战老师求解的给定对称矩阵A\boldsymbol {A}AA14[5−3−35] \boldsymbol {A}\frac{1}{4}\begin{bmatrix} 5 -3 \\ -3 5 \end{bmatrix}A41​[5−3​−35​]一、求特征值特征方程∣λI−A∣0|\lambda \boldsymbol {I} - \boldsymbol {A}| 0∣λI−A∣0∣λI−A∣∣λ−543434λ−54∣(λ−54)2−(34)20 |\lambda \boldsymbol {I} - \boldsymbol {A}| \begin{vmatrix} \lambda - \frac{5}{4} \frac{3}{4} \\ \frac{3}{4} \lambda-\frac{5}{4} \end{vmatrix} \left(\lambda-\frac{5}{4}\right)^2 - \left(\frac{3}{4}\right)^2 0∣λI−A∣​λ−45​43​​43​λ−45​​​(λ−45​)2−(43​)20展开求解(λ−5)29 ⟹ λ−54±34 (\lambda-5)^2 9 \implies \lambda-\frac{5}{4} \pm\frac{3}{4}(λ−5)29⟹λ−45​±43​得特征值λ10.5,λ22 \lambda_1 0.5,\quad \lambda_2 2λ1​0.5,λ2​2二、求特征向量单位化对应λ10.5\lambda_10.5λ1​0.5(0.5I−A)x0 ⟹ [−343434−34][x1x2]0 (0.5\boldsymbol {I} - \boldsymbol {A})\boldsymbol{x} \boldsymbol{0} \implies \begin{bmatrix} -\frac{3}{4} \frac{3}{4} \\ \frac{3}{4} -\frac{3}{4} \end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} \boldsymbol{0}(0.5I−A)x0⟹[−43​43​​43​−43​​][x1​x2​​]0解得x1x2x_1x_2x1​x2​基础解系ξ1[11]\boldsymbol{\xi}_1\begin{bmatrix}1\\1\end{bmatrix}ξ1​[11​]单位化特征向量u112[11] \boldsymbol{u}_1 \frac{1}{\sqrt{2}}\begin{bmatrix}1\\1\end{bmatrix}u1​2​1​[11​]对应λ22\lambda_22λ2​2(2I−A)x0 ⟹ [34343434][x1x2]0 (2\boldsymbol {I} - \boldsymbol {A})\boldsymbol{x} \boldsymbol{0} \implies \begin{bmatrix} \frac{3}{4} \frac{3}{4} \\ \frac{3}{4} \frac{3}{4} \end{bmatrix} \begin{bmatrix}x_1\\x_2\end{bmatrix} \boldsymbol{0}(2I−A)x0⟹[43​43​​43​43​​][x1​x2​​]0解得x1−x2x_1-x_2x1​−x2​基础解系ξ2[−11]\boldsymbol{\xi}_2\begin{bmatrix}-1\\1\end{bmatrix}ξ2​[−11​]单位化特征向量u212[−11] \boldsymbol{u}_2 \frac{1}{\sqrt{2}}\begin{bmatrix}-1\\1\end{bmatrix}u2​2​1​[−11​]三、正交对角化分解AUΛU⊤\boldsymbol {A}\boldsymbol {U}\boldsymbol {\varLambda} \boldsymbol {U}^{\top}AUΛU⊤实对称矩阵它总是可以对角化的并且其特征向量可以构成一个正交矩阵。Λdiag⁡(λ1,λ2),U[u1,u2] \boldsymbol {\varLambda} {\operatorname {diag}}(\lambda_1,\lambda_2), \boldsymbol {U}[\boldsymbol {u}_1, \boldsymbol {u}_2]Λdiag(λ1​,λ2​),U[u1​,u2​]AUΛU⊤ \boldsymbol {A}\boldsymbol {U}\boldsymbol {\varLambda} \boldsymbol {U}^{\top}AUΛU⊤正交矩阵U\boldsymbol {U}U列向量为单位特征向量U[12−121212] \boldsymbol {U} \begin{bmatrix} \dfrac{1}{\sqrt{2}} -\dfrac{1}{\sqrt{2}} \\ \dfrac{1}{\sqrt{2}} \dfrac{1}{\sqrt{2}} \end{bmatrix}U​2​1​2​1​​−2​1​2​1​​​对角矩阵Λ\boldsymbol {\varLambda}Λ对角线为特征值Λ[0.5002] \boldsymbol {\varLambda} \begin{bmatrix} 0.5 0 \\ 0 2 \end{bmatrix}Λ[0.50​02​]