Maths Analysis II - Section 8.3

Author： Bi_Gu
发布时间：February 23, 2022
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# 8.3 Quadrics

It is recommended to study [**Section 8.4**](https://dabigu.xyz/index.php/archives/92/) first before reading this because the knowledge of coordinate transformation is needed.

</div>

## General form of a quadratic function:

$F(x,y,z)=a_{11}x^2+a_{22}y^2+a_{33}z^2+2a_{12}xy+2a_{13}xz+2a_{23}yz+b_1x+b_2y+b_3z+c$

{$F(x,y,z)=0$} is called a $\textcolor[RGB]{30,144,255}{\text{quadric (surface)}}$. [二次曲面]

## Classification of quadrics

The quadratic equation can be simplified by being written in the matrix form:

$F(x,y,z)=\vec{x}^TA\vec{x}+\vec{b}^T\vec{x}+c$, where

$$
A=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix},\ \vec{b}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix},\
\vec{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}.
$$

So that

$$
\begin{align}
\vec{b}&=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix},\vec{b}^T\vec{x}=\begin{pmatrix}b_1&b_2&b_3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=b_1x+b_2y+b_3z;\\
\vec{x}^TA\vec{x}&=\begin{pmatrix}x&y&z\end{pmatrix}
\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}\\
&=\sum\limits_{j=1}^{3}(\vec{x}^TA)_{1j}x_j=\sum\limits_{j=1}^{3}(\sum\limits_{k=1}^{3}(\vec{x}^T)_{1k}a_{kj})x_j=\sum\limits_{j,k=1}^{3}x^T_{1k}a_{kj}x_j.
\end{align}
$$

> For $\vec{x}^TA\vec{x}$, $A$ is the matrix for the quadratic form.

Let $\vec{x}=Q\vec{x'}+\vec{v}$ for some $Q\in SO(3),\vec{v}\in\mathbb{R}^3$,

$$
\begin{align}
F(\vec{x})&=F(Q\vec{x'}+\vec{v})=(Q\vec{x'}+\vec{v})^TA(Q\vec{x'}+\vec{v})+\vec{b}^T(Q\vec{x'}+\vec{v})+c\\
&=\vec{x'}^TQ^TAQ\vec{x'}\\&+\vec{v}^TAQ\vec{x'}+\vec{x'}^TQ^TA\vec{v}+\vec{b}^TQ\vec{x'}[=(2Q^TA\vec{v}+Q^T\vec{b})^T\vec{x'}]\\
&+\vec{v}^TA\vec{v}+\vec{b}^T\vec{v}+c
\end{align}
$$

> That's a coordinate transformation from $\vec{x}(x,y,z)$ to $\vec{x'}(x',y',z')$.
> 
> The goal is to find the $\textcolor{blue}{\text{right transformation}}$ (find the suitable $Q$ and $\vec{v}$) so that the equation can be simplified, $\mathrm{i.e.}$ the degree 2 part only has three terms left, which are $x^2,y^2$ and $z^2$.

### $\textcolor{purple}{\text{Question}}$

Can we choose suitable $Q$ and $\vec{v}$ $\mathrm{s.t.}$

$(1)$ $Q^TAQ$ is diagonal?

$(2)$ $Q^TA\vec{v}+Q^T\vec{b}=\textbf{0}$?

### $\textcolor[RGB]{0,206,209}{\text{Solution}}$

As we have the **Spectrum Theorem**:

There exists $P\in SO(3)$ $\mathrm{s.t.}$ $P^TAP=\Lambda=\text{diag}(\lambda_1,\lambda_2,\lambda_3)\Leftrightarrow AP=P\Lambda$, where

$$
\text{diag}(\lambda_1,\lambda_2,\lambda_3)=\begin{bmatrix}\lambda_1&0&0\\0&\lambda_2&0\\0&0&\lambda_3\end{bmatrix}.
$$

Let $Q=P$, we can simplify the equation to

$F(\vec{x})=\lambda_1x'^2+\lambda_2y'^2+\lambda_3z'^2+F(\vec{v})$

Let $\vec{v}=-\frac{1}{2}A^{-1}\vec{b}$, then for $\vec{x}=P\vec{x'}+\vec{v},$ we know "$\lambda_1x'^2+\lambda_2y'^2+\lambda_3z'^2+F(\vec{v})=0$".

> 这里$F(\vec{v})$是平移的部分; $\vec{v}$的坐标通过$\vec{v}=-\frac{1}{2}A^{-1}\vec{b}$计算得出；
> 
> 将坐标代入$F(x,y,z)$,由于三个坐标都已知，计算结果为0次实数，即简化后方程中的常数项。

Now for classification, consider the sign of $(\lambda_1,\lambda_2,\lambda_3,F(\vec{v}))$:

$[1]$ $\text{rank}(A)=3$ ($\Leftrightarrow \lambda_1,\lambda_2,\lambda_3\neq0$)

If the sign of $(\lambda_1,\lambda_2,\lambda_3,F(\vec{v}))$ is

$(1)$ $(+,+,+,-)$, then $\vec{x'}$ lies on a set of the shape $\textcolor[RGB]{30,144,255}{\text{Ellipsoid}}$ [椭球面],

$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}-1=0\ (a>0,b>0,c>0).
$$

$(2)$ $(+,+,+,0)$, a point.

*The equation only holds when $x=y=z=0$*.

$(3)$ $(+,+,+,+)$, $\emptyset$.

*No solution in $\mathbb{R}^3.$*

$(4)$ $(+,+,-,+)$, called a $\textcolor[RGB]{30,144,255}{\text{Hyperboloid of two sheets}}$. [双叶双曲面]

$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}+1=0\ (a>0,b>0,c>0).
$$

> $\textcolor{red}{\text{Also holds}}$ for $(-,+,+,+)$ and $(+,-,+,+)$.
> 
> The image below is for $(-,+,+,+)$:

$(5)$ $(+,+,-,-)$ A $\textcolor[RGB]{30,144,255}{\text{Hyperboloid of one sheet}}$. [单叶双曲面]

$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}-1=0\ (a>0,b>0,c>0).
$$

> $\textcolor{red}{\text{Also holds}}$ for $(-,+,+,-)$ and $(+,-,+,-)$.

$(6)$ $(+,+,-,0)$ A $\textcolor[RGB]{30,144,255}{\text{Quadric conical surface}}$ (Double cone) [二次锥面]

$$
\frac{x^2}{a^2}+\frac{y^2}{b^2}-\frac{z^2}{c^2}=0\ (a>0,b>0,c>0).
$$

> $\textcolor{red}{\text{Also holds}}$ for $(-,+,+,0)$ and $(+,-,+,0)$.

$[2]$ $\text{rank}(A)=2$ ($\Leftrightarrow\lambda_3=0,\lambda_1,\lambda_2\neq0$)

$\lambda_1x'^2+\lambda_2y'^2+\lambda_3z'^2+b_1'x'+b_2'y'+b_3'z'+c'=0$

Let

$$
\left\{
\begin{array}{}x''=x'+\frac{b_1'}{2\lambda_1}\\y''=y'+\frac{b_2}{2\lambda_2}\end{array}
\right.
$$

then it can be siplified:

$\rightsquigarrow$ $\lambda_1x''^2+\lambda_2y''^2+b_3''z+c''=0$

> 通过适当的平移（配方）消去$x,y$的一次项。

* $b_3''\neq 0$, $\lambda_1\lambda_2>0$, we get a $\textcolor[RGB]{30,144,255}{\text{Elliptical Paraboloid}}$. [椭圆抛物面]
  
  $$
  \frac{x^2}{a^2}+\frac{y^2}{b^2}=z\ (a>0,b>0).
  $$
  
  <img src="https://cdn.dabigu.xyz/QQ图片20220225083042.png" width="600px">
* $b_3''\neq 0$, $\lambda_1\lambda_2<0$, we get a $\textcolor[RGB]{30,144,255}{\text{Hyperbolic Paraboloid}}$ / $\textcolor{green}{\text{Saddle surface}}$.
  
  [双曲抛物面/马鞍面]
  
  $$
  \frac{x^2}{a^2}-\frac{y^2}{b^2}=z
  $$
* $b_3''=0$ $\rightsquigarrow$ $\lambda_1x'^2+\lambda_2y'^2+c'=0$, we get a $\textcolor{blue}{\text{Cylindrical Surface}}$. [柱状平面]
  
  For a cylindrical surface, let $L$ be a **curve** on the $xOy$ plane whose function is
  
  $$
  \left\{
  \begin{array}{}F(x,y)=0\\z=0\end{array}
  \right.
  $$
  
  $L$ is called the $\textcolor[RGB]{30,144,255}{\text{directrix}}$ [准线] of the surface. Take any line that is parallel to $z$-axis and intersects with $L$, such line is called the $\textcolor[RGB]{30,144,255}{\text{generatrix}}$ [母线] of the surface.
  
  We classify cylindrical surfaces by looking at the shape of their **intersection** with any plane that is $\textcolor{green}{\text{perpendicular}}$ to the generatrix.
  
  Sorting by the shape of the **directrix**, we have:
  
  > format: shape of directrix - shape of the corresponding quadric
  
  * $\emptyset$ - $\emptyset$
  * a point - a line
  * two intersecting lines - two intersecting planes
  * ellipse - $\textcolor[RGB]{30,144,255}{\text{Elliptical Cylinder}}$ [椭圆柱面]
    
    <img src="https://cdn.dabigu.xyz/QQ截图20220303212547.png" width="600px">
    
    $$
    \frac{x^2}{a^2}+\frac{y^2}{b^2}=1.
    $$
    
  * hyperbola - $\textcolor[RGB]{30,144,255}{\text{Hyperbolic Cylinder}}$ [双曲柱面]
    
    <img src="https://cdn.dabigu.xyz/QQ截图20220303212649.png" width="600px">
    
    $$
    \frac{x^2}{a^2}-\frac{y^2}{b^2}=1.
    $$

$[3]$ $\text{Rank}(A)=1$ $\rightsquigarrow$ $\lambda_1x'^2+b_2'y'+b_3'z'+c'=0$

Make a further rotation on $y'O'z'$ plane

$$
\begin{pmatrix}y'\\z'\end{pmatrix}=
\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}
\begin{pmatrix}y''\\z''\end{pmatrix}
$$

> 此处旋转的目的在于消去其中的一个一次项。

* $b_2'^2+b_3'^2\neq 0$ $\xrightarrow{\text{Rotation}}$ $\lambda_1x''^2+b_2''y''+c''=0(b_2''\neq 0)$, a $\textcolor[RGB]{30,144,255}{\text{Parabolic Cylinder}}$. [抛物柱面]
  
  $$
  x^2=ay.
  $$
* $b_2'=b_3'=0$ $\rightsquigarrow$ $\lambda_1x''^2+c''=0$
  
  * $c''<0$ Two parallel planes
  * $c''=0$ Plane (with multiplicity 2)
  * $c''>0$ $\emptyset$.

## General Classification of Quadrics

## Symmetries & Cross sections [截影] on quadrics

Let $S:F(\vec{x})=0$ be a quadric in $\mathbb{R}^3$; $\pi$ be a plane, consider their intersection:

$$
\let\{
\begin{array}{}
S:x^2+y^2+z^2=1\\\pi:x+y+z=0
\end{array}
\right.
$$

> 相交部分是三维空间中的一条曲线，因而该方程组便是该曲线的一般方程表达；为了研究其形状，我们对$z$进行消元。注意，消元也即研究该曲线在$xOy$平面上的<font color=red>投影</font>。

$\Rightarrow$ $x^2+y^2+(x+y)^2=1$ $\Rightarrow$ $2x^2+2y^2+2xy=1$

In order to study the shape of $2x^2+2y^2+2xy=1$:

$$
\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}-1=0
$$

$(1)$ Orthogonally diagonalize

$$
\begin{gather}
A=\begin{pmatrix}2&1\\1&2\end{pmatrix},|\lambda I-A|=\begin{vmatrix}\lambda-2&-1\\-1&\lambda-2\end{vmatrix}=(\lambda-2)^2-1=0\\
\Rightarrow\lambda_1=3,\lambda_2=1\\
\lambda_1I-A=\begin{pmatrix}1&-1\\-1&1\end{pmatrix}\xrightarrow{\text{Gaussian Elimination}}\begin{pmatrix}1&-1\\0&0\end{pmatrix}\\\rightsquigarrow\vec{u_1}=\begin{pmatrix}\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{pmatrix}\text{ is the eigenvalue of }\lambda_1;\\
\lambda_2I-A=\begin{pmatrix}-1&-1\\-1&-1\end{pmatrix}\xrightarrow{\text{Gaussian Elimination}}\begin{pmatrix}1&1\\0&0\end{pmatrix}\\\rightsquigarrow\vec{u_2}=\begin{pmatrix}-\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}\end{pmatrix}\text{ is the eigenvalue of }\lambda_2.\\
\Rightarrow\begin{pmatrix}2&1\\1&2\end{pmatrix}\begin{pmatrix}\vec{u_1}&\vec{u_2}\end{pmatrix}=\begin{pmatrix}\lambda_1\vec{u_1}&\lambda_2\vec{u_2}\end{pmatrix}=\begin{pmatrix}\vec{u_1}&\vec{u_2}\end{pmatrix}\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}\\
\begin{pmatrix}2&1\\1&2\end{pmatrix}=\begin{pmatrix}\vec{u_1}&\vec{u_2}\end{pmatrix}\begin{pmatrix}3&0\\0&1\end{pmatrix}\begin{pmatrix}\vec{u_1}^T\\\vec{u_2}^T\end{pmatrix}\\
\rightsquigarrow\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}\vec{u_1}&\vec{u_2}\end{pmatrix}\begin{pmatrix}3&0\\0&1\end{pmatrix}\begin{pmatrix}\vec{u_1}^T\\\vec{u_2}^T\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}-1=0
\end{gather}
$$

Now, define

$$
\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\vec{u_1}^T\\\vec{u_2}^T\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}\frac{\sqrt{2}}{2}(x+y)\\\frac{\sqrt{2}}{2}(-x+y)\end{pmatrix}
$$

as new coordinate, then

$$
\begin{gather}
\begin{pmatrix}x'&y'\end{pmatrix}\begin{pmatrix}3&0\\0&1\end{pmatrix}\begin{pmatrix}x'\\y'\end{pmatrix}-1=0\\
\mathrm{i.e.}\quad 3x'^2+y'^2=1.
\end{gather}
$$

> 在这个过程中，我们使用了本节开头提到的对角化方法找到合适的坐标变换，使得截影的方程转换成标准形式，从而得以研究其形状。

> 然而，因为得到的是在平面上的投影，因而其形状是失真的。为了得到原来的形状，我们继续进行操作：

To get the real shape,

$$
\begin{bmatrix}\vec{i'}&\vec{j'}&\vec{k'}\end{bmatrix}=\begin{bmatrix}\vec{i}&\vec{j}&\vec{k}\end{bmatrix}\begin{bmatrix}\frac{\sqrt{3}}{3}&\frac{\sqrt{2}}{2}&\frac{\sqrt{6}}{6}\\\frac{\sqrt{3}}{3}&-\frac{\sqrt{2}}{2}&\frac{\sqrt{6}}{6}\\\frac{\sqrt{3}}{3}&0&-\frac{\sqrt{6}}{3}\end{bmatrix}
$$

$$
\begin{gather}
\vec{x}=Q\vec{x'};\vec{x}^T\vec{x}=1\rightarrow\vec{x'}^TQ^TQ\vec{x'}=1\\
\mathrm{i.e.}\quad x'^2+y'^2+z'^2=1
\end{gather}
$$

For $\pi:x+y+z=0,x'=0$, then the shape is a circle.

> 这里是通过旋转坐标系使得坐标平面$yOz$落在平面$\pi$上，从而通过平行$x$轴方向观察即是该曲线的实际形状。

### $\textcolor{purple}{\text{Definition}}$ Symmetry

A pair $(P,\vec{v})$ where $P$ is a $\textcolor{blue}{\text{orthogonal matrix}}$ in $\mathbb{R}^3$, $\vec{v}\in\mathbb{R}^3$ is called a symmetry of the quadric if the transformation $\vec{x}=P\vec{x'}+\vec{v}$ leaves the quadric invariant.

> 即二次曲面的方程经过由$(P,\vec{v})$表示的坐标变换后保持不变；例如上述例子中，方程$x^2+y^2+z^2=1$经过变换后得到$x'^2+y'^2+z'^2=1$，我们就说该坐标变换是对于该二次曲面对称的。

注意：对称与二次曲面有关(symmetry "of the quadric")。即该坐标变换对于其它二次曲面可能不是对称的。

</div>

## Parametric Interpretations of Quadrics

$(1)$ For the hyperboloid of one sheet $x^2+y^2-z^2=1$,

its parameter equation

$$
\left\{
\begin{array}{}x=\cosh u\cdot\cos v\\y=\cosh u\cdot\sin v\\z=\sinh u\
\end{array},\quad (u,v)\in\mathbb{R}\times[0,2\pi).
\right.
$$

**Another interpretation:**

Let a line be $L:\vec{r}(t)=(0,1,0)+t(1,0,1)=(t,1,t)$. A hyperboloid of one sheet forms as a result of $L$ rotating around the $z$-axis.

$$
\begin{gather}
\begin{pmatrix}x'\\y'\end{pmatrix}=\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\begin{pmatrix}t\\1\end{pmatrix}\\
\Rightarrow\left\{
\begin{array}{}x=t\cos\theta-\sin\theta\\y=t\sin\theta+\cos\theta\\z=t\end{array}
\right.,\quad(t,\theta)\in\mathbb{R}\times[0,2\pi).
\end{gather}
$$

$(2)$ For the circle $(y-2)^2+z^2=1$,make it rotate around the $z$-axis and we get a torus.

$$
\begin{gather}
\left\{
\begin{array}{}y=2+\cos\theta\\z=\sin\theta\end{array}
\right.,\quad\theta\in[0,2\pi)\\
\Rightarrow\left\{
\begin{array}{}x=0\cdot\cos\phi-\sin\phi(2+\cos\theta)\\
y=0\cdot\sin\phi+\cos\phi(2+\cos\theta)\\z=\sin\theta\end{array}
\right.,\quad\theta,\phi\in[0,2\pi).
\end{gather}
$$

$(3)$ For a hyperbola $y^2-z^2=1$, the parametric equation

$$
\left\{
\begin{array}{}y=\cosh u\\z=\sinh u\end{array}
\right.,\quad u\in\mathbb{R}
$$

shows only the right branch of it.(Since $\cosh u>0$ for all $u\in\mathbb{R}$) For the left branch, we need another set of equations

$$
\left\{
\begin{array}{}y=-\cosh u\\z=\sinh u\end{array}
\right.,\quad u\in\mathbb{R}.
$$

Last modification：March 3, 2022

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Maths Analysis II - Section 8.3

Bi_Gu • 2022 年 02 月 23 日