東京大学 新領域創成科学研究科 メディカル情報生命専攻 2016年8月実施 問題8

标签：

• Tokyo-University
• Projection-Operator
• Eigenvalues-and-Eigenvectors

Author

zephyr

Description

Answer the following questions about linear algebra.

(1) Denote by $\mathbf{o}$ the zero vector. Let $\mathbf{a}$ denote a two-dimensional vector that is not $\mathbf{o}$. $T_{\mathbf{a}}(\mathbf{x})$ is the orthogonal projection of a point $\mathbf{x}$ on $\mathbf{a}$. Prove the following propositions.

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

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

(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.

(3) Denote by $\mathbf{a_1}, \mathbf{a_2}$ the column vectors corresponding to the bases of a two-dimensional subspace of the three dimensional space. Describe the projection matrix to the subspace using $\mathbf{A} = [\mathbf{a_1}, \mathbf{a_2}]$.

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回答以下有关线性代数的问题。

(1) 用 $\mathbf{o}$ 表示零向量。设 $\mathbf{a}$ 表示一个二维向量，它不是 $\mathbf{o}$。$T_{\mathbf{a}}(\mathbf{x})$ 是点 $\mathbf{x}$ 在 $\mathbf{a}$ 上的正交投影。证明以下命题。

• (1.1) 对于任意二维点 $\mathbf{x}$，$T_{\mathbf{a}}(T_{\mathbf{a}}(\mathbf{x})) = T_{\mathbf{a}}(\mathbf{x})$。

• (1.2) 对于任意非零二维向量 $\mathbf{b}$，它与 $\mathbf{a}$ 正交，$T_{\mathbf{b}}(T_{\mathbf{a}}(\mathbf{x})) = \mathbf{o}$。

(2) 假设一个实对称矩阵 $\mathbf{P}$ 满足 $\mathbf{P}^2 = \mathbf{P}$。证明 $\mathbf{P}$ 的特征值要么是 0，要么是 1。

(3) 用 $\mathbf{a_1}, \mathbf{a_2}$ 表示对应于三维空间的二维子空间基的列向量。用 $\mathbf{A} = [\mathbf{a_1}, \mathbf{a_2}]$ 描述该子空间的投影矩阵。

题目描述

回答下列线性代数问题：

1. 记零向量为 $\mathbf o$，取非零二维向量 $\mathbf a$，并令 $T_{\mathbf a}(\mathbf x)$ 表示点 $\mathbf x$ 在 $\mathbf a$ 所张成直线上的正交投影。结合上图证明：
   1. 对任意二维点 $\mathbf x$，
      $$T_{\mathbf a}(T_{\mathbf a}(\mathbf x))=T_{\mathbf a}(\mathbf x);$$
   2. 对任意与 $\mathbf a$ 正交的非零二维向量 $\mathbf b$，
      $$T_{\mathbf b}(T_{\mathbf a}(\mathbf x))=\mathbf o.$$
2. 设实对称矩阵 $\mathbf P$ 满足 $\mathbf P^2=\mathbf P$，证明其特征值只能是 $0$ 或 $1$。
3. 在三维空间中，列向量 $\mathbf a_1,\mathbf a_2$ 构成某二维子空间的一组基。令
   $$\mathbf A=[\mathbf a_1,\mathbf a_2],$$
   用 $\mathbf A$ 写出到该子空间的正交投影矩阵。

考点

• 正交投影算子：用投影的几何意义或内积公式证明幂等性，以及正交方向上的再次投影为零。
• 投影矩阵的谱性质：把 $\mathbf P^2=\mathbf P$ 作用于特征向量，推出特征值满足 $\lambda^2=\lambda$。
• 非正交基下的投影矩阵：由基矩阵 $\mathbf A$ 与 Gram 矩阵构造到其列空间的正交投影。

Kai

(1)

(1.1)

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)

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:

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

(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 $\lambda$:

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

Applying $\mathbf{P}$ again:

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

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

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

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:

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

(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 $\mathbf{A}$:

$$\mathbf{P}\mathbf{x} = c_1 \mathbf{a_1} + c_2 \mathbf{a_2}$$

In matrix form, we write:

$$\mathbf{P}\mathbf{x} = \mathbf{A}\mathbf{c}$$

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

$$\mathbf{c} = \begin{bmatrix} c_1 \\ c_2 \end{bmatrix}$$

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:

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

This equation implies:

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

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:

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

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

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

Knowledge

对称矩阵 特征值和特征向量 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.