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19 , 试求一个正交的相似变换矩阵 , 将下列对称阵化为对角阵:

(1) [2−20−21−20−20]\begin{bmatrix} 2 & - 2 & 0 \\ - 2 & 1 & - 2 \\ 0 & - 2 & 0 \end{bmatrix} (2) [22−225−4−2−45]\begin{bmatrix} 2 & 2 & - 2 \\ 2 & 5 & - 4 \\ - 2 & - 4 & 5 \end{bmatrix}

解:(1)先求特征值:

|A -λE|=|2−λ−20−21−λ−20−2−λ|\left| \begin{matrix} 2 - \lambda & - 2 & 0 \\ - 2 & 1 - \lambda & - 2 \\ 0 & - 2 & - \lambda \end{matrix} \right|=-λ(1-λ)(2-λ)-4(2-λ)+4λ=0

特征值为λ1=-2 , λ2=1 , λ3=4

再求特征向量:

对应λ1=-2 , 解方程(A+2E)x=[4−20−23−20−22]\begin{bmatrix} 4 & - 2 & 0 \\ - 2 & 3 & - 2 \\ 0 & - 2 & 2 \end{bmatrix}x=0

得同解方程组：{x1−12x3=0x2−x3=0\left\{ \begin{array}{r} x_{1} - \frac{1}{2}x_{3} = 0 \\ x_{2} - x_{3} = 0 \end{array} \right.\

得参数形式：{x1=12x3x2=x3\left\{ \begin{array}{r} x_{1} = \frac{1}{2}x_{3} \\ x_{2} = x_{3} \end{array} \right.\ (x3 可任意取值)

得参数形式：{x1=12cx2=cx3=c\left\{ \begin{matrix} x_{1} = {\frac{1}{2}c} \\ x_{2} = c \\ x_{3} = c \end{matrix} \right.\ (其中x3=c )

得齐次通解：[x1x2x3]=[12ccc]=12c[122]\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} {\frac{1}{2}c} \\ c \\ c \end{bmatrix} = \frac{1}{2}c\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} (c为任意实数)

得特征向量p1=[122]\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}\

对应λ2=1 , 解方程(A-E)x=[1−20−20−20−2−1]\begin{bmatrix} 1 & - 2 & 0 \\ - 2 & 0 & - 2 \\ 0 & - 2 & - 1 \end{bmatrix}x=0

得同解方程组：{x1+x3=0x2+12x3=0\left\{ \begin{array}{r} x_{1} + x_{3} = 0 \\ x_{2} + \frac{1}{2}x_{3} = 0 \end{array} \right.\

得参数形式：{x1=−x3x2=−12x3\left\{ \begin{array}{r} x_{1} = - x_{3} \\ x_{2} = \frac{- 1}{2}x_{3} \end{array} \right.\ (x3 可任意取值)

得参数形式：{x1=−cx2=−12cx3=c\left\{ \begin{matrix} x_{1} = {- c} \\ x_{2} = \frac{- 1}{2}c \\ x_{3} = c \end{matrix} \right.\ (其中x3=c )

得齐次通解：[x1x2x3]=[−12−ccc]=12c[21−2]\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \left\lbrack \frac{- 1}{2}\begin{matrix} {- c} \\ c \\ c \end{matrix} \right\rbrack = \frac{1}{2}c\begin{bmatrix} 2 \\ 1 \\ - 2 \end{bmatrix} (c为任意实数)

得特征向量p2=[−−212]\left\lbrack - \begin{matrix} {- 2} \\ 1 \\ 2 \end{matrix} \right\rbrack

对应λ3=4 , 解方程(A-4E)x=[−2−20−2−3−20−2−4]\begin{bmatrix} - 2 & - 2 & 0 \\ - 2 & - 3 & - 2 \\ 0 & - 2 & - 4 \end{bmatrix}x=0

得同解方程组：{x1−2x3=0x2+2x3=0\left\{ \begin{array}{r} x_{1} - 2x_{3} = 0 \\ x_{2} + 2x_{3} = 0 \end{array} \right.\

得参数形式：{x1=2x3x2=−2x3\left\{ \begin{array}{r} x_{1} = 2x_{3} \\ x_{2} = - 2x_{3} \end{array} \right.\ (x3 可任意取值)

得参数形式：{x1=2cx2=−2cx3=c\left\{ \begin{matrix} x_{1} = {2c} \\ x_{2} = - 2c \\ x_{3} = c \end{matrix} \right.\ (其中x3=c )

得齐次通解：[x1x2x3]=[2c−2cc]=c[−221]\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} {2c} \\ - 2c \\ c \end{bmatrix} = c\left\lbrack - \begin{matrix} 2 \\ 2 \\ 1 \end{matrix} \right\rbrack (c为任意实数)

得特征向量p3=[−221]\left\lbrack - \begin{matrix} 2 \\ 2 \\ 1 \end{matrix} \right\rbrack

正交单位化特征向量

对称矩阵不同特征值对应的特征向量已正交（ξ₁与ξ₂、ξ₁与ξ₃、ξ₂与ξ₃内积为0），只需单位化：

单位化ξ₁: η1=ξ1|ξ1|=112+22+22[122]=13[122]\eta_{1} = \frac{\xi_{1}}{|\xi_{1}|} = \frac{1}{\sqrt{1^{2} + 2^{2} + 2^{2}}}\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 1 \\ 2 \\ 2 \end{bmatrix}

单位化ξ₂: η2=ξ2|ξ2|=122+12+(−2)2[21−2]=13[21−2]\eta_{2} = \frac{\xi_{2}}{|\xi_{2}|} = \frac{1}{\sqrt{2^{2} + 1^{2} + ( - 2)^{2}}}\begin{bmatrix} 2 \\ 1 \\ - 2 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 2 \\ 1 \\ - 2 \end{bmatrix}

单位化ξ₃: ξ3|ξ3|=122+(−2)2+12[2−21]=13[2−21]\frac{\xi_{3}}{|\xi_{3}|} = \frac{1}{\sqrt{2^{2} + ( - 2)^{2} + 1^{2}}}\begin{bmatrix} 2 \\ - 2 \\ 1 \end{bmatrix} = \frac{1}{3}\begin{bmatrix} 2 \\ - 2 \\ 1 \end{bmatrix}

令P=(p1 , p2 , p3)=13[12221−22−21]\frac{1}{3}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{bmatrix} , 则P是正交阵

且 有P-1AP=PTAP=19[12221−22−21][2−20−21−20−20][12221−22−21]\frac{1}{9}\begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{bmatrix}\begin{bmatrix} 2 & - 2 & 0 \\ - 2 & 1 & - 2 \\ 0 & - 2 & 0 \end{bmatrix}\ \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & - 2 \\ 2 & - 2 & 1 \end{bmatrix}=[−214]\begin{bmatrix} - 2 & & \\ & 1 & \\ & & 4 \end{bmatrix}

(2) det(A -λE)=|2−λ2−225−λ−4−2−45−λ|\left| \begin{matrix} 2 - \lambda & 2 & - 2 \\ 2 & 5 - \lambda & - 4 \\ - 2 & - 4 & 5 - \lambda \end{matrix} \right|=-(1-λ)2(λ-10)

所以A的特征值为λ1=10 , λ2=λ3=1(二重根)

对应λ1=10 , 解方程(A -10E)x=[−82−22−5−4−2−4−5]\begin{bmatrix} - 8 & 2 & - 2 \\ 2 & - 5 & - 4 \\ - 2 & - 4 & - 5 \end{bmatrix}x=0

得同解方程组：{x1+12x3=0x2+x3=0\left\{ \begin{array}{r} x_{1} + \frac{1}{2}x_{3} = 0 \\ x_{2} + x_{3} = 0 \end{array} \right.\

得参数形式：{x1=−12x3x2=−x3\left\{ \begin{array}{r} x_{1} = \frac{- 1}{2}x_{3} \\ x_{2} = - x_{3} \end{array} \right.\ (x3 可任意取值)

得参数形式：{x1=−12cx2=−cx3=c\left\{ \begin{matrix} x_{1} = {\frac{- 1}{2}c} \\ x_{2} = - c \\ x_{3} = c \end{matrix} \right.\ (其中x3=c )

得齐次通解：[x1x2x3]=[−12c−cc]=c[−−1211]\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} {\frac{- 1}{2}c} \\ - c \\ c \end{bmatrix} = c\left\lbrack - \begin{matrix} \frac{- 1}{2} \\ 1 \\ 1 \end{matrix} \right\rbrack (c为任意实数)

得特征向量p1=[−1−22]\begin{bmatrix} {- 1} \\ - 2 \\ 2 \end{bmatrix} 得单位特征向量p1=13[−1−22]\frac{1}{3}\begin{bmatrix} - 1 \\ - 2 \\ 2 \end{bmatrix}

对应λ2=λ3=1 , 解方程(A-E)x=[12−224−4−2−44]\begin{bmatrix} 1 & 2 & - 2 \\ 2 & 4 & - 4 \\ - 2 & - 4 & 4 \end{bmatrix}x=0

得同解方程组： x1+2x2-2x3=0

得参数形式： x1=-2x2+2x3 (x2 , x3可任意取值)

得参数形式：{x1=−2c1+2c2x2=c1x3=c2\left\{ \begin{matrix} x_{1} = {- 2c_{1} + 2c_{2}} \\ x_{2} = c_{1} \\ x_{3} = c_{2} \end{matrix} \right.\ (其中x2=c1, x3=c2 )

得齐次通解：[x1x2x3]=[−2c1+2c2c1c2]=c1[−210]+c2[201]\begin{bmatrix} x_{1} \\ x_{2} \\ x_{3} \end{bmatrix} = \begin{bmatrix} {- 2c_{1} + 2c_{2}} \\ c_{1} \\ c_{2} \end{bmatrix} = c_{1}\begin{bmatrix} {- 2} \\ 1 \\ 0 \end{bmatrix} + c_{2}\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} (c1, c2为任意实数)

得线性无关特征向量: a1=[−210]\begin{bmatrix} - 2 \\ 1 \\ 0 \end{bmatrix} , a2=[201]\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix}

将a1和a2正交化得 , b1=a1=[−210]b_{1} = a_{1} = \begin{bmatrix} - 2 \\ 1 \\ 0 \end{bmatrix}

b2=a2- [b1,a2][b1,b1]\frac{\left\lbrack b_{1}\ ,\ a_{2} \right\rbrack}{\left\lbrack b_{1}\ ,\ b_{1} \right\rbrack}b1=[201]−45[−210]=[2−85451]=[25451]\begin{bmatrix} 2 \\ 0 \\ 1 \end{bmatrix} - \frac{4}{5}\begin{bmatrix} - 2 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 - \frac{8}{5} \\ \frac{4}{5} \\ 1 \end{bmatrix} = \begin{bmatrix} \frac{2}{5} \\ \frac{4}{5} \\ 1 \end{bmatrix}

再分别单位化得: p2=15[−210]\frac{1}{\sqrt{5}}\begin{bmatrix} - 2 \\ 1 \\ 0 \end{bmatrix} ,

p3=1(25)2+(45)2+12[25451]=54+16+25[25451]=145[245]\frac{1}{\sqrt{\left( \frac{2}{5} \right)^{2} + \left( \frac{4}{5} \right)^{2} + 1^{2}}}\begin{bmatrix} \frac{2}{5} \\ \frac{4}{5} \\ 1 \end{bmatrix} = \frac{5}{\sqrt{4 + 16 + 25}}\begin{bmatrix} \frac{2}{5} \\ \frac{4}{5} \\ 1 \end{bmatrix} = \frac{1}{\sqrt{45}}\begin{bmatrix} 2 \\ 4 \\ 5 \end{bmatrix}

令P=(p1 , p2 , p3)=[−13−25235−2315435230535]\begin{bmatrix} - \frac{1}{3} & - \frac{2}{\sqrt{5}} & \frac{2}{3\sqrt{5}} \\ - \frac{2}{3} & \frac{1}{\sqrt{5}} & \frac{4}{3\sqrt{5}} \\ \frac{2}{3} & 0 & \frac{5}{3\sqrt{5}} \end{bmatrix} , 则P是正交阵

且有P-1AP=PTAP =[−13−2323−25150235435535][22−225−4−2−45][−13−25235−2315435230535]\begin{bmatrix} - \frac{1}{3} & - \frac{2}{3} & \frac{2}{3} \\ - \frac{2}{\sqrt{5}} & \frac{1}{\sqrt{5}} & 0 \\ \frac{2}{3\sqrt{5}} & \frac{4}{3\sqrt{5}} & \frac{5}{3\sqrt{5}} \end{bmatrix}\begin{bmatrix} 2 & 2 & - 2 \\ 2 & 5 & - 4 \\ - 2 & - 4 & 5 \end{bmatrix}\begin{bmatrix} - \frac{1}{3} & - \frac{2}{\sqrt{5}} & \frac{2}{3\sqrt{5}} \\ - \frac{2}{3} & \frac{1}{\sqrt{5}} & \frac{4}{3\sqrt{5}} \\ \frac{2}{3} & 0 & \frac{5}{3\sqrt{5}} \end{bmatrix}

=[1110]\begin{bmatrix} 1 & & \\ & 1 & \\ & & 10 \end{bmatrix}

20 , 设矩阵A=[1−2−4−2x−2−4−21]\begin{bmatrix} 1 & - 2 & - 4 \\ - 2 & x & - 2 \\ - 4 & - 2 & 1 \end{bmatrix}与Λ=[5−4y]\begin{bmatrix} 5 & & \\ & - 4 & \\ & & y \end{bmatrix}相似 , 求x , y

并求一个正交阵P , 使P-1AP=Λ

解：

根据相似矩阵的性质：

1. 迹相等：tr(A)=tr(Λ)tr(A) = tr(\Lambda)

2. 行列式相等：|A|=|Λ||A| = |\Lambda|

步骤 1：利用迹相等

tr(A)=1+x+1=x+2,tr(Λ)=5+(−4)+y=y+1tr(A) = 1 + x + 1 = x + 2,\quad tr(\Lambda) = 5 + ( - 4) + y = y + 1

令其相等：x+2=y+1⇒y=x+1x + 2 = y + 1\quad \Rightarrow \quad y = x + 1

步骤 2：利用行列式相等

|A|=|1−2−4−2x−2−4−21|=−15x−40|A| = \left| \begin{matrix} 1 & - 2 & - 4 \\ - 2 & x & - 2 \\ - 4 & - 2 & 1 \end{matrix} \right| = - 15x - 40

|Λ|=5⋅(−4)⋅y=−20y|\Lambda| = 5 \cdot ( - 4) \cdot y = - 20y

令行列式相等：−15x−40=−20y- 15x - 40 = - 20y

代入 y=x+1y = x + 1：

−15x−40=−20(x+1)⇒5x=20得x=4,y=x+1=5- 15x - 40 = - 20(x + 1)\quad \Rightarrow \quad 5x = 20\quad 得\quad x = 4,\quad y = x + 1 = 5

二、求正交矩阵 PP，使得 P−1AP=ΛP^{- 1}AP = \Lambda

此时：A=[1−2−4−24−2−4−21],Λ=[5−45]A = \begin{bmatrix} 1 & - 2 & - 4 \\ - 2 & 4 & - 2 \\ - 4 & - 2 & 1 \end{bmatrix},\quad\Lambda = \begin{bmatrix} 5 & & \\ & - 4 & \\ & & 5 \end{bmatrix}

1. 求特征值 λ=5\lambda = 5 对应的正交单位特征向量

解 (A−5E)x=[−4−2−4−2−1−2−4−2−4]x=0(A - 5E)x = \begin{bmatrix} - 4 & - 2 & - 4 \\ - 2 & - 1 & - 2 \\ - 4 & - 2 & - 4 \end{bmatrix}x = 0：

得同解方程：2x1+x2+2x3=02x_{1} + x_{2} + 2x_{3} = 0

得基础解系：ξ1=[−101],ξ3=[−120]\xi_{1} = \begin{bmatrix} - 1 \\ 0 \\ 1 \end{bmatrix},\quad\xi_{3} = \begin{bmatrix} - 1 \\ 2 \\ 0 \end{bmatrix}

正交化：

β1=ξ1=[−101]\beta_{1} = \xi_{1} = \begin{bmatrix} - 1 \\ 0 \\ 1 \end{bmatrix}

β3=ξ3−⟨ξ3,β1⟩⟨β1,β1⟩β1=[−120]−12[−101]=[−122−12]\beta_{3} = \xi_{3} - \frac{\langle\xi_{3},\beta_{1}\rangle}{\langle\beta_{1},\beta_{1}\rangle}\beta_{1} = \begin{bmatrix} - 1 \\ 2 \\ 0 \end{bmatrix} - \frac{1}{2}\begin{bmatrix} - 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} - \frac{1}{2} \\ 2 \\ - \frac{1}{2} \end{bmatrix}

单位化：

p1=β1∥β1∥=12[−101]=[−22022]p_{1} = \frac{\beta_{1}}{\parallel \beta_{1} \parallel} = \frac{1}{\sqrt{2}}\begin{bmatrix} - 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} - \frac{\sqrt{2}}{2} \\ 0 \\ \frac{\sqrt{2}}{2} \end{bmatrix}

p3=β3∥β3∥=232[−122−12]=[−26223−26]p_{3} = \frac{\beta_{3}}{\parallel \beta_{3} \parallel} = \frac{2}{3\sqrt{2}}\begin{bmatrix} - \frac{1}{2} \\ 2 \\ - \frac{1}{2} \end{bmatrix} = \begin{bmatrix} - \frac{\sqrt{2}}{6} \\ \frac{2\sqrt{2}}{3} \\ - \frac{\sqrt{2}}{6} \end{bmatrix}

2. 求特征值 λ=−4\lambda = - 4 对应的单位特征向量

解 (A+4E)x=[5−2−4−28−2−4−25]x=0(A + 4E)x = \begin{bmatrix} 5 & - 2 & - 4 \\ - 2 & 8 & - 2 \\ - 4 & - 2 & 5 \end{bmatrix}x = 0

得同解方程组：{x1−x3=0x2−12x3=0\left\{ \begin{matrix} x_{1} - x_{3} = 0 \\ x_{2} - \frac{1}{2}x_{3} = 0 \end{matrix} \right.\

得基础解系：ξ2=[212]\xi_{2} = \begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix}

单位化：p2=13[212]p_{2} = \frac{1}{3}\begin{bmatrix} 2 \\ 1 \\ 2 \end{bmatrix}

3. 构造正交矩阵 PP

P=[p1p2p3]=[−2223−260132232223−26]P = \begin{bmatrix} p_{1} & p_{2} & p_{3} \end{bmatrix} = \begin{bmatrix} - \frac{\sqrt{2}}{2} & \frac{2}{3} & - \frac{\sqrt{2}}{6} \\ 0 & \frac{1}{3} & \frac{2\sqrt{2}}{3} \\ \frac{\sqrt{2}}{2} & \frac{2}{3} & - \frac{\sqrt{2}}{6} \end{bmatrix}

由于 p1,p2,p3p_{1},p_{2},p_{3} 是两两正交的单位向量，PP 是正交矩阵，

求正交矩阵 PP 的逆矩阵 P−1P^{- 1}

根据正交矩阵的性质：

若 PP 为正交矩阵，则其逆矩阵等于其转置矩阵，即P−1=PT.P^{- 1} = P^{T}.

对矩阵 PP 进行转置运算：P−1=PT=[−22022231323−26223−26].P^{- 1} = P^{T} = \begin{bmatrix} - \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ - \frac{\sqrt{2}}{6} & \frac{2\sqrt{2}}{3} & - \frac{\sqrt{2}}{6} \end{bmatrix}.

验证 P−1AP=ΛP^{- 1}AP = \Lambda

由于 P−1=PTP^{- 1} = P^{T}，只需验证 PTAP=ΛP^{T}AP = \Lambda 即可。

PTAP=[−22022231323−26223−26][1−2−4−24−2−4−21][−2223−260132232223−26]P^{T}AP = \begin{bmatrix} - \frac{\sqrt{2}}{2} & 0 & \frac{\sqrt{2}}{2} \\ \frac{2}{3} & \frac{1}{3} & \frac{2}{3} \\ - \frac{\sqrt{2}}{6} & \frac{2\sqrt{2}}{3} & - \frac{\sqrt{2}}{6} \end{bmatrix}\begin{bmatrix} 1 & - 2 & - 4 \\ - 2 & 4 & - 2 \\ - 4 & - 2 & 1 \end{bmatrix}\begin{bmatrix} - \frac{\sqrt{2}}{2} & \frac{2}{3} & - \frac{\sqrt{2}}{6} \\ 0 & \frac{1}{3} & \frac{2\sqrt{2}}{3} \\ \frac{\sqrt{2}}{2} & \frac{2}{3} & - \frac{\sqrt{2}}{6} \end{bmatrix}

=[5000−40005]=Λ.= \begin{bmatrix} 5 & 0 & 0 \\ 0 & - 4 & 0 \\ 0 & 0 & 5 \end{bmatrix} = \Lambda.

经计算验证，P−1AP=ΛP^{- 1}AP = \Lambda 成立。

故该正交相似对角化结果正确。

最终答案：x=4,y=5x = 4,\ y = 5，P=[−2223−260132232223−26]P = \begin{bmatrix} - \frac{\sqrt{2}}{2} & \frac{2}{3} & - \frac{\sqrt{2}}{6} \\ 0 & \frac{1}{3} & \frac{2\sqrt{2}}{3} \\ \frac{\sqrt{2}}{2} & \frac{2}{3} & - \frac{\sqrt{2}}{6} \end{bmatrix}