华东师范大学 2017年高等代数第3题

计算特征多项式 $|\lambda I - A| = \begin{vmatrix} \lambda-4 & -1 & -1 \\ -1 & \lambda-4 & -1 \\ -1 & -1 & \lambda-4 \end{vmatrix}$。令 $t = \lambda-4$，则行列式化为 $\begin{vmatrix} t & -1 & -1 \\ -1 & t & -1 \\ -1 & -1 & t \end{vmatrix} = t^3 - 3t - 2 = (t-2)(t+1)^2$。因此特征值为 $\lambda_1 = 6$（单根），$\lambda_2 = 3$（二重根）。

对于 $\lambda_1 = 6$，解 $(6I - A)x = 0$：$\begin{pmatrix} 2 & -1 & -1 \\ -1 & 2 & -1 \\ -1 & -1 & 2 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$。解得 $x_1 = x_2 = x_3$，取基础解系 $\alpha_1 = (1,1,1)^T$。

对于 $\lambda_2 = 3$，解 $(3I - A)x = 0$：$\begin{pmatrix} -1 & -1 & -1 \\ -1 & -1 & -1 \\ -1 & -1 & -1 \end{pmatrix} \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix} = 0$，即 $x_1 + x_2 + x_3 = 0$。取基础解系 $\alpha_2 = (1,-1,0)^T$，$\alpha_3 = (1,0,-1)^T$。

对 $\alpha_2, \alpha_3$ 进行施密特正交化：令 $\beta_2 = \alpha_2 = (1,-1,0)^T$。计算 $\beta_3 = \alpha_3 - \frac{(\alpha_3,\beta_2)}{(\beta_2,\beta_2)} \beta_2 = (1,0,-1) - \frac{1}{2}(1,-1,0) = (\frac12, \frac12, -1)^T$。

将三个特征向量单位化：$p_1 = \frac{\alpha_1}{\|\alpha_1\|} = \frac{1}{\sqrt{3}}(1,1,1)^T$，$p_2 = \frac{\beta_2}{\|\beta_2\|} = \frac{1}{\sqrt{2}}(1,-1,0)^T$，$p_3 = \frac{\beta_3}{\|\beta_3\|} = \frac{1}{\sqrt{6}}(1,1,-2)^T$。

取正交矩阵 $T = (p_1, p_2, p_3)$，即 $T = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & -\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{3}} & 0 & -\frac{2}{\sqrt{6}} \end{pmatrix}$。则 $T^{-1}AT = T^TAT = \operatorname{diag}(6,3,3)$。