CBMS-2017-08

题目来源：[[文字版题库/CBMS/2017#Problem 8|2017#Question 8]] 日期：2024-07-29 题目主题：Math-线性代数-正交投影与特征值

解题思路

此问题涉及正交投影、对称矩阵的特征值及矩阵的基向量变换。通过理解正交投影的定义和性质，可以证明所要求的命题。对称矩阵的特征值性质和幺模矩阵的特性，将有助于证明矩阵 $\mathbf{P}$ 的特征值。最后，通过矩阵运算和基变换，可以得到投影矩阵。

Solution

Problem (1)

(1.1) $T_{\mathbf{a}}(T_{\mathbf{a}}(\mathbf{x}))=T_{\mathbf{a}}(\mathbf{x})$ for any two-dimensional point $\mathbf{x}$

The orthogonal projection of $\mathbf{x}$ on $\mathbf{a}$ is given by:

$$T_{\mathbf{a}}(\mathbf{x})=\frac{\mathbf{a}\cdot \mathbf{x}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}$$

To prove the proposition, we apply $T_{\mathbf{a}}$ again on $T_{\mathbf{a}}(\mathbf{x})$:

$$T_{\mathbf{a}}(T_{\mathbf{a}}(\mathbf{x}))=T_{\mathbf{a}}\left(\frac{\mathbf{a}\cdot \mathbf{x}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}\right)$$

Using the definition of orthogonal projection:

$$T_{\mathbf{a}}\left(\frac{\mathbf{a}\cdot \mathbf{x}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}\right)=\frac{\mathbf{a}\cdot \left(\frac{\mathbf{a}\cdot \mathbf{x}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}\right)}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}$$

Simplifying the dot products:

$$=\frac{\frac{(\mathbf{a}\cdot \mathbf{x})(\mathbf{a}\cdot \mathbf{a})}{(\mathbf{a}\cdot \mathbf{a})}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}=\frac{\mathbf{a}\cdot \mathbf{x}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}$$

Thus:

$$T_{\mathbf{a}}(T_{\mathbf{a}}(\mathbf{x}))=T_{\mathbf{a}}(\mathbf{x})$$

(1.2) $T_{\mathbf{b}}(T_{\mathbf{a}}(\mathbf{x}))=\mathbf{o}$ for any non-zero two-dimensional vector $\mathbf{b}$ orthogonal to $\mathbf{a}$

Given $\mathbf{b}\cdot \mathbf{a}=0$, we need to show:

$$T_{\mathbf{b}}(T_{\mathbf{a}}(\mathbf{x}))=T_{\mathbf{b}}\left(\frac{\mathbf{a}\cdot \mathbf{x}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}\right)$$

Using the definition of orthogonal projection:

$$T_{\mathbf{b}}\left(\frac{\mathbf{a}\cdot \mathbf{x}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}\right)=\frac{\mathbf{b}\cdot \left(\frac{\mathbf{a}\cdot \mathbf{x}}{\mathbf{a}\cdot \mathbf{a}}\mathbf{a}\right)}{\mathbf{b}\cdot \mathbf{b}}\mathbf{b}$$

Since $\mathbf{b}\cdot \mathbf{a}=0$:

$$=\frac{\frac{(\mathbf{a}\cdot \mathbf{x})(\mathbf{b}\cdot \mathbf{a})}{(\mathbf{a}\cdot \mathbf{a})}}{\mathbf{b}\cdot \mathbf{b}}\mathbf{b}=\frac{(\mathbf{a}\cdot \mathbf{x})\cdot 0}{(\mathbf{a}\cdot \mathbf{a})\cdot (\mathbf{b}\cdot \mathbf{b})}\mathbf{b}=0$$

Thus:

$$T_{\mathbf{b}}(T_{\mathbf{a}}(\mathbf{x}))=\mathbf{o}$$

Problem (2)

Assume that a real symmetric matrix $\mathbf{P}$ satisfies ${\mathbf{P}}^2=\mathbf{P}$. Prove that the eigenvalues of $\mathbf{P}$ are either 0 or 1.

Let $\mathbf{P}$ be a real symmetric matrix. Therefore, it is diagonalizable. Let $\mathbf{v}$ be an eigenvector of $\mathbf{P}$ with eigenvalue $\lambda$:

$$\mathbf{Pv}=\lambda \mathbf{v}$$

Applying $\mathbf{P}$ again:

$${\mathbf{P}}^2\mathbf{v}=\mathbf{P}(\mathbf{Pv})=\mathbf{P}(\lambda \mathbf{v})=\lambda \mathbf{Pv}=\lambda (\lambda \mathbf{v})={\lambda }^2\mathbf{v}$$

Since ${\mathbf{P}}^2=\mathbf{P}$, we have:

$${\mathbf{P}}^2\mathbf{v}=\mathbf{Pv}=\lambda \mathbf{v}$$

Thus:

$${\lambda }^2\mathbf{v}=\lambda \mathbf{v}$$

Since $\mathbf{v}$ is a non-zero vector, we can conclude:

$${\lambda }^2=\lambda$$

Thus, the eigenvalues $\lambda$ must satisfy:

$$\lambda (\lambda -1)=0$$

Therefore:

$$\lambda =0\text{or}\lambda =1$$

Problem (3)

Given the matrix $\mathbf{A}$ formed by two column vectors ${\mathbf{a}}_1$ and ${\mathbf{a}}_2$, which represent the basis of a two-dimensional subspace in three-dimensional space, we want to find the projection matrix $\mathbf{P}$ that projects any vector in ${\mathbb{R}}^3$ onto this subspace.

Matrix $\mathbf{A}$ is:

$$\mathbf{A}=[{\mathbf{a}}_1,{\mathbf{a}}_2]$$

where $\mathbf{A}$ is a $3\times 2$ matrix.

Derivation of the Projection Matrix

1. Projection of a Vector

The projection of a vector $\mathbf{x}$ onto the subspace spanned by the columns of $\mathbf{A}$ can be expressed as a linear combination of the columns of $\mathbf{A}$:

$$\mathbf{Px}=c_1{\mathbf{a}}_1+c_2{\mathbf{a}}_2$$

In matrix form, we write:

$$\mathbf{Px}=\mathbf{Ac}$$

where $\mathbf{c}$ is a column vector of coefficients:

$$\mathbf{c}=\left[\begin{array}{c}{c}_1\\ c_2\end{array}\right]$$

2. Finding the Coefficients

To determine the coefficients $\mathbf{c}$, we use the property that the projection minimizes the distance to the subspace. This can be formulated as:

$${\mathbf{A}}^T(\mathbf{x}-\mathbf{Ac})=0$$

This equation implies:

$${\mathbf{A}}^T\mathbf{x}={\mathbf{A}}^T\mathbf{Ac}$$

Assuming ${\mathbf{A}}^T\mathbf{A}$ is invertible, we solve for $\mathbf{c}$:

$$\mathbf{c}=({\mathbf{A}}^T\mathbf{A}{)}^{-1}{\mathbf{A}}^T\mathbf{x}$$

3. Constructing the Projection Matrix

Substituting $\mathbf{c}$ back into the projection formula, we have:

$$\mathbf{Px}=\mathbf{A}({\mathbf{A}}^T\mathbf{A}{)}^{-1}{\mathbf{A}}^T\mathbf{x}$$

Since this holds for any vector $\mathbf{x}$, the projection matrix $\mathbf{P}$ can be identified as:

$$\mathbf{P}=\mathbf{A}({\mathbf{A}}^T\mathbf{A}{)}^{-1}{\mathbf{A}}^T$$

知识点

对称矩阵特征值投影矩阵

重点词汇

• Orthogonal projection 正交投影
• Symmetric matrix 对称矩阵
• Eigenvalue 特征值
• Column vector 列向量
• Subspace 子空间
• Projection matrix 投影矩阵

参考资料

1. Gilbert Strang, “Linear Algebra and Its Applications,” Chap. 3, 5.
2. David C. Lay, “Linear Algebra and Its Applications,” Chap. 6.