利用正交变换判断二次曲面类型
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正交变换是欧式空间保持向量内积不变的线性变换。不仅保持向量的长度不变,而且还保持向量
的夹角不变。二维或三维空间中的旋转变换、关于某一条直线或平面的对称变换都是正交变换.投影变换、平移变换不是正交变换. -
正交变它从实内积空间 V V V 映射到 V V V自 身,保持变换前后内积不变.它应用在几何学上就是保持变换前后图形的不变性,这是正交变换的优势,从而达到了判断二次曲面类型、辨明二次曲面形状的目的.
任意一个实二次型
f ( x 1 , x 2 , ⋯ x n ) = ∑ i = 1 n ∑ j = 1 n a i j x i x j = X T A X , ( a i j = a j i ) f\left(x_{1}, x_{2}, \cdots x_{n}\right)=\sum_{i=1}^{n} \sum_{j=1}^{n} a_{i j} x_{i} x_{j}=X^{T} A X,\left(a_{i j}=a_{j i}\right) f(x1,x2,⋯xn)=i=1∑nj=1∑naijxixj=XTAX,(aij=aji)
都可以经过正交的线性替换变成平方和 λ 1 y 1 2 + λ 2 y 2 2 + ⋯ + λ n y n 2 \lambda_{1} y_{1}^{2}+\lambda_{2} y_{2}^{2}+\cdots+\lambda_{n} y_{n}^{2} λ1y12+λ2y22+⋯+λnyn2 ,其中 λ i ( i = 1 , 2 , … , n ) λ_i( i = 1,2,…,n) λi(i=1,2,…,n) 就是矩阵 A A A的特征多项式全部的根.(高代第八,九章)
前置结论:
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Q1. 方程
x 1 2 − 2 x 2 2 − 2 x 3 2 − 4 x 1 x 2 + 4 x 1 x 3 + 8 x 2 x 3 = 1 x_{1}^{2}-2 x_{2}^{2}-2 x_{3}^{2}-4 x_{1} x_{2}+4 x_{1} x_{3}+8 x_{2} x_{3}=1 x12−2x22−2x32−4x1x2+4x1x3+8x2x3=1
表示何种二次曲面?
解: 首先利用正交的线性替换将实二次型
f ( x 1 , x 2 , x 3 ) = x 1 2 − 2 x 2 2 − 2 x 3 2 − 4 x 1 x 2 + 4 x 1 x 3 + 8 x 2 x 3 f\left(x_{1}, x_{2}, x_{3}\right)=x_{1}^{2}-2 x_{2}^{2}-2 x_{3}^{2}-4 x_{1} x_{2}+4 x_{1} x_{3}+8 x_{2} x_{3} f(x1,x2,x3)=x12−2x22−2x32−4x1x2+4x1x3+8x2x3
化为标准形
A = ( 1 − 2 2 − 2 − 2 4 2 4 − 2 ) , X = ( x 1 x 2 x 3 ) A=\left(\begin{array}{ccc} 1 & -2 & 2 \\ -2 & -2 & 4 \\ 2 & 4 & -2 \end{array}\right), X=\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right) A= 1−22−2−2424−2 ,X= x1x2x3
A A A 的特征多项式为
f ( λ ) = ∣ λ I − A ∣ = ∣ λ − 1 2 − 2 2 λ + 2 − 4 − 2 − 4 λ + 2 ∣ = ( λ + 7 ) ( λ − 2 ) 2 , f(\lambda)=|\lambda I-A|=\left|\begin{array}{ccc} \lambda-1 & 2 & -2 \\ 2 & \lambda+2 & -4 \\ -2 & -4 & \lambda+2 \end{array}\right|=(\lambda+7)(\lambda-2)^{2}, f(λ)=∣λI−A∣= λ−12−22λ+2−4−2−4λ+2 =(λ+7)(λ−2)2,
A
A
A 的特征值为
λ
1
=
λ
2
=
2
,
λ
3
=
−
7.
\lambda_{1}=\lambda_{2}=2, \lambda_{3}=-7 .
λ1=λ2=2,λ3=−7.
可求得对应的
λ
1
=
λ
2
=
2
\lambda_{1}=\lambda_{2}=2
λ1=λ2=2 特征向量分别为
p
1
=
(
−
2
,
1
,
0
)
T
,
p
2
=
(
2
,
0
,
1
)
T
p_{1}=(-2,1,0) ^{T}, p_{2}=(2,0,1)^{T}
p1=(−2,1,0)T,p2=(2,0,1)T, 将其正交化
α 1 = p 1 = ( − 2 , 1 , 0 ) T , α 2 = p 2 − ( p 1 , α 1 ) ( α 1 , α 1 ) α 1 = ( 2 3 , 4 5 , 1 ) T , \alpha_{1}=p_{1}=(-2,1,0) ^T, \alpha_{2}=p_{2}-\frac{\left(p_{1}, \alpha_{1}\right)}{\left(\alpha_{1}, \alpha_{1}\right)} \alpha_{1}=\left(\frac{2}{3}, \frac{4}{5}, 1\right)^{T} \text { , } α1=p1=(−2,1,0)T,α2=p2−(α1,α1)(p1,α1)α1=(32,54,1)T ,
再单位化得
q 1 = ( − 2 5 , 1 5 , 0 ) T , q 2 = ( 2 3 5 , 4 3 5 , 5 3 5 ) T , q_{1}=\left(-\frac{2}{\sqrt{5}}, \frac{1}{\sqrt{5}}, 0\right)^{T}, q_{2}=\left(\frac{2}{3 \sqrt{5}}, \frac{4}{3 \sqrt{5}}, \frac{5}{3 \sqrt{5}}\right)^{T}, q1=(−52,51,0)T,q2=(352,354,355)T,
又对应 λ 3 = − 7 \lambda_{3}=-7 λ3=−7 的特征向量为 p 3 = ( − 1 , − 2 , 2 ) T p_{3}=(-1,-2,2)^{T} p3=(−1,−2,2)T , 单位化得 q 3 = ( − 1 3 , − 2 3 , 2 3 ) T q_{3}=\left(-\frac{1}{3},-\frac{2}{3}, \frac{2}{3}\right)^T q3=(−31,−32,32)T , 故正交变换
( x 1 x 2 x 3 ) = ( − 2 5 2 3 5 − 1 3 1 5 4 3 5 − 2 3 0 5 3 5 2 3 ) ( y 1 y 2 y 3 ) \left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{ccc} \frac{-2}{\sqrt{5}} & \frac{2}{3 \sqrt{5}} & -\frac{1}{3} \\ \frac{1}{\sqrt{5}} & \frac{4}{3 \sqrt{5}} & -\frac{2}{3} \\ 0 & \frac{5}{3 \sqrt{5}} & \frac{2}{3} \end{array}\right)\left(\begin{array}{l} y_{1} \\ y_{2} \\ y_{3} \end{array}\right) x1x2x3 = 5−2510352354355−31−3232 y1y2y3
将实二次型 f ( x 1 , x 2 , x 3 ) f\left(x_{1}, x_{2}, x_{3}\right) f(x1,x2,x3) 化为标准形
f = 2 y 1 2 + 2 y 2 2 − 7 y 3 2 , f=2 y_{1}^{2}+2 y_{2}^{2}-7 y_{3}^{2}, f=2y12+2y22−7y32,
可知方程 f ( x 1 , x 2 , x 3 ) = x 1 2 − 2 x 2 2 − 2 x 3 2 − 4 x 1 x 2 + 4 x 1 x 3 + 8 x 2 x 3 = 1 f\left(x_{1}, x_{2}, x_{3}\right)=x_{1}^{2}-2 x_{2}^{2}-2 x_{3}^{2}-4 x_{1} x_{2}+4 x_{1} x_{3}+8 x_{2} x_{3}=1 f(x1,x2,x3)=x12−2x22−2x32−4x1x2+4x1x3+8x2x3=1 表示的曲面为单叶双曲面.
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Q2.
判断二次曲面 2 x 2 + 2 y 2 + 3 z 2 + 4 x y + 2 x z + 2 y z − 4 z + 6 y − 2 z + 3 = 0 2 x^{2}+2 y^{2}+3 z^{2}+4 x y+2 x z+2 y z-4 z+6 y-2 z+3=0 2x2+2y2+3z2+4xy+2xz+2yz−4z+6y−2z+3=0 的形状.
分析: 可先通过正交变换再通过平移变换, 将二次曲面方程化成标准形式的方程.
解:
A = ( 2 2 1 2 2 1 1 1 3 ) , B = ( − 4 6 − 2 ) , X = ( x y z ) A=\left(\begin{array}{lll} 2 & 2 & 1 \\ 2 & 2 & 1 \\ 1 & 1 & 3 \end{array}\right), B=\left(\begin{array}{c} -4 \\ 6 \\ -2 \end{array}\right), X=\left(\begin{array}{l} x \\ y \\ z \end{array}\right) A= 221221113 ,B= −46−2 ,X= xyz
则二次曲面方程为
X
T
A
X
+
B
T
X
+
3
=
0.
X^{T} A X+B^{T} X+3=0 .
XTAX+BTX+3=0.
A
A
A 的特征值为
λ
=
2
,
5
,
0
\lambda=2,5,0
λ=2,5,0, 对应的单位特征向量为
(
1
6
1
6
−
2
6
)
,
(
1
3
1
3
1
3
)
,
(
−
1
2
1
2
0
)
,
\left(\begin{array}{c} \frac{1}{\sqrt{6}} \\ \frac{1}{\sqrt{6}} \\ \frac{-2}{\sqrt{6}} \end{array}\right),\left(\begin{array}{c} \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{3}} \end{array}\right),\left(\begin{array}{c} \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} \\ 0 \end{array}\right),
61616−2
,
313131
,
2−1210
,
则经正交变换
X
=
C
Y
X=CY
X=CY , 即
(
x
y
z
)
=
(
1
6
1
3
−
1
2
1
6
1
3
1
2
−
2
6
1
3
0
)
Y
,
\left(\begin{array}{l} x \\ y \\ z \end{array}\right)=\left(\begin{array}{ccc} \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} & \frac{-1}{\sqrt{2}} \\ \frac{1}{\sqrt{6}} & \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} \\ \frac{-2}{\sqrt{6}} & \frac{1}{\sqrt{3}} & 0 \end{array}\right) Y,
xyz
=
61616−23131312−1210
Y,
二次曲面方程化为
Y T ( C T A C ) Y + B C Y + 3 = 0 , Y^{T} (C^{T} A C) Y+B C Y+3=0, YT(CTAC)Y+BCY+3=0,
即
2 x 2 + 5 y 2 + 6 x ′ + 5 2 z ′ + 3 = 0 , 2 x^{2}+5 y^{2}+\sqrt{6} x^{\prime}+5 \sqrt{2} z^{\prime}+3=0, 2x2+5y2+6x′+52z′+3=0,
配方,得
2
(
x
′
+
6
4
)
2
+
5
y
2
=
−
5
2
(
z
′
+
9
2
40
)
,
2\left(x^{\prime}+\frac{\sqrt{6}}{4}\right)^{2}+5 y^{2}=-5 \sqrt{2}\left(z^{\prime}+\frac{9 \sqrt{2}}{40}\right),
2(x′+46)2+5y2=−52(z′+4092),
再经平移变换
{ x ′ ′ = x ′ + 6 4 y ′ ′ = y ′ z ′ ′ = z ′ + 9 2 40 \left\{\begin{array}{c} x^{\prime \prime}=x^{\prime}+\frac{\sqrt{6}}{4} \\ y^{\prime \prime}=y^{\prime} \\ z^{\prime \prime}=z^{\prime}+\frac{9 \sqrt{2}}{40} \end{array}\right. ⎩ ⎨ ⎧x′′=x′+46y′′=y′z′′=z′+4092文章来源:https://www.toymoban.com/news/detail-475217.html
二次曲面方程化为 2 x ′ ′ 2 + 5 y ′ ′ 2 = − 5 2 z ′ ′ 2 x^{\prime \prime 2}+5 y^{\prime \prime 2}=-5 \sqrt{2} z^{\prime \prime} 2x′′2+5y′′2=−52z′′ , 是椭圆抛物面.文章来源地址https://www.toymoban.com/news/detail-475217.html
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