IS Math-2022-01

题目来源：Problem 1 日期：2024-06-22 题目主题：Math-线性代数-矩阵变换

具体题目

Consider the following multiple conditions on $x,y,z\in \mathbb{R}$.

$$\{\begin{array}{l}0<z-xy<1\\ 0<z-(x+y{)}^2<-xy\end{array}$$

Let $\Omega$ be the set of points $(x,y)$ for which at least one $z$ exists satisfying the above conditions. Note that the set $\Omega$ can be seen in the three-dimensional Cartesian coordinate system as the orthogonal projection of points $(x,y,z)$ satisfying the above conditions onto the $xy$-plane. Answer the following questions.

1. Find the inequalities on $x$ and $y$ representing $\Omega$.

2. Draw a figure of $\Omega$ in the $xy$-plane. If the boundary of $\Omega$ intersects with the $x$-axis or the $y$-axis, write down the coordinates at each intersection.

3. The curved segments of the boundary of $\Omega$ correspond to the linear transformation of arcs of the unit circle with a matrix $\mathbf{X}$. Find one such $\mathbf{X}$. Note that the point $(1,0)$ on the unit circle must be transformed to a point where the curvature is maximized in the curved segments.

4. Calculate the determinant of $\mathbf{X}$ found in (3).

5. Calculate the area of the set $\Omega$. Note that the absolute value of the determinant of a matrix is the area scale factor of the transformation with that matrix.

Solution

Question 1: Find the inequalities on $x$ and $y$ representing $\Omega$

Given the inequalities:

$$\{\begin{array}{l}0<z-xy<1\\ 0<z-(x+y{)}^2<-xy\end{array}$$

We start by analyzing the given inequalities.

1. From the first inequality:

   $$0<z-xy<1⟹xy<z<xy+1$$

2. From the second inequality:

   $$0<z-(x+y{)}^2<-xy⟹(x+y{)}^2<z<(x+y{)}^2-xy$$

To find the region $\Omega$ in the $xy$-plane, we subtract the first inequality from the second:

$$0<(z-(x+y{)}^2)-(z-xy)=xy-(x+y{)}^2<-xy-1$$

Simplifying the expression:

$$xy-(x^2+2xy+y^2)=-x^2-xy-y^2$$

Thus, we have:

$$0<-x^2-xy-y^2<-xy-1$$

We can split this into two conditions:

$$x^2+xy+y^2>0$$

and

$$x^2+xy+y^2<1$$

Therefore, $\Omega$ is defined by:

$$x^2+xy+y^2<1\text{and}xy>0$$

Question 2: Draw a figure of $\Omega$ in the $xy$-plane. If the boundary of $\Omega$ intersects with the $x$-axis or the $y$-axis, write down the coordinates at each intersection

The boundary of $\Omega$ is given by the equation:

$$x^2+xy+y^2=1$$

This equation represents an ellipse. Additionally, the condition $xy>0$ restricts $\Omega$ to the first and third quadrants.

To find the intersection points with the $x$ and $y$ axes:

• When $y=0$: $x^2=1$, so $x=\pm 1$.
• When $x=0$: $y^2=1$, so $y=\pm 1$.

However, considering $xy>0$, we only keep the intersections in the first and third quadrants:

• The ellipse intersects the $x$-axis at $(1,0)$.
• The ellipse intersects the $y$-axis at $(0,1)$.
• Also, $(x=-1,y=0)$ and $(x=0,y=-1)$ are valid.

3. Matrix transforming unit circle to boundary of $\Omega$

We can consider the transformation that maps $(1,0)$ on the unit circle to $(1,-1)$ on the ellipse, and $(0,1)$ to $(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$. Let’s call this transformation matrix $\mathbf{X}$. Then:

$$\mathbf{X}\left[\begin{array}{c}1\\ 0\end{array}\right]=\left[\begin{array}{c}1\\ -1\end{array}\right],\mathbf{X}\left[\begin{array}{c}0\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}\end{array}\right]$$

Writing this in matrix form:

$$\mathbf{X}\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}1& \frac{1}{\sqrt{3}}\\ -1& \frac{1}{\sqrt{3}}\end{array}\right]$$

Therefore, the matrix $\mathbf{X}$ is:

$$\mathbf{X}=\left[\begin{array}{cc}1& \frac{1}{\sqrt{3}}\\ -1& \frac{1}{\sqrt{3}}\end{array}\right]$$

This transformation indeed maps the unit circle to the boundary of $\Omega$, with $(1,0)$ mapped to the point of maximum curvature $(1,-1)$.

4. Determinant of $\mathbf{X}$

$$\det (\mathbf{X})=\det \left[\begin{array}{cc}1& \frac{1}{\sqrt{3}}\\ -1& \frac{1}{\sqrt{3}}\end{array}\right]=\frac{2}{\sqrt{3}}$$

Question 5: Area of $\Omega$

To calculate the area of the set $\Omega$, we need to carefully consider the sector of the unit circle that corresponds to the region of the ellipse defined by the condition $xy<0$ after the transformation.

Step 1: Identify the Sector of the Unit Circle

Given the transformation matrix:

$$\mathbf{X}=\left[\begin{array}{cc}1& \frac{1}{\sqrt{3}}\\ -1& \frac{1}{\sqrt{3}}\end{array}\right]$$

we first apply the inverse transformation ${\mathbf{X}}^{-1}$ to map the boundary points of the ellipse $x^2+xy+y^2=1$ back to the unit circle. The inverse matrix is:

$${\mathbf{X}}^{-1}=\frac{1}{2}\left[\begin{array}{cc}1& -1\\ \sqrt{3}& \sqrt{3}\end{array}\right]$$

Using this inverse matrix, we find the transformed points on the unit circle corresponding to key points on the ellipse:

• The point $(1,0)$ on the ellipse maps to $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ on the unit circle.
• The point $(0,1)$ on the ellipse maps to $\left(-\frac{1}{2},\frac{\sqrt{3}}{2}\right)$ on the unit circle.

These points correspond to angles ${\theta }_1=\frac{\pi }{3}$ and ${\theta }_2=\frac{2\pi }{3}$ on the unit circle.

Step 2: Identify the Relevant Sector for $xy<0$

Since the region of interest after the transformation is where $xy<0$, we focus on the sector of the unit circle between the angles ${\theta }_2=\frac{2\pi }{3}$ and ${\theta }_3=\frac{4\pi }{3}$ (which is the angle opposite to ${\theta }_1=\frac{\pi }{3}$). This sector lies in the upper left (second quadrant) and lower right (fourth quadrant) of the unit circle, corresponding to the $xy<0$ condition.

The angle span for this sector is:

$$\Delta \theta =\frac{4\pi }{3}-\frac{2\pi }{3}=\frac{2\pi }{3}$$

Step 3: Calculate the Sector Area on the Unit Circle

The area of this sector of the unit circle is:

$$A_{\text{sector}}=\frac{1}{2}{r}^2\Delta \theta =\frac{1}{2}\times \frac{2\pi }{3}=\frac{\pi }{3}$$

Step 4: Scale the Area by the Determinant of the Transformation

Finally, we apply the determinant of $\mathbf{X}$ to scale this area. The determinant of $\mathbf{X}$ is:

$$\det (\mathbf{X})=\frac{2}{\sqrt{3}}$$

Thus, the area of $\Omega$ is:

$$A_{\Omega }=A_{\text{sector}}\times \left|\det (\mathbf{X})\right|=\frac{\pi }{3}\times \frac{2}{\sqrt{3}}=\frac{2\pi }{3\sqrt{3}}$$

知识点

线性变换行列式特征值和特征向量

Linear Transformation

解题技巧和信息

• 多重约束条件下确定可行域的关键是找出约束之间的联系，本题中通过比较两个不等式的上下界得到 $\Omega$ 的不等式表示。
• 二次型曲线（如椭圆）可写成矩阵形式 ${\mathbf{x}}^T\mathbf{Ax}=1$，其中 $\mathbf{A}$ 是对称矩阵。若 ${\mathbf{X}}^T\mathbf{X}=\mathbf{A}$，则 $\mathbf{X}$ 将单位圆变换为该曲线。
• 行列式的绝对值是线性变换的面积伸缩因子。对角化 $\mathbf{A}$ 求特征值，其几何意义是椭圆的长半轴和短半轴的平方。

重点词汇

• inequality 不等式
• constraint 约束
• ellipse 椭圆
• linear transformation 线性变换
• determinant 行列式
• eigenvalue 特征值
• diagonalization 对角化

参考资料

1. Gilbert Strang. Introduction to Linear Algebra. 5th ed. Ch. 6.
2. Lay et al. Linear Algebra and Its Applications. 5th ed. Ch. 5.