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 2022年8月実施 問題8

标签：

• Tokyo-University
• Eigenvalues-and-Eigenvectors
• Positive-Semidefinite-Matrix
• Diagonal-Dominance

Author

zephyr

Description

(1) Describe the eigenvalues and eigenvectors of the following matrix. ($\lambda$ is a real value.)

$$\mathbf{A}_\lambda = \begin{pmatrix} \lambda & 1 & 1 \\ 1 & \lambda & 1 \\ 1 & 1 & \lambda \end{pmatrix}$$

(2) What is the range of $\lambda$ such that $\mathbf{A}_\lambda$ is positive semidefinite.

(3) Consider an $n \times n$ symmetric matrix where all diagonal elements are $b$ and all non-diagonal elements are $a$. Show that this matrix is non-singular when $|b| > |(n - 1)a|$.

题目描述

1. 对实参数 $\lambda$，求矩阵
   $$\mathbf A_\lambda= \begin{pmatrix} \lambda&1&1\\ 1&\lambda&1\\ 1&1&\lambda \end{pmatrix}$$
   的全部特征值及相应特征向量。
2. 求使 $\mathbf A_\lambda$ 为半正定矩阵的 $\lambda$ 取值范围。
3. 设一个 $n\times n$ 对称矩阵的全部对角元均为 $b$，全部非对角元均为 $a$。证明当
   $$|b|>|(n-1)a|$$
   时该矩阵非奇异。

考点

• 高对称矩阵的特征分解：把全一方向与其正交补分开，求常对角、常非对角矩阵的两个特征值及特征空间。
• 半正定判定：利用实对称矩阵全部特征值非负的充要条件，确定参数 $\lambda$ 的范围。
• 严格对角占优：由 $|b|>(n-1)|a|$ 或等价谱表达证明零不可能成为特征值，从而得到非奇异性。

Kai

(1) Eigenvalues and Eigenvectors

To find the eigenvalues $\mu_i$ of the matrix $\mathbf{A}_\lambda$, we need to solve the characteristic equation $\det(\mathbf{A}_\lambda - \mu \mathbf{I}) = 0$.

$$\mathbf{A}_\lambda - \mu \mathbf{I} = \begin{pmatrix} \lambda - \mu & 1 & 1 \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{pmatrix}$$

Using the row addition method for simplification, we can add all rows to the first row:

$$\begin{vmatrix} \lambda - \mu & 1 & 1 \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{vmatrix} = \begin{vmatrix} \lambda - \mu + 1 + 1 & 1 + (\lambda - \mu) + 1 & 1 + 1 + (\lambda - \mu) \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{vmatrix} = \begin{vmatrix} \lambda - \mu + 2 & \lambda - \mu + 2 & \lambda - \mu + 2 \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{vmatrix}$$

Now, we can absorb the common factor $(\lambda - \mu + 2)$ from the first row:

$$= (\lambda - \mu + 2) \begin{vmatrix} 1 & 1 & 1 \\ 1 & \lambda - \mu & 1 \\ 1 & 1 & \lambda - \mu \end{vmatrix} = (\lambda - \mu + 2) \begin{vmatrix} 1 & 1 & 1 \\ 0 & \lambda - \mu - 1 & 0 \\ 0 & 0 & \lambda - \mu - 1 \end{vmatrix} = (\lambda - \mu + 2)(\lambda - \mu - 1)^2$$

Setting the determinant to zero:

So, the eigenvalues are:

$$\mu_1 = \lambda + 2, \quad \mu_2 = \lambda - 1 \quad \text{(with multiplicity 2)}$$

Eigenvectors

For $\mu_1 = \lambda + 2$:

$$\mathbf{A}_\lambda - (\lambda + 2) \mathbf{I} = \begin{pmatrix} -2 & 1 & 1 \\ 1 & -2 & 1 \\ 1 & 1 & -2 \end{pmatrix}$$

Solving

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

, we find:

$$\mathbf{v}_1 = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$$

For $\mu_2 = \lambda - 1$:

$$\mathbf{A}_\lambda - (\lambda - 1) \mathbf{I} = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$$

We need to solve $\mathbf{A}_\lambda \mathbf{v} = (\lambda - 1)\mathbf{v}$, which gives the eigenvectors corresponding to $\mu_2$. This generally results in two linearly independent eigenvectors orthogonal to $\mathbf{v}_1$:

$$\mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}, \quad \mathbf{v}_3 = \begin{pmatrix} 0 \\ 1 \\ -1 \end{pmatrix}$$

(2) Positive Semi-definiteness

A matrix is positive semidefinite if all its eigenvalues are non-negative. For $\mathbf{A}_\lambda$ to be positive semidefinite:

$$\mu_1 = \lambda + 2 \geq 0$$

$$\mu_2 = \lambda - 1 \geq 0$$

Solving these inequalities:

$$\lambda + 2 \geq 0 \implies \lambda \geq -2$$

$$\lambda - 1 \geq 0 \implies \lambda \geq 1$$

The most restrictive condition is $\lambda \geq 1$.

Thus, the range of $\lambda$ such that $\mathbf{A}_\lambda$ is positive semidefinite is:

$$\lambda \geq 1$$

(3) Non-Singularity of Symmetric Matrix

Consider an $n \times n$ symmetric matrix $\mathbf{B}$ where all diagonal elements are $b$ and all non-diagonal elements are $a$.

$$\mathbf{B} = \begin{pmatrix} b & a & a & \cdots & a \\ a & b & a & \cdots & a \\ a & a & b & \cdots & a \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ a & a & a & \cdots & b \end{pmatrix}$$

The matrix $\mathbf{B}$ can be written as:

$$\mathbf{B} = b\mathbf{I} + a\mathbf{J}$$

where $\mathbf{J}$ is the $n \times n$ matrix with all elements equal to 1.

The eigenvalues of $\mathbf{J}$ are $n$ (with multiplicity 1) and $0$ (with multiplicity $n-1$). Thus, the eigenvalues of $\mathbf{B}$ are:

$$\lambda_1 = b + (n-1)a \quad \text{and} \quad \lambda_2 = b - a \quad (\text{with multiplicity } n-1)$$

The matrix $\mathbf{B}$ is non-singular if all its eigenvalues are non-zero:

$$b + (n-1)a \neq 0$$

$$b - a \neq 0$$

This holds if:

$$|b| > |(n-1)a|$$

since $|b| > (n-1)|a|$ implies $|b| > |a|$ when $n \geq 2$.

Thus, the matrix $\mathbf{B}$ is non-singular when $|b| > |(n-1)a|$.

Knowledge

特征值和特征向量 正定矩阵

难点解题思路

1. 通过求解特征方程来找到特征值。
2. 根据特征值的符号判断矩阵的半正定性。
3. 使用矩阵特征值的性质判断矩阵的非奇异性。

解题技巧和信息

1. 计算特征方程时，使用行列式和代数余子式。
2. 确定半正定矩阵时，所有特征值必须为非负数。
3. 判断矩阵是否非奇异，可以通过特征值是否全非零来实现。

重点词汇

eigenvalue 特征值

eigenvector 特征向量

positive semidefinite 正半定

non-singular 非奇异

参考资料

1. Linear Algebra and Its Applications by Gilbert Strang, Chap. 6
2. Introduction to Linear Algebra by Gilbert Strang, Chap. 7