広島大学 先進理工系科学研究科 情報科学プログラム 2022年8月実施 専門科目I 問題1

标签：

• Hiroshima-University
• Eigenvalues-and-Eigenvectors
• Matrix-Diagonalization
• Orthogonal-Matrix

Author

祭音Myyura

Description

(1) Find all the eigenvalues and the corresponding eigenvectors of the 2-dimensional square matrix $A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$.

(2) Let $x, y, z$ be real. Then find all the eigenvalues of the 2-dimensional real symmetric matrix: $A = \begin{bmatrix} x & y \\ y & z \end{bmatrix}$ and show that the eigenvalues are real.

(3) Let $\lambda_1, \lambda_2 , (\lambda_1 \geq \lambda_2)$ be the eigenvalues of the real symmetric matrix $A$. Then the matrix $A$ can be diagonalized by using the orthogonal matrix:

$$R(\theta) = \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}$$

where $R(\theta)^T$ denotes the transpose of the matrix $R(\theta)$. Express $\cos 2\theta$ and $\sin 2\theta$ using $x, y, z$.

题目描述

1. 求二阶方阵

   $$A=\begin{bmatrix}1&2\\2&3\end{bmatrix}$$

   的全部特征值及相应特征向量。

2. 设 $x,y,z$ 为实数。求二阶实对称矩阵

   $$A=\begin{bmatrix}x&y\\y&z\end{bmatrix}$$

   的全部特征值，并证明这些特征值均为实数。

3. 设 $\lambda_1,\lambda_2$ 是上述实对称矩阵 $A$ 的特征值，且 $\lambda_1\ge\lambda_2$。矩阵 $A$ 可用正交矩阵

   $$R(\theta)= \begin{bmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{bmatrix}$$

   对角化，其中 $R(\theta)^T$ 表示其转置。用 $x,y,z$ 表示 $\cos2\theta$ 和 $\sin2\theta$。

考点

• 特征值与特征向量：求具体及一般二阶实对称矩阵的谱，并由判别式证明特征值为实数。
• 矩阵对角化：把实对称矩阵化为按 $\lambda_1\ge\lambda_2$ 排列的对角形式，并从非对角元为零推导旋转角。
• 正交矩阵：利用 $R(\theta)^{-1}=R(\theta)^T$ 表达正交相似变换及 $\cos2\theta,\sin2\theta$。

Kai

(1)

For the given matrix:

$$A = \begin{bmatrix} 1 & 2 \\ 2 & 3 \end{bmatrix}$$

To find the eigenvalues and eigenvectors, we need to solve the characteristic equation:

$$\text{det}(A - \lambda I) = \lambda^2 - 4\lambda - 1 = 0$$

the eigenvalues and eigenvectors are

$$\lambda_1 = 2+\sqrt{5},\ \lambda_2= 2- \sqrt{5}$$

$$v_1 = \left(\frac{1}{2}(-1 + \sqrt{5}), 1\right), \ v_2 = \left(\frac{1}{2}(-1 - \sqrt{5}), 1\right)$$

(2)

For this matrix, the characteristic equation is:

$$\begin{vmatrix} x - \lambda & y \\ y & z - \lambda \end{vmatrix} = 0$$

This results in the quadratic equation:

$$(x - \lambda)(z - \lambda) - y^2 = 0$$

Expanding this gives:

$$\lambda^2 - (x+z)\lambda + (xz - y^2) = 0$$

The eigenvalues are the roots of this equation,

$$\lambda_1 = \frac{1}{2} (\sqrt{x^2 - 2 x z + 4 y^2 + z^2} + x + z)$$

$$\lambda_2 = \frac{1}{2} (-\sqrt{x^2 - 2 x z + 4 y^2 + z^2} + x + z)$$

which are real because the discriminant is non-negative:

$$\Delta = (x+z)^2 - 4(xz - y^2) = (x - z)^2 + y^2$$

(3)

$$\begin{aligned} R^{-1} A R &= \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x & y \\ y & z \end{bmatrix} \begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix} \\ &= \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} x \cos \theta + y \sin \theta & -x \sin \theta + y \cos \theta \\ y \cos \theta + z \sin \theta & -y \sin \theta + z \cos \theta \end{bmatrix}\\ &= \begin{bmatrix} x \cos^2 \theta + 2c \cos \theta \sin \theta + z \sin^2 \theta & (z - x) \cos \theta \sin \theta + y (\cos^2 \theta - \sin^2 \theta) \\ (z - x) \cos \theta \sin \theta + y (\cos^2 \theta - \sin^2 \theta) & x \sin^2 \theta - 2c \cos \theta \sin \theta + z \cos^2 \theta \end{bmatrix} \\ &= \begin{bmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{bmatrix} \end{aligned}$$

Then we have

$$\begin{cases} \lambda_1 + \lambda_2 = x + z \\ \lambda_1 - \lambda_2 = (x - z)\cos 2\theta + 2y \sin 2\theta \\ (z - x) \cos \theta \sin \theta + y (\cos^2 \theta - \sin^2 \theta) = 0 \end{cases}$$

which can be simplified to

$$\begin{cases} \sqrt{(x - z)^2 +4y^2} = (x - z)\cos 2\theta + 2y \sin 2\theta \\ \tan 2\theta = \displaystyle \frac{2y}{x-z} \Rightarrow x-z = \frac{2y \cos 2\theta}{\sin 2\theta} \end{cases}$$

Hence we have

$$(x-z)\cos 2\theta + 2y \sin 2\theta = 2y \frac{\cos^2 2\theta}{\sin 2\theta} +2y \sin 2\theta = \frac{2y}{\sin 2\theta}$$

Substituting this result back into

$$\sqrt{(x - z)^2 +4y^2} = (x - z)\cos 2\theta + 2y \sin 2\theta$$

we find:

$$\sin 2\theta = \frac{2y}{\sqrt{(x - z)^2 +4y^2}}$$

similarly,

$$\cos 2\theta = \frac{x-z}{\sqrt{(x - z)^2 +4y^2}}$$