线代强化NO17|特征值与特征向量|抽象型-优快云博客

抽象型矩阵的特征值与特征向量

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$\begin{aligned} &|A|=0,0是特征值 \\ &A(\alpha + \beta) = \alpha + \beta,\ 1\ \text{是特征值}. \\ &A(\alpha - \beta) = \beta - \alpha = (-1)(\alpha - \beta),\ -1\ \text{是特征值}.\\ &A的特征值:-1, 0, 1. \quad A + 2E的特征值: 1, 2, 3 \\ &(A + 2E)^* = |A + 2E|(A + 2E)^{-1} \\ &|A+2E|=1 \cdot 2 \cdot 3 = 6\\&(A+2E)^{-1}的特征值为1,\frac{1}{2},\frac{1}{3}\\&(A+2E)^{*}的特征值为6\cdot1,6\cdot\frac{1}{2},6\cdot\frac{1}{3}=6,3,2\\&\text{tr}((A + 2E)^*) = 6+3+2=11 \end{aligned}$
$\boxed{【小结】本题可作为结论来记，若有\ A\alpha = \beta,\ A\beta = \alpha，则\ A\ 的特征值含1和-1。}$
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$\begin{aligned} &\text{解：}A(\beta_1, \beta_2, \beta_3) = 2(\beta_1, \beta_2, \beta_3),\ A\beta_1 = 2\beta_1,\ A\beta_2 = 2\beta_2,\ A\beta_3 = 2\beta_3 \\ &2\ \text{是}A\ \text{的特征值，}r(B)=2,\ 2\ \text{的重数} \geq 2 \\ &(E+A)\boldsymbol{x}=0\ \text{有非零解，}-1\ \text{为其特征值} \\ &A\ \text{的特征值：}-1,\ 2,\ 2 \\ &A^* = |A|A^{-1}\ \text{的特征值：}4,\ -2,\ -2 \\ &A^* - E\ \text{的特征值：}3,\ -3,\ -3 \\ &|A^* - E| = 27 \end{aligned}$
$\begin{aligned} \\&【小结】\\&(1)\ \text{如果}\ AB = C，\text{则可以检验}\ B\ \text{和}\ C\ \text{对应的列是否是成比例的，如果是，则可以} \\ &\text{通过将矩阵}\ B,C\ \text{列分块凑出特征值和特征向量的定义。特别地，如果}\ AB = kB，\text{则} \\ &B\ \text{的每一列（非零的）都是矩阵}\ A\ \text{属于特征值}\ k\ \text{的特征向量。} \\ & \\ &(2)\ \text{如果}\ r(A - kE) < n，\text{则}\ k\ \text{为矩阵}\ A\ \text{的特征值。} \end{aligned}$
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$\begin{aligned} &\text{解：}A\ \text{的特征值为}\ 1,\ 1,\ 7. \\ &1\ \text{的特征向量为}\ k_1\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} + k_2\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix},\ k_1,k_2\text{不全为}0； \\ &7\ \text{的特征向量为}\ k_3\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},\ k_3\neq0. \\ & \\ &A^* = |A|A^{-1}\ \text{的特征值为}\ 7,\ 7,\ 1. \\ &7\ \text{的特征向量为}\ k_1\begin{pmatrix} -1 \\ 1 \\ 0 \end{pmatrix} + k_2\begin{pmatrix} -1 \\ 0 \\ 1 \end{pmatrix},\ k_1,k_2\text{不全为}0； \\ &1\ \text{的特征向量为}\ k_3\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix},\ k_3\neq0. \end{aligned}$
$\begin{aligned} A\alpha &= \lambda\alpha \\ B(P^{-1}\alpha) &= P^{-1}APP^{-1}\alpha = P^{-1}A\alpha = P^{-1}\lambda\alpha = \lambda P^{-1}\alpha \end{aligned}$
$\begin{aligned} &\begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \to \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 0 & 1 & -1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \\ & \\ &\begin{pmatrix} 0 & 1 & -1 \\ 1 & 0 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} -1 & -1 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 1 \end{pmatrix} = \begin{pmatrix} 1 & -1 & 0 \\ -1 & -1 & 1 \\ 0 & 1 & 1 \end{pmatrix} \\ & \\ &\text{故}\ B + 2E\ \text{的特征值为}\ 9,\ 9,\ 3. \\ &9\ \text{的特征向量为}\ k_1\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix} + k_2\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix},\ k_1,k_2\ \text{不全为}0. \\ &3\ \text{的特征向量为}\ k_3\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix},\ k_3 \neq 0. \end{aligned}$
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选B。1和4
【小结】
$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline \text{矩阵} & A & kA & A^n & f(A) & A^{-1} & A^* & f(A^{-1}) & P^{-1}AP & A^T \\ \hline \text{特征值} & \lambda & k\lambda & \lambda^n & f(\lambda) & \lambda^{-1} & |A|\lambda^{-1} & f(\lambda^{-1}) & \lambda & \lambda \\ \hline \text{特征向量} & \boldsymbol{\alpha} & \boldsymbol{\alpha} & \boldsymbol{\alpha} & \boldsymbol{\alpha} & \boldsymbol{\alpha} & \boldsymbol{\alpha} & \boldsymbol{\alpha} & P^{-1}\boldsymbol{\alpha} &? \\ \hline \end{array}$
$\begin{aligned} &A\alpha = \lambda\alpha, \quad kA\alpha = k\lambda\alpha, \quad kA\alpha = (k\lambda)\alpha \\ & \\ &A^n\alpha = A^{n-1}A\alpha = A^{n-1}\lambda\alpha = \lambda A^{n-1}\alpha = \lambda A^{n-2}A\alpha = \lambda^2 A^{n-2}\alpha \\ &\quad = \cdots = \lambda^n\alpha \\ & \\ &A\alpha = \lambda\alpha, \quad A^{-1}A\alpha = A^{-1}\lambda\alpha, \quad \alpha = \lambda A^{-1}\alpha, \quad A^{-1}\alpha = \frac{1}{\lambda}\alpha \\ & \\ &|A^T - \lambda E| = |(A - \lambda E)^T| = |A - \lambda E| \end{aligned}$