2, \\ \\ 4\boldsymbol{E} - x\boldsymbol{B} - (1 - x)\boldsymbol{B} &= 4\boldsymbol{E} - x\boldsymbol{B} - \frac{1 - x}{3}\boldsymbol{B}^2, \\ \\ \Delta &= x^2 + \frac{16}{3}(1 - x) = 0, \\ 3x^2 - 16x + 16 &= 0, \\ (3x - 4)(x - 4) &= 0, \\ x &= 4\ \text{或}\ x = \frac{4}{3}. \end{aligned}$
$\text{令}\boldsymbol{C} = 2\boldsymbol{E} - \frac{\boldsymbol{B}}{3} = \begin{pmatrix} \frac{5}{3} & -\frac{1}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{5}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} & \frac{5}{3} \end{pmatrix}$
$故C为正定矩阵.\begin{aligned} \\ &\text{又}\boldsymbol{C}\text{的特征值全为正，且}\boldsymbol{C}^T=\boldsymbol{C},\ \text{故}\boldsymbol{C}\text{为正定矩阵.}\end{aligned}$
公式推导
$\begin{aligned} &\text{tr}(\boldsymbol{B}) = 1,\ \boldsymbol{B} = \alpha\boldsymbol{\beta}^T, \\ &\boldsymbol{B}^n = \alpha\boldsymbol{\beta}^T \cdot \alpha\boldsymbol{\beta}^T \cdot \ldots \cdot \alpha\boldsymbol{\beta}^T, \\ &= \alpha \cdot \text{tr}(\boldsymbol{B}) \cdot \text{tr}(\boldsymbol{B}) \cdot \ldots \cdot \text{tr}(\boldsymbol{B}) \cdot \boldsymbol{\beta}^T, \\ &= (\text{tr}(\boldsymbol{B}))^{n-1} \alpha\boldsymbol{\beta}^T = (\text{tr}(\boldsymbol{B}))^{n-1} \boldsymbol{B}. \end{aligned}$
$\begin{aligned} &\text{【小结】运用本题的方法可以将任何一个正定矩阵化为另一个正定矩阵的平方，思路有两个：} \\ &①\ \text{利用正交变换；} \\ &②\ \text{利用因式分解，凑成完全平方式。} \end{aligned}$
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$\begin{aligned} &(\text{A})\quad \text{等价} \\ &(\text{B})\quad \text{相似} \\ &(\text{C})\quad =Q\Lambda Q^T\ \text{相似且合同}\quad Q^T=Q^{-1} \\ &(\text{D})\quad \text{合同} \end{aligned}$
$\begin{aligned} &\text{解：(A) 若}A\text{的特征值为}1,-1,0,\text{则}A\text{可以相似对角化}.r(A)=2, \\ &r(\Lambda)=2,\text{故}A\text{与}\Lambda\text{等价}. \\ &\text{若存在可逆阵}P,Q,\text{使得}A=P\Lambda Q,\text{则}A\text{与}\Lambda\text{等价, 不能得到} \\ &A\text{的特征值与}\Lambda\text{的特征值相同}. \\ &\text{反例：}A=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix},r(A)=2.\text{但}A\text{的特征值不是}1,-1,0.\\&必要非充分 \end{aligned}$
$\begin{aligned} &(\text{B})\ A\text{的特征值为}1,-1,0,\text{故}A\text{可相似对角化}. \\ &\text{则由于}\Lambda\text{的特征值均为}1,-1,0,\text{故}A\text{与}\Lambda相似 \\ &\text{若}A\text{与}\Lambda\text{相似，则}A\text{与}\Lambda\text{的特征值相同}. \\ &\text{故}A\text{的特征值为}1,-1,0.\ \text{选}(\text{B}) \end{aligned}$
$\begin{aligned} &(C)\ \text{若存在正交矩阵}Q,\text{使得}A=Q\Lambda Q^{-1},\text{则}A\text{与}\Lambda\text{相似}, \\ &\Lambda\text{的特征值为}1,-1,0,\text{故}A\text{的特征值为}1,-1,0. \\ &\text{若}A\text{的特征值为}1,-1,0,\text{只能说明存在可逆矩阵}P \\ &\text{使得}A=P\Lambda P^{-1},\text{不一定存在一个正交矩阵}Q\text{满足}A=Q\Lambda Q^{-1}. \\ \\ &\text{若}A\text{为实对称矩阵，则一定存在一个正交矩阵}Q\text{使得}A=Q\Lambda Q^{-1}. \\ &\text{若}A\text{不是实对称矩阵，则一定不存在一个正交矩阵}Q\text{使得}A=Q\Lambda Q^{-1}. \\ \\ &\text{若存在一个正交矩阵}Q,\text{且}Q Q^T=E,\ Q^T=Q^{-1}, \\ &\text{使得}A=Q\Lambda Q^{-1}=Q\Lambda Q^T, \\ &A^T=(Q\Lambda Q^T)^T=(Q^T)^T\Lambda^T Q^T=Q\Lambda Q^T=A, \\ &A\text{为实对称矩阵}. \end{aligned}$
$\begin{aligned} &(\text{D})\ A\text{的特征值为}1,-1,0,\text{则不一定存在可逆矩阵}P, \\ &\text{使得}A=P\Lambda P^T. \\ &\text{若存在可逆矩阵}P\text{使得}A=P\Lambda P^T,\ A\text{为实对称矩阵}. \\ &\text{若存在可逆矩阵}P\text{使得}A=P\Lambda P^T,\text{得不到}A\text{的任何特征值}. \\ \\ &P=\begin{pmatrix}1&0&0\\0&1&1\\0&2&1\end{pmatrix},\ A=\begin{pmatrix}1&0&0\\0&1&1\\0&2&1\end{pmatrix}\begin{pmatrix}1&0&0\\0&-1&0\\0&0&0\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&1\\0&2&1\end{pmatrix}=\begin{pmatrix}1&0&0\\0&-1&-2\\0&-2&-4\end{pmatrix}. \\ &\text{tr}(A)=-4,\ \text{tr}(\Lambda)=0,\ A\text{与}\Lambda\text{不相似}. \end{aligned}$
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