九州大学 システム情報科学府 情報理工学専攻・電気電子工学専攻 2024年1月実施 線形代数

标签：

• Kyushu-University
• Rayleigh-Quotient
• Eigenvalues-and-Eigenvectors

Author

祭音Myyura (assisted by ChatGPT 5.4 Thinking)

Description

Let $A$ be an $n \times n$ real matrix and $\mathbf{x}$ be an $n$-dimensional nonzero real column vector. Define

$$R_A(\mathbf{x})=\frac{\mathbf{x}^T A\mathbf{x}}{\|\mathbf{x}\|^2},$$

where $\mathbf{x}^T$ is the transpose of a vector $\mathbf{x}$. Answer the following questions.

(1) For

$$B= \begin{bmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix} \quad \text{and} \quad \mathbf{z}= \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix},$$

find $R_B(\mathbf{z})$.

(2) Find all the eigenvalues of $B$. For each eigenvalue of $B$, find its eigenspace.

(3) Suppose that $A$ is a symmetric matrix. Show

$$\max_{\mathbf{x} \neq 0} R_A(\mathbf{x})=\lambda_{\max},$$

where $\lambda_{\max}$ is the largest eigenvalue of $A$. Use the fact that the symmetric matrix $A$ has eigenvectors $\mathbf{q}_1,\dots,\mathbf{q}_n \in \mathbb{R}^n$ that form an orthonormal basis of $\mathbb{R}^n$.

题目描述

设 $A$ 为 $n\times n$ 实矩阵，$\mathbf x$ 为非零的 $n$ 维实列向量，定义 Rayleigh 商

$$R_A(\mathbf x)=\frac{\mathbf x^\top A\mathbf x}{\|\mathbf x\|^2}.$$

1. 对
   $$B=\begin{bmatrix}2&0&1\\0&1&0\\1&0&2\end{bmatrix}, \qquad \mathbf z=\begin{bmatrix}1\\1\\0\end{bmatrix},$$
   求 $R_B(\mathbf z)$。
2. 求 $B$ 的全部特征值以及每个特征值的特征空间。
3. 设 $A$ 为对称矩阵，证明
   $$\max_{\mathbf x\ne0}R_A(\mathbf x)=\lambda_{\max},$$
   其中 $\lambda_{\max}$ 是 $A$ 的最大特征值。可使用如下事实：对称矩阵 $A$ 存在构成 $\mathbb R^n$ 一组标准正交基的特征向量 $\mathbf q_1,\ldots,\mathbf q_n$。

Kai

(1)

We are given

$$B= \begin{bmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix}, \qquad \mathbf{z}= \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix}.$$

By definition,

$$R_B(\mathbf{z}) = \frac{\mathbf{z}^T B\mathbf{z}}{\|\mathbf{z}\|^2}.$$

First compute $B\mathbf{z}$:

$$B\mathbf{z} = \begin{bmatrix} 2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 2 \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}.$$

Then

$$\mathbf{z}^T B\mathbf{z} = \begin{bmatrix} 1 & 1 & 0 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix} = 3.$$

Also,

$$\|\mathbf{z}\|^2 = 1^2+1^2+0^2 = 2.$$

Therefore,

$$R_B(\mathbf{z}) = \frac{3}{2}.$$

Hence,

$$\boxed{R_B(\mathbf{z})=\frac{3}{2}}.$$

(2)

To find the eigenvalues, solve

$$\det(B-\lambda I)=0.$$

We have

$$B-\lambda I = \begin{bmatrix} 2-\lambda & 0 & 1 \\ 0 & 1-\lambda & 0 \\ 1 & 0 & 2-\lambda \end{bmatrix}.$$

Thus,

$$\det(B-\lambda I) = (1-\lambda) \det \begin{bmatrix} 2-\lambda & 1 \\ 1 & 2-\lambda \end{bmatrix}.$$

Now compute the $2\times 2$ determinant:

$$\det(B-\lambda I) = (1-\lambda)\left((2-\lambda)^2-1\right).$$

Since

$$(2-\lambda)^2-1 = (2-\lambda-1)(2-\lambda+1) = (1-\lambda)(3-\lambda),$$

we get

$$\det(B-\lambda I) = (1-\lambda)^2(3-\lambda).$$

Therefore,

$$(1-\lambda)^2(3-\lambda)=0.$$

So the eigenvalues are

$$\boxed{\lambda=1,3}.$$

The eigenvalue $\lambda=1$ has algebraic multiplicity $2$, and the eigenvalue $\lambda=3$ has algebraic multiplicity $1$.

Eigenspace for $\lambda=3$

Solve

$$(B-3I)\mathbf{x}=0.$$

We have

$$B-3I = \begin{bmatrix} -1 & 0 & 1 \\ 0 & -2 & 0 \\ 1 & 0 & -1 \end{bmatrix}.$$

Let

$$\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}.$$

Then the system is

$$-x+z=0,$$

$$-2y=0,$$

$$x-z=0.$$

Hence,

$$z=x, \qquad y=0.$$

Therefore,

$$\mathbf{x} = \begin{bmatrix} x \\ 0 \\ x \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix}.$$

So the eigenspace corresponding to $\lambda=3$ is

$$\boxed{ E_3= \operatorname{span} \left\{ \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} \right\} }.$$

Eigenspace for $\lambda=1$

Solve

$$(B-I)\mathbf{x}=0.$$

We have

$$B-I = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 0 & 0 \\ 1 & 0 & 1 \end{bmatrix}.$$

Let

$$\mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}.$$

The system gives

$$x+z=0.$$

Thus,

$$z=-x.$$

Here $x$ and $y$ are free variables. Therefore,

$$\mathbf{x} = \begin{bmatrix} x \\ y \\ -x \end{bmatrix} = x \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix} + y \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix}.$$

So the eigenspace corresponding to $\lambda=1$ is

$$\boxed{ E_1= \operatorname{span} \left\{ \begin{bmatrix} 1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \right\} }.$$

(3)

Suppose $A$ is a symmetric matrix.

Since $A$ is symmetric, it has an orthonormal basis of eigenvectors

$$\mathbf{q}_1,\mathbf{q}_2,\dots,\mathbf{q}_n$$

for $\mathbb{R}^n$.

Let their corresponding eigenvalues be

$$\lambda_1,\lambda_2,\dots,\lambda_n.$$

That is,

$$A\mathbf{q}_i=\lambda_i\mathbf{q}_i, \qquad i=1,2,\dots,n.$$

Let

$$\lambda_{\max} = \max\{\lambda_1,\lambda_2,\dots,\lambda_n\}.$$

Since $\mathbf{q}_1,\dots,\mathbf{q}_n$ form an orthonormal basis, any nonzero vector $\mathbf{x}\in\mathbb{R}^n$ can be written as

$$\mathbf{x} = c_1\mathbf{q}_1+c_2\mathbf{q}_2+\cdots+c_n\mathbf{q}_n.$$

Equivalently,

$$\mathbf{x} = \sum_{i=1}^n c_i\mathbf{q}_i.$$

Because $\mathbf{x}\neq 0$, we have

$$\sum_{i=1}^n c_i^2>0.$$

Now compute $A\mathbf{x}$:

$$A\mathbf{x} = A\left(\sum_{i=1}^n c_i\mathbf{q}_i\right).$$

By linearity,

$$A\mathbf{x} = \sum_{i=1}^n c_i A\mathbf{q}_i.$$

Since

$$A\mathbf{q}_i=\lambda_i\mathbf{q}_i,$$

we get

$$A\mathbf{x} = \sum_{i=1}^n c_i\lambda_i\mathbf{q}_i.$$

Next,

$$\mathbf{x}^T A\mathbf{x} = \left(\sum_{i=1}^n c_i\mathbf{q}_i\right)^T \left(\sum_{j=1}^n c_j\lambda_j\mathbf{q}_j\right).$$

Expanding this expression gives

$$\mathbf{x}^T A\mathbf{x} = \sum_{i=1}^n\sum_{j=1}^n c_i c_j \lambda_j \mathbf{q}_i^T\mathbf{q}_j.$$

Because the vectors $\mathbf{q}_1,\dots,\mathbf{q}_n$ are orthonormal,

$$\mathbf{q}_i^T\mathbf{q}_j = \begin{cases} 1, & i=j,\\ 0, & i\neq j. \end{cases}$$

Therefore, all cross terms vanish, and we obtain

$$\mathbf{x}^T A\mathbf{x} = \sum_{i=1}^n \lambda_i c_i^2.$$

Similarly,

$$\|\mathbf{x}\|^2 = \mathbf{x}^T\mathbf{x}.$$

Thus,

$$\|\mathbf{x}\|^2 = \left(\sum_{i=1}^n c_i\mathbf{q}_i\right)^T \left(\sum_{j=1}^n c_j\mathbf{q}_j\right).$$

Using orthonormality again, we get

$$\|\mathbf{x}\|^2 = \sum_{i=1}^n c_i^2.$$

Therefore,

$$R_A(\mathbf{x}) = \frac{\mathbf{x}^T A\mathbf{x}}{\|\mathbf{x}\|^2} = \frac{\sum_{i=1}^n \lambda_i c_i^2}{\sum_{i=1}^n c_i^2}.$$

Since

$$\lambda_i\leq \lambda_{\max} \qquad \text{for all } i,$$

we have

$$\lambda_i c_i^2 \leq \lambda_{\max}c_i^2.$$

Therefore,

$$\sum_{i=1}^n \lambda_i c_i^2 \leq \sum_{i=1}^n \lambda_{\max}c_i^2.$$

Hence,

$$\sum_{i=1}^n \lambda_i c_i^2 \leq \lambda_{\max}\sum_{i=1}^n c_i^2.$$

Since

$$\sum_{i=1}^n c_i^2>0,$$

we may divide both sides by $\sum_{i=1}^n c_i^2$. This gives

$$R_A(\mathbf{x}) = \frac{\sum_{i=1}^n \lambda_i c_i^2}{\sum_{i=1}^n c_i^2} \leq \lambda_{\max}.$$

So for every nonzero vector $\mathbf{x}$,

$$R_A(\mathbf{x})\leq \lambda_{\max}.$$

Thus,

$$\max_{\mathbf{x}\neq 0} R_A(\mathbf{x}) \leq \lambda_{\max}.$$

Now we show that this upper bound is achieved.

Let $\mathbf{q}_k$ be an eigenvector corresponding to the largest eigenvalue $\lambda_{\max}$. Then

$$A\mathbf{q}_k=\lambda_{\max}\mathbf{q}_k.$$

Choose

$$\mathbf{x}=\mathbf{q}_k.$$

Then

$$R_A(\mathbf{q}_k) = \frac{\mathbf{q}_k^T A\mathbf{q}_k}{\|\mathbf{q}_k\|^2}.$$

Using

$$A\mathbf{q}_k=\lambda_{\max}\mathbf{q}_k,$$

we get

$$R_A(\mathbf{q}_k) = \frac{\mathbf{q}_k^T \lambda_{\max}\mathbf{q}_k}{\|\mathbf{q}_k\|^2}.$$

Thus,

$$R_A(\mathbf{q}_k) = \frac{\lambda_{\max}\mathbf{q}_k^T\mathbf{q}_k}{\|\mathbf{q}_k\|^2}.$$

Since

$$\mathbf{q}_k^T\mathbf{q}_k = \|\mathbf{q}_k\|^2,$$

we have

$$R_A(\mathbf{q}_k) = \lambda_{\max}.$$

Therefore, the maximum value is actually attained, and we conclude that

$$\boxed{ \max_{\mathbf{x}\neq 0} R_A(\mathbf{x})=\lambda_{\max} }.$$