IS Math-2023-01

题目来源：Problem 1 日期：2024-08-13 题目主题: Math-Linear Algebra-Determinants and Interpolation

解题思路

这组问题涉及行列式、线性方程组和多项式插值，重点在于理解这些概念之间的联系。主要步骤包括：

1. 行列式的几何意义：理解行列式与平面上直线方程的关系。这需要展开行列式并重新排列项，以得到直线的标准形式。

2. Vandermonde 行列式的计算：利用行列式的性质（如因式分解）来简化计算。关键是识别出行列式中的模式，并巧妙地提取公因子。

3. 多项式插值的唯一性：利用线性代数的知识证明插值多项式的存在性和唯一性。这涉及到 Vandermonde 矩阵的可逆性。

4. Lagrange 插值多项式的构造：理解 Lagrange 基函数的定义和性质，并利用这些知识来表示插值多项式。

5. 高阶插值多项式的推广：将前面问题中的方法推广到更高阶的情况，理解随着插值点数量增加，系数表达式如何变化。

对于每个问题，我们需要灵活运用线性代数和多项式理论的知识，并注意不同概念之间的联系。特别是要认识到 Vandermonde 行列式、线性方程组和多项式插值这三者之间的密切关系。

Solution

Question 1

We are given the function $f(x,y)$ defined as:

$$f(x,y)=\left|\begin{array}{ccc}1& x_1& y_1\\ 1& x_2& y_2\\ 1& x& y\end{array}\right|.$$

We need to show that the set of solutions to $f(x,y)=0$ is a line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$ on the $xy$ plane, where $x_1\ne x_2$.

The determinant can be expanded as follows:

$$f(x,y)=1\cdot \left|\begin{array}{cc}{x}_2& y_2\\ x& y\end{array}\right|-x_1\cdot \left|\begin{array}{cc}1& y_2\\ 1& y\end{array}\right|+y_1\cdot \left|\begin{array}{cc}1& x_2\\ 1& x\end{array}\right|$$

Expanding these $2\times 2$ determinants gives:

$$f(x,y)=1\cdot (x_{2}y-x{y}_2)-x_1\cdot (y_2-y)+y_1\cdot (x_2-x)$$

Simplifying, we get:

$$f(x,y)=x_{2}y-x{y}_2-x_1{y}_2+x_{1}y+y_1{x}_2-y_{1}x$$

This can be rearranged as:

$$f(x,y)=(x_2-x_1)y+(y_1-y_2)x+(x_1{y}_2-y_1{x}_2)$$

The set of points $(x,y)$ that satisfy $f(x,y)=0$ is given by:

$$(x_2-x_1)y+(y_1-y_2)x+(x_1{y}_2-y_1{x}_2)=0$$

This is the equation of a line. To check whether this line passes through the points $(x_1,y_1)$ and $(x_2,y_2)$, substitute $(x_1,y_1)$ into the equation:

$$(x_2-x_1)y_1+(y_1-y_2)x_1+(x_1{y}_2-y_1{x}_2)=(x_2-x_1)y_1+(y_1-y_2)x_1+(x_1{y}_2-y_1{x}_2)=0$$

Similarly, substituting $(x_2,y_2)$:

$$(x_2-x_1)y_2+(y_1-y_2)x_2+(x_1{y}_2-y_1{x}_2)=(x_2-x_1)y_2+(y_1-y_2)x_2+(x_1{y}_2-y_1{x}_2)=0$$

Therefore, the set of solutions to $f(x,y)=0$ indeed represents the line passing through the points $(x_1,y_1)$ and $(x_2,y_2)$.

Question 2

We are asked to find the value of the determinant:

$$\left|\begin{array}{ccc}1& x_1& x_1^2\\ 1& x_2& x_2^2\\ 1& x_3& x_3^2\end{array}\right|.$$

This is a $3\times 3$ determinant. We will expand it using cofactor expansion:

$$\left|\begin{array}{ccc}1& x_1& x_1^2\\ 1& x_2& x_2^2\\ 1& x_3& x_3^2\end{array}\right|=1\cdot \left|\begin{array}{cc}{x}_2& x_2^2\\ x_3& x_3^2\end{array}\right|-x_1\cdot \left|\begin{array}{cc}1& x_2^2\\ 1& x_3^2\end{array}\right|+x_1^2\cdot \left|\begin{array}{cc}1& x_2\\ 1& x_3\end{array}\right|$$

Now, each $2\times 2$ determinant can be calculated:

$$\left|\begin{array}{cc}{x}_2& x_2^2\\ x_3& x_3^2\end{array}\right|=x_2{x}_3^2-x_3{x}_2^2=x_2{x}_3(x_3-x_2)$$ $$\left|\begin{array}{cc}1& x_2^2\\ 1& x_3^2\end{array}\right|=x_3^2-x_2^2=(x_3-x_2)(x_3+x_2)$$ $$\left|\begin{array}{cc}1& x_2\\ 1& x_3\end{array}\right|=x_3-x_2$$

Substituting these into the original determinant expression:

$$\text{Determinant}=x_2{x}_3(x_3-x_2)-x_1(x_3-x_2)(x_3+x_2)+x_1^2(x_3-x_2)$$

Factor out $(x_3-x_2)$:

$$\text{Determinant}=(x_3-x_2)\left[x_2{x}_3-x_1(x_3+x_2)+x_1^2\right]$$

Simplify the expression inside the brackets:

$$\text{Determinant}=(x_3-x_2)\left[x_2{x}_3-x_1{x}_3-x_1{x}_2+x_1^2\right]=(x_3-x_2)(x_1-x_2)(x_1-x_3)$$

Thus, the determinant in factored form is:

$$\left|\begin{array}{ccc}1& x_1& x_1^2\\ 1& x_2& x_2^2\\ 1& x_3& x_3^2\end{array}\right|=(x_3-x_2)(x_2-x_1)(x_1-x_3).$$

Question 3

We need to show that there is a unique curve $y=a_0+a_{1}x+a_2{x}^2$ passing through three points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ on the $xy$ plane, where $a_0,a_1,a_2$ are constants and $x_1,x_2,x_3$ are all distinct.

To pass through the given points, the curve must satisfy the system of equations:

$$\begin{array}{rl}{y}_1& =a_0+a_1{x}_1+a_2{x}_1^2,\\ y_2& =a_0+a_1{x}_2+a_2{x}_2^2,\\ y_3& =a_0+a_1{x}_3+a_2{x}_3^2.\end{array}$$

This system can be written in matrix form as:

$$\left(\begin{array}{ccc}1& x_1& x_1^2\\ 1& x_2& x_2^2\\ 1& x_3& x_3^2\end{array}\right)\left(\begin{array}{c}{a}_0\\ a_1\\ a_2\end{array}\right)=\left(\begin{array}{c}{y}_1\\ y_2\\ y_3\end{array}\right).$$

To show that there is a unique solution $(a_0,a_1,a_2)$, we need to check that the coefficient matrix is invertible. The determinant of the coefficient matrix is:

$$\left|\begin{array}{ccc}1& x_1& x_1^2\\ 1& x_2& x_2^2\\ 1& x_3& x_3^2\end{array}\right|$$

From Question 2, we know that this determinant is $(x_3-x_2)(x_2-x_1)(x_1-x_3)$, which is non-zero because $x_1,x_2,x_3$ are distinct. Since the determinant is non-zero, the matrix is invertible, and there exists a unique solution for $a_0,a_1,a_2$. Hence, the curve $y=a_0+a_{1}x+a_2{x}^2$ passing through the three points is unique.

Question 4: Inverse of a $3\times 3$ Vandermonde Matrix

We need to express the curve $y=a_0+a_{1}x+a_2{x}^2$ passing through the points $(x_1,y_1)$, $(x_2,y_2)$, $(x_3,y_3)$ in the form $y=c_1{y}_1+c_2{y}_2+c_3{y}_3$, where $c_1,c_2,c_3$ do not depend on $y_1,y_2,y_3$.

Let’s start by considering the Vandermonde matrix $\mathbf{V}$ corresponding to the points $x_1,x_2,x_3$:

$$\mathbf{V}=\left(\begin{array}{ccc}1& x_1& x_1^2\\ 1& x_2& x_2^2\\ 1& x_3& x_3^2\end{array}\right)$$

Our goal is to find the inverse of $\mathbf{V}$, denoted as ${\mathbf{V}}^{-1}$. To derive this, we use the relationship between the Vandermonde matrix and Lagrange interpolation polynomials.

Step 1: Lagrange Interpolation Polynomials

The Lagrange interpolation polynomial for the $i$-th point, denoted by $L_i(x)$, is defined as:

$$L_i(x)=\prod _{\begin{array}{c}j=1\\ j\ne i\end{array}}^3\frac{x-x_j}{x_i-x_j}$$

For each $i$, this polynomial has the following property:

$$L_i(x_j)=\{\begin{array}{ll}1& \text{if }j=i\\ 0& \text{if }j\ne i\end{array}$$

Explicitly, the polynomials are:

$$L_1(x)=\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)},L_2(x)=\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)},L_3(x)=\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}$$

These polynomials form a basis for the space of polynomials of degree at most 2, and any such polynomial can be written as a linear combination of these basis polynomials.

Step 2: Relating Lagrange Polynomials to the Vandermonde Inverse

Now, consider the Lagrange interpolation condition applied to our Vandermonde matrix problem. The original system of linear equations based on the curve passing through the points is:

$$\mathbf{V}\left(\begin{array}{c}{a}_0\\ a_1\\ a_2\end{array}\right)=\left(\begin{array}{c}{y}_1\\ y_2\\ y_3\end{array}\right)$$

To solve for the coefficients $a_0,a_1,a_2$, we can multiply both sides by the inverse of $\mathbf{V}$:

$$\left(\begin{array}{c}{a}_0\\ a_1\\ a_2\end{array}\right)={\mathbf{V}}^{-1}\left(\begin{array}{c}{y}_1\\ y_2\\ y_3\end{array}\right)$$

However, since $y_i=L_i(x)$ for $i=1,2,3$, the entries of the inverse matrix ${\mathbf{V}}^{-1}$ are derived by evaluating derivatives of these Lagrange polynomials at $x=0$. Specifically, the $(j-1)$th derivative of $L_i(x)$ evaluated at $x=0$ gives us the corresponding element in the inverse matrix:

$$({\mathbf{V}}^{-1}{)}_{ji}=\frac{L_i^{(j-1)}(0)}{(j-1)!}$$

This relationship arises because each row of ${\mathbf{V}}^{-1}$ corresponds to the coefficients of the polynomial expansion of the Lagrange basis polynomial evaluated at the corresponding point.

Step 3: Explicit Calculation of ${\mathbf{V}}^{-1}$

Let’s compute the derivatives of the Lagrange polynomials at $x=0$ to form ${\mathbf{V}}^{-1}$:

For $L_1(x)$:

• $L_1(x)=\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}$
• $L_1(0)=\frac{(-x_2)(-x_3)}{(x_1-x_2)(x_1-x_3)}=\frac{x_2{x}_3}{(x_1-x_2)(x_1-x_3)}$
• The first derivative $\frac{\mathrm{d}{L}_1(x)}{\mathrm{d}x}{|}_{x=0}$ results in the coefficient of $x$ term.

Step 4: Computing the Coefficients $c_1$, $c_2$, $c_3$

To find the coefficients $c_1$, $c_2$, and $c_3$ in the expression $y=c_1{y}_1+c_2{y}_2+c_3{y}_3$, we use the Lagrange interpolation polynomials.

Coefficient $c_1$

The coefficient $c_1$ corresponds to the Lagrange polynomial $L_1(x)$ for the point $x_1$:

$$c_1=L_1(x)=\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)}$$

Coefficient $c_2$

Similarly, the coefficient $c_2$ corresponds to the Lagrange polynomial $L_2(x)$ for the point $x_2$:

$$c_2=L_2(x)=\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)}$$

Coefficient $c_3$

Finally, the coefficient $c_3$ corresponds to the Lagrange polynomial $L_3(x)$ for the point $x_3$:

$$c_3=L_3(x)=\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}$$

Final Expression

Thus, the coefficients $c_1$, $c_2$, and $c_3$ in the expression $y=c_1{y}_1+c_2{y}_2+c_3{y}_3$ are given by:

$$\begin{array}{rl}{c}_1& =\frac{(x-x_2)(x-x_3)}{(x_1-x_2)(x_1-x_3)},\\ c_2& =\frac{(x-x_1)(x-x_3)}{(x_2-x_1)(x_2-x_3)},\\ c_3& =\frac{(x-x_1)(x-x_2)}{(x_3-x_1)(x_3-x_2)}.\end{array}$$

These expressions provide the explicit forms of $c_1$, $c_2$, and $c_3$ in terms of $x$ and the points $x_1$, $x_2$, and $x_3$. This completes the interpolation and gives us the coefficients needed for the expression of the curve passing through the points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$.

Question 5

To solve for $c_1$ in the expression $y=c_1{y}_1+c_2{y}_2+c_3{y}_3+c_4{y}_4+c_5{y}_5$, we will use the Lagrange interpolation method. Specifically, $c_1$ will be the Lagrange polynomial corresponding to the point $(x_1,y_1)$ evaluated at the point $x$.

Step 1: Lagrange Interpolation Polynomials

The Lagrange interpolation polynomial $L_1(x)$ corresponding to the point $x_1$ is defined as:

$$L_1(x)=\prod _{\begin{array}{c}j=2\end{array}}^5\frac{x-x_j}{x_1-x_j}$$

This polynomial can be expanded as follows:

$$L_1(x)=\frac{(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_1-x_5)}$$

Step 2: Compute $c_1$

The coefficient $c_1$ is exactly $L_1(x)$. Thus, $c_1$ is given by:

$$c_1=\frac{(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_1-x_5)}$$

This expression provides the value of $c_1$ explicitly in terms of $x$ and the known points $x_1,x_2,x_3,x_4,x_5$.

Summary

So, the coefficient $c_1$ in the linear combination $y=c_1{y}_1+c_2{y}_2+c_3{y}_3+c_4{y}_4+c_5{y}_5$ is:

$$c_1=\frac{(x-x_2)(x-x_3)(x-x_4)(x-x_5)}{(x_1-x_2)(x_1-x_3)(x_1-x_4)(x_1-x_5)}$$

This form of $c_1$ is derived directly from the Lagrange interpolation polynomial and is the required value. The same method can be applied to find $c_2,c_3,c_4,c_5$ by simply shifting the index of the corresponding Lagrange polynomials.

知识点

行列式插值多项式线性代数线性方程组范德蒙行列式

难点思路

本题目中最具挑战性的部分是第 3 问和第 4 问。第 3 问需要理解 Vandermonde 矩阵的性质及其与多项式插值的关系，这需要对线性代数和多项式理论有较深入的了解。第 4 问要求将插值多项式表示为 Lagrange 形式，这需要熟悉 Lagrange 插值多项式的构造方法和性质。

解题技巧和信息

1. 在处理行列式问题时，尝试使用行列式的性质（如展开、因式分解）来简化计算。
2. 对于插值问题，考虑使用 Vandermonde 矩阵和 Lagrange 插值多项式等工具。
3. 在解决线性方程组时，关注系数矩阵的性质（如是否可逆）。
4. 熟悉常见的代数结构（如多项式空间）对解决此类问题很有帮助。
5. 在处理高阶多项式时，考虑使用分治法，将问题分解为处理低阶多项式的子问题。

重点词汇

• determinant 行列式
• Vandermonde matrix 范德蒙矩阵
• interpolation 插值
• Lagrange polynomial 拉格朗日多项式
• linear system 线性系统
• polynomial space 多项式空间

参考资料

1. Strang, G. (2006). Linear Algebra and Its Applications, 4th Edition. Brooks Cole. Chapter 5: Determinants.
2. Trefethen, L. N., & Bau III, D. (1997). Numerical Linear Algebra. SIAM. Chapter 4: Interpolation and Approximation.
3. Horn, R. A., & Johnson, C. R. (2012). Matrix Analysis, 2nd Edition. Cambridge University Press. Chapter 0: Preliminaries.
4. Burden, R. L., & Faires, J. D. (2010). Numerical Analysis, 9th Edition. Brooks Cole. Chapter 3: Interpolation and Polynomial Approximation.
5. Golub, G. H., & Van Loan, C. F. (2013). Matrix Computations, 4th Edition. Johns Hopkins University Press. Chapter 1: Matrix Multiplication.