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 2024年1月実施 問題8

标签：

• Tokyo-University
• Singular-Value-Decomposition
• Linear-Transformation
• Moore-Penrose-Pseudoinverse
• Orthogonal-Decomposition

Author

zephyr

Description

Suppose that the eigenvalues and the corresponding eigenvectors of an $n \times n$ square matrix $\mathbf{A}$ are $\lambda_1, \dots, \lambda_n$ and $\mathbf{\alpha}_1, \dots, \mathbf{\alpha}_n$ respectively.

Suppose that $\mathbf{I}_n$ is the $n \times n$ identity matrix, and the inverse matrix of an invertible matrix $\mathbf{C}$ is $\mathbf{C}^{-1}$.

Answer the following questions.

1. Show all the eigenvalues and the corresponding eigenvectors of $\mathbf{A}^2$.

2. If $\lambda_1, \dots, \lambda_n$ are mutually different, show that $\mathbf{P}^{-1} \mathbf{A} \mathbf{P}$ is a diagonal matrix, using $\mathbf{P} = (\mathbf{\alpha}_1, \dots, \mathbf{\alpha}_n)$ that is a matrix of concatenated eigenvectors.

3. Show all the eigenvalues and the corresponding eigenvectors of $\mathbf{B}$.

$$\mathbf{B} = \begin{pmatrix} 3 & 0 & 0 \\ -2 & 3 & 2 \\ 0 & 0 & 1 \end{pmatrix}$$

4. Suppose that $\mu$ is the maximum eigenvalue of $\mathbf{B}$, and $\mathbf{\gamma} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$. Calculate $\mathbf{\delta} = (\mathbf{B} - \mu \mathbf{I}_3)\mathbf{\gamma}$.

5. Suppose that $\beta$ is the eigenvector of $\mathbf{B}$ corresponding to the minimum eigenvalue. Calculate $\mathbf{Q}^{-1}\mathbf{B}\mathbf{Q}$ using $\mathbf{Q} = (\mathbf{\delta}, \mathbf{\gamma}, \beta)$ that is a matrix concatenating $\mathbf{\delta}, \mathbf{\gamma}, \beta$.

6. Suppose that $m$ is an arbitrary positive integer. Calculate $\mathbf{B}^m$.

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假设 $n \times n$ 方阵 $\mathbf{A}$ 的特征值及相应的特征向量分别为 $\lambda_1, \dots, \lambda_n$ 和 $\mathbf{\alpha}_1, \dots, \mathbf{\alpha}_n$。

假设 $\mathbf{I}_n$ 是 $n \times n$ 的单位矩阵，并且可逆矩阵 $\mathbf{C}$ 的逆矩阵为 $\mathbf{C}^{-1}$。

回答以下问题。

1. 展示 $\mathbf{A}^2$ 的所有特征值及相应的特征向量。

2. 如果 $\lambda_1, \dots, \lambda_n$ 是互不相同的，证明 $\mathbf{P}^{-1} \mathbf{A} \mathbf{P}$ 是一个对角矩阵，其中 $\mathbf{P} = (\mathbf{\alpha}_1, \dots, \mathbf{\alpha}_n)$ 是由特征向量构成的矩阵。

3. 展示 $\mathbf{B}$ 的所有特征值及相应的特征向量。

$$\mathbf{B} = \begin{pmatrix} 3 & 0 & 0 \\ -2 & 3 & 2 \\ 0 & 0 & 1 \end{pmatrix}$$

4. 假设 $\mu$ 是 $\mathbf{B}$ 的最大特征值，并且 $\mathbf{\gamma} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$。
   计算 $\mathbf{\delta} = (\mathbf{B} - \mu \mathbf{I}_3)\mathbf{\gamma}$。

5. 假设 $\beta$ 是 $\mathbf{B}$ 对应于最小特征值的特征向量。计算 $\mathbf{Q}^{-1}\mathbf{B}\mathbf{Q}$，其中 $\mathbf{Q} = (\mathbf{\delta}, \mathbf{\gamma}, \beta)$ 是由 $\mathbf{\delta}, \mathbf{\gamma}, \beta$ 构成的矩阵。

6. 假设 $m$ 是任意正整数。计算 $\mathbf{B}^m$。

题目描述

设 $n\times n$ 方阵 $\mathbf A$ 的特征值及对应特征向量分别为

$$\lambda_1,\ldots,\lambda_n,\qquad \boldsymbol\alpha_1,\ldots,\boldsymbol\alpha_n.$$

$\mathbf I_n$ 表示单位矩阵，可逆矩阵 $\mathbf C$ 的逆记为 $\mathbf C^{-1}$。回答：

1. 列出 $\mathbf A^2$ 的全部特征值及对应特征向量。
2. 若 $\lambda_1,\ldots,\lambda_n$ 两两不同，令
   $$\mathbf P=(\boldsymbol\alpha_1,\ldots,\boldsymbol\alpha_n),$$
   证明 $\mathbf P^{-1}\mathbf A\mathbf P$ 为对角矩阵。
3. 求
   $$\mathbf B= \begin{pmatrix} 3&0&0\\ -2&3&2\\ 0&0&1 \end{pmatrix}$$
   的全部特征值与相应特征向量。
4. 设 $\mu$ 为 $\mathbf B$ 的最大特征值，
   $$\boldsymbol\gamma=\begin{pmatrix}1\\0\\0\end{pmatrix},$$
   计算
   $$\boldsymbol\delta=(\mathbf B-\mu\mathbf I_3)\boldsymbol\gamma.$$
5. 设 $\boldsymbol\beta$ 是 $\mathbf B$ 最小特征值对应的特征向量，并令
   $$\mathbf Q=(\boldsymbol\delta,\boldsymbol\gamma,\boldsymbol\beta),$$
   计算 $\mathbf Q^{-1}\mathbf B\mathbf Q$。
6. 对任意正整数 $m$，计算 $\mathbf B^m$。

考点

• 特征值与矩阵幂：由 $\mathbf A\boldsymbol\alpha_i=\lambda_i\boldsymbol\alpha_i$ 推出 $\mathbf A^2$ 的谱，并直接求具体三阶矩阵的特征空间。
• 对角化条件：利用互异特征值对应特征向量线性无关，构造特征向量矩阵完成相似对角化。
• 广义特征向量与 Jordan 形：识别 $\mathbf B$ 的重特征值缺少足够普通特征向量，以 $\boldsymbol\gamma,\boldsymbol\delta$ 构造 Jordan 链并据此求 $\mathbf B^m$。

Kai

1. Positive Eigenvalues and Normalized Eigenvectors of $\mathbf{A}^T \mathbf{A}$

Given the singular value decomposition (SVD) of $\mathbf{A}$ as $\mathbf{A} = \mathbf{U} \mathbf{\Sigma} \mathbf{V}^T$, we can express $\mathbf{A}^T \mathbf{A}$ as follows:

$$\mathbf{A}^T \mathbf{A} = (\mathbf{U} \mathbf{\Sigma} \mathbf{V}^T)^T (\mathbf{U} \mathbf{\Sigma} \mathbf{V}^T) = \mathbf{V} \mathbf{\Sigma}^T \mathbf{U}^T \mathbf{U} \mathbf{\Sigma} \mathbf{V}^T = \mathbf{V} \mathbf{\Sigma}^2 \mathbf{V}^T$$

The matrix $\mathbf{\Sigma}^2$ is diagonal with the diagonal elements $\sigma_k^2$ ($k = 1, \ldots, r$). Thus, the positive eigenvalues of $\mathbf{A}^T \mathbf{A}$ are exactly the $\sigma_k^2$, and the associated normalized eigenvectors are the columns of $\mathbf{V}$.

2. Surjectivity and Injectivity of $T_{\mathbf{A}}$

Surjective (onto): The mapping $T_{\mathbf{A}}: \mathbb{R}^m \to \mathbb{R}^n$ is surjective if the range of $\mathbf{A}$ spans $\mathbb{R}^n$, i.e., $\mathbf{A}$ has full row rank. This occurs when $r = n \leq m$.

Injective (one-to-one): The mapping $T_{\mathbf{A}}$ is injective if the kernel of $\mathbf{A}$ contains only the zero vector, i.e., $\mathbf{A}$ has full column rank. This occurs when $r = m \leq n$.

3. Image of $T_{\mathbf{B}}$ and Kernel of $T_{\mathbf{A}}$

The pseudoinverse $\mathbf{A}^+$ is defined as $\mathbf{A}^+ = \mathbf{V} \mathbf{\Sigma}^{-1} \mathbf{U}^T$. Consider $\mathbf{B} = \mathbf{I}_m - \mathbf{A}^+ \mathbf{A}$.

We need to show that $\mathrm{Im}(T_{\mathbf{B}})$ is isomorphic to $\mathrm{Ker}(T_{\mathbf{A}})$. Observe the following:

$$\mathbf{B} \mathbf{A} = (\mathbf{I}_m - \mathbf{A}^+ \mathbf{A}) \mathbf{A} = \mathbf{A} - \mathbf{A}^+ \mathbf{A} \mathbf{A} = \mathbf{A} - \mathbf{A} = \mathbf{0}$$

Thus, $\mathrm{Im}(\mathbf{B}) \subseteq \mathrm{Ker}(\mathbf{A})$.

Now, consider $\mathbf{x} \in \mathrm{Ker}(\mathbf{A})$. Then $\mathbf{A} \mathbf{x} = \mathbf{0}$, and

$$\mathbf{B} \mathbf{x} = (\mathbf{I}_m - \mathbf{A}^+ \mathbf{A}) \mathbf{x} = \mathbf{x}$$

Thus, $\mathbf{x} \in \mathrm{Im}(\mathbf{B})$. Therefore, $\mathrm{Im}(\mathbf{B}) = \mathrm{Ker}(\mathbf{A})$.

4. Orthogonal Decomposition

Given $\mathbf{x} = \mathbf{x}_1 + \mathbf{x}_2$ where $\mathbf{x}_1 = \mathbf{B} \mathbf{x}$ and $\mathbf{x}_2 = \mathbf{x} - \mathbf{x}_1$:

$$\mathbf{x}_2 = \mathbf{x} - \mathbf{B} \mathbf{x} = \mathbf{x} - (\mathbf{I}_m - \mathbf{A}^+ \mathbf{A}) \mathbf{x} = \mathbf{A}^+ \mathbf{A} \mathbf{x}$$

To show orthogonality:

$$\mathbf{x}_1^T \mathbf{x}_2 = (\mathbf{B} \mathbf{x})^T (\mathbf{A}^+ \mathbf{A} \mathbf{x}) = \mathbf{x}^T \mathbf{B}^T \mathbf{A}^+ \mathbf{A} \mathbf{x}$$

Since $\mathbf{B}$ is symmetric ($\mathbf{B} = \mathbf{I}_m - \mathbf{A}^+ \mathbf{A}$):

$$\mathbf{x}^T (\mathbf{I}_m - \mathbf{A}^+ \mathbf{A}) \mathbf{A}^+ \mathbf{A} \mathbf{x} = \mathbf{x}^T (\mathbf{A}^+ \mathbf{A} - \mathbf{A}^+ \mathbf{A}) \mathbf{x} = \mathbf{0}$$

Thus, $\mathbf{x}_1$ and $\mathbf{x}_2$ are orthogonal.

5. Minimizing $(\mathbf{A} \mathbf{x} - \mathbf{b})^T (\mathbf{A} \mathbf{x} - \mathbf{b})$

Let $\mathbf{x}_0 = \mathbf{A}^+ \mathbf{b}$. We need to show that $\mathbf{x} = \mathbf{x}_0$ minimizes the expression.

Consider the error:

$$\mathbf{A} \mathbf{x} - \mathbf{b} = \mathbf{A} (\mathbf{x} - \mathbf{x}_0) + (\mathbf{A} \mathbf{x}_0 - \mathbf{b})$$

Since $\mathbf{x}_0 = \mathbf{A}^+ \mathbf{b}$, we have $\mathbf{A} \mathbf{x}_0 = \mathbf{b}$, thus:

$$\mathbf{A} \mathbf{x} - \mathbf{b} = \mathbf{A} (\mathbf{x} - \mathbf{x}_0)$$

The norm to be minimized is:

$$(\mathbf{A} \mathbf{x} - \mathbf{b})^T (\mathbf{A} \mathbf{x} - \mathbf{b}) = (\mathbf{A} (\mathbf{x} - \mathbf{x}_0))^T (\mathbf{A} (\mathbf{x} - \mathbf{x}_0))$$

This is minimized when $\mathbf{x} = \mathbf{x}_0$ since $\mathbf{A} \mathbf{x}_0 = \mathbf{b}$ and $\mathbf{A} (\mathbf{x} - \mathbf{x}_0) = \mathbf{0}$.

Knowledge

重点词汇

• singular value decomposition (SVD) 奇异值分解
• pseudoinverse 广义逆
• surjective 满射
• injective 单射
• orthogonal decomposition 正交分解

参考资料

1. "Linear Algebra and Its Applications" by Gilbert Strang, Chapter 7: The Singular Value Decomposition (SVD)
2. "Matrix Computations" by Gene H. Golub and Charles F. Van Loan, Chapter 2: Matrix Analysis