CBMS-2016-08

题目来源：[[文字版题库/CBMS/2016#Problem 8|2016#Problem 8]] 日期：2024-07-31 题目主题：Math-线性代数-矩阵与特征值

解题思路

1. 求逆矩阵：首先，我们使用 $2\times 2$ 矩阵的公式来求其逆矩阵。
2. 方差-协方差矩阵：
   • 计算给定数据点的均值。
   • 计算方差和协方差。
   • 构造方差-协方差矩阵。
3. 特征值和特征向量：
   • 计算方差-协方差矩阵的特征值和特征向量。
4. 证明：利用特征值和特征向量的性质，证明矩阵 $\mathbf{A}$ 的逆矩阵 ${\mathbf{A}}^{-1}$ 的特征值为 $1/{\lambda }_i$。

Solution

Question 1: Inverse Matrix

To find the inverse of the matrix

$$\mathbf{A}=\left(\begin{array}{cc}1& 2\\ 2& 5\end{array}\right),$$

we use the formula for the inverse of a $2\times 2$ matrix:

$${\mathbf{A}}^{-1}=\frac{1}{\det (\mathbf{A})}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right),$$

where $\mathbf{A}=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $\det (\mathbf{A})=ad-bc$.

For our matrix,

$$a=1,b=2,c=2,d=5.$$

First, compute the determinant:

$$\det (\mathbf{A})=(1\cdot 5)-(2\cdot 2)=5-4=1.$$

Then, the inverse is

$${\mathbf{A}}^{-1}=\frac{1}{1}\left(\begin{array}{cc}5& -2\\ -2& 1\end{array}\right)=\left(\begin{array}{cc}5& -2\\ -2& 1\end{array}\right).$$

Question 2: Variance-Covariance Matrix and Eigenvalues

A: Variance-Covariance Matrix

Given data points $(-2,-2),(2,2),(1,-1),(-1,1)$, we first compute the mean values:

$$\bar{x}=\frac{-2+2+1-1}{4}=0,\bar{y}=\frac{-2+2-1+1}{4}=0.$$

Next, we compute the variances and covariances:

$${\sigma }_{xx}=\frac{1}{4}\sum _{i=1}^4(x_i-\bar{x}{)}^2=\frac{1}{4}[(2{)}^2+(2{)}^2+(1{)}^2+(1{)}^2]=\frac{10}{4}=2.5,$$ $${\sigma }_{yy}=\frac{1}{4}\sum _{i=1}^4(y_i-\bar{y}{)}^2=\frac{1}{4}[(2{)}^2+(2{)}^2+(1{)}^2+(1{)}^2]=\frac{10}{4}=2.5,$$ $${\sigma }_{xy}=\frac{1}{4}\sum _{i=1}^4(x_i-\bar{x})(y_i-\bar{y})=\frac{1}{4}[(-2)(-2)+(2)(2)+(1)(-1)+(-1)(1)]=\frac{6}{4}=\mathrm{1.5.}$$

Thus, the variance-covariance matrix is:

$$\mathbf{C}=\left(\begin{array}{cc}2.5& 1.5\\ 1.5& 2.5\end{array}\right).$$

B: Eigenvalues and Eigenvectors

To find the eigenvalues $\lambda$ of $\mathbf{C}$, solve the characteristic equation:

$$\det (\mathbf{C}-\lambda \mathbf{I})=0.$$

For our matrix $\mathbf{C}$,

$$\mathbf{C}-\lambda \mathbf{I}=\left(\begin{array}{cc}2.5-\lambda & 1.5\\ 1.5& 2.5-\lambda \end{array}\right),$$

the determinant is:

$$\det (\mathbf{C}-\lambda \mathbf{I})=(2.5-\lambda )(2.5-\lambda )-(1.5)(1.5)={\lambda }^2-5\lambda +4=0.$$

Solving for $\lambda$, we get:

$${\lambda }^2-5\lambda +4=0⟹(\lambda -4)(\lambda -1)=0⟹{\lambda }_1=4,{\lambda }_2=1.$$

To find the eigenvectors corresponding to the eigenvalues ${\lambda }_1=4$ and ${\lambda }_2=1$, we solve the equation $(\mathbf{C}-\lambda \mathbf{I})\mathbf{v}=0$.

For ${\lambda }_1=4$

$$\mathbf{C}-4\mathbf{I}=\left(\begin{array}{cc}2.5-4& 1.5\\ 1.5& 2.5-4\end{array}\right)=\left(\begin{array}{cc}-1.5& 1.5\\ 1.5& -1.5\end{array}\right).$$

The equation $(\mathbf{C}-4\mathbf{I})\mathbf{v}=0$ becomes:

$$\left(\begin{array}{cc}-1.5& 1.5\\ 1.5& -1.5\end{array}\right)\left(\begin{array}{c}{v}_1\\ v_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$$

This gives us the system of equations:

$$\begin{gathered} -1.5v_1+1.5v_2=0, \\ 1.5v_1-1.5v_2=0. \end{gathered}$$

From the first equation, we obtain $v_1=v_2$. Therefore, an eigenvector corresponding to ${\lambda }_1=4$ is:

$${\mathbf{v}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right).$$

For ${\lambda }_2=1$

$$\mathbf{C}-1\mathbf{I}=\left(\begin{array}{cc}2.5-1& 1.5\\ 1.5& 2.5-1\end{array}\right)=\left(\begin{array}{cc}1.5& 1.5\\ 1.5& 1.5\end{array}\right).$$

The equation $(\mathbf{C}-1\mathbf{I})\mathbf{v}=0$ becomes:

$$\left(\begin{array}{cc}1.5& 1.5\\ 1.5& 1.5\end{array}\right)\left(\begin{array}{c}{v}_1\\ v_2\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right).$$

This gives us the system of equations:

$$\begin{gathered} 1.5v_1+1.5v_2=0, \\ 1.5v_1+1.5v_2=0. \end{gathered}$$

From the first equation, we obtain $v_1=-v_2$. Therefore, an eigenvector corresponding to ${\lambda }_2=1$ is:

$${\mathbf{v}}_2=\left(\begin{array}{c}1\\ -1\end{array}\right).$$

Thus, the eigenvectors corresponding to the eigenvalues ${\lambda }_1=4$ and ${\lambda }_2=1$ are ${\mathbf{v}}_1=\left(\begin{array}{c}1\\ 1\end{array}\right)$ and ${\mathbf{v}}_2=\left(\begin{array}{c}1\\ -1\end{array}\right)$, respectively.

Question 3: Prove Eigenvalues of ${\mathbf{A}}^{-1}$

Let $\mathbf{A}$ be a regular matrix with eigenvalues ${\lambda }_1,{\lambda }_2,\dots ,{\lambda }_n$ and corresponding eigenvectors ${\mathbf{v}}_1,{\mathbf{v}}_2,\dots ,{\mathbf{v}}_n$. By definition, we have:

$$\mathbf{A}{\mathbf{v}}_i={\lambda }_i{\mathbf{v}}_i⟹{\mathbf{A}}^{-1}\mathbf{A}{\mathbf{v}}_i={\mathbf{A}}^{-1}({\lambda }_i{\mathbf{v}}_i)⟹{\mathbf{v}}_i={\lambda }_i{\mathbf{A}}^{-1}{\mathbf{v}}_i⟹{\mathbf{A}}^{-1}{\mathbf{v}}_i=\frac{1}{{\lambda }_i}{\mathbf{v}}_i.$$

Thus, the eigenvalues of ${\mathbf{A}}^{-1}$ are $\frac{1}{{\lambda }_i}$ for $i=1,2,\dots ,n$.

知识点

矩阵逆方差协方差矩阵特征值特征向量

解题技巧和信息

1. 计算逆矩阵时，确保熟记 $2\times 2$ 矩阵的逆矩阵公式。
2. 计算方差-协方差矩阵时，需准确计算均值、方差和协方差。
3. 找特征值和特征向量时，熟悉特征值方程和特征向量的计算方法。
4. 证明部分注意利用特征值和特征向量的定义和性质。

重点词汇

• Inverse matrix: 逆矩阵
• Variance-Covariance matrix: 方差-协方差矩阵
• Eigenvalue: 特征值
• Eigenvector: 特征向量

参考资料

1. Gilbert Strang, Linear Algebra and Its Applications, Chapter 3.
2. Axler, Sheldon, Linear Algebra Done Right, Chapter 5.