京都大学 情報学研究科 知能情報学専攻 2020年2月実施 情報学基礎 F1-1

标签：

• 京都大学
• 特征多项式
• 特征值与特征向量
• 内积与正交性
• 行列式

Author

思齐塾, 祭音Myyura

Description

Let $a$ and $b$ be real numbers. We define a matrix

$$A = \begin{pmatrix} 0 & a & b \\ a & 0 & 1 \\ b & 1 & 0 \end{pmatrix}$$

In all of the questions (1), (2), (3), and (4), assume that an eigenvalue (characteristic value) of $A$ is 1.

(1) What is the relation between $a$ and $b$ ?

(2) When $a = 0$ , list all eigenvalues of $A$ .

In questions (3) and (4), let us additionally assume that $a > 0$ and also that the characteristic polynomial of $A$ has a double (multiple) root of an integer.

(3) Give the eigenvector $v$ corresponding to the smallest eigenvalue of $A$ .

(4) Give a pair of vectors $u$ and $w$ satisfying all of the following conditions:

• both $u$ and $w$ are eigenvectors corresponding to the eigenvalue which is the double root of the characteristic polynomial of $A$ ,
• $u$ is orthogonal to both $e = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}$ and $w$ , and
• $||u|| = ||w|| = 1$ .

题目描述

设 $a,b$ 为实数，并定义

$$A= \begin{pmatrix} 0&a&b\\ a&0&1\\ b&1&0 \end{pmatrix}.$$

在以下第 1–4 问中，始终假设 $1$ 是 $A$ 的一个特征值。

1. 求 $a$ 与 $b$ 之间必须满足的关系。
2. 当 $a=0$ 时，列出 $A$ 的全部特征值。

在第 3、4 问中，进一步假设 $a>0$，并且 $A$ 的特征多项式有一个整数二重根。

3. 求与 $A$ 的最小特征值对应的一个特征向量 $\boldsymbol{v}$。

4. 求一对向量 $\boldsymbol{u},\boldsymbol{w}$，使其同时满足：

   • $\boldsymbol{u}$ 与 $\boldsymbol{w}$ 都是特征多项式二重根所对应的特征向量；
   • 令

$$\boldsymbol{e} = \begin{pmatrix} 0\\0\\1 \end{pmatrix},$$

则 $\boldsymbol{u}$ 同时与 $\boldsymbol{e}$、$\boldsymbol{w}$ 正交；

• $\|\boldsymbol{u}\|=\|\boldsymbol{w}\|=1$。

Kai

(1) Since 1 is an eigenvalue of A, we have $\det(A - I) = 0$ , where $I$ is the identity matrix. Therefore,

$$\det(A - I) = \det\begin{pmatrix} -1 & a & b \\ a & -1 & 1 \\ b & 1 & -1 \end{pmatrix} = -1(1-1) - a(-a-b) + b(a+b) = a^2 + ab + ab + b^2 = a^2 + 2ab + b^2 = (a+b)^2 = 0$$

Thus, $a+b = 0$ , so $b = -a$ .

(2) When $a=0$ , we have $b = -a = 0$ . Then the matrix $A$ becomes

$$A = \begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0 \end{pmatrix}$$

The characteristic polynomial is given by

$$\det(A - \lambda I) = \det\begin{pmatrix} -\lambda & 0 & 0 \\ 0 & -\lambda & 1 \\ 0 & 1 & -\lambda \end{pmatrix} = -\lambda(\lambda^2 - 1) = -\lambda(\lambda - 1)(\lambda + 1) = 0$$

The eigenvalues are $\lambda_1 = 0, \lambda_2 = 1, \lambda_3 = -1$ .

(3) We are given $a>0$ and that the characteristic polynomial has a double root. Also $b=-a$ , so $A = \begin{pmatrix} 0 & a & -a \\ a & 0 & 1 \\ -a & 1 & 0 \end{pmatrix}$ . The characteristic polynomial is

$$\det(A - \lambda I) = \det\begin{pmatrix} -\lambda & a & -a \\ a & -\lambda & 1 \\ -a & 1 & -\lambda \end{pmatrix} = -\lambda(\lambda^2 - 1) - a(-a\lambda + a) - a(a - a\lambda) = -\lambda^3 + \lambda + a^2\lambda - a^2 - a^2 + a^2\lambda = -\lambda^3 + (1 + 2a^2)\lambda - 2a^2 = 0$$

Dividing by $\lambda - 1$ , we get

$$-(\lambda - 1)(\lambda^2 + \lambda - 2a^2) = 0$$

The other roots are $\lambda = \frac{-1 \pm \sqrt{1 + 8a^2}}{2}$ . The two roots of the quadratic cannot coincide because $1+8a^2>0$ . Hence a double root occurs only when one of them is equal to the already known root 1. Thus $1+1-2a^2=0$ , and $a=1$ because $a>0$ . Therefore the eigenvalues are $1,1,-2$ . For the smallest eigenvalue $-2$ ,

$$(A+2I)v=0$$

is satisfied by

$$v=\begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}.$$

(4) The eigenspace for the double eigenvalue 1 is given by

$$-x+y-z=0.$$

The unit vector in this eigenspace that is orthogonal to $e$ can be chosen as

$$u=\frac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}.$$

Within the same eigenspace, a vector orthogonal to $u$ is $(1,-1,-2)^T$ . Hence we may choose

$$w=\frac{1}{\sqrt{6}}\begin{pmatrix} 1 \\ -1 \\ -2 \end{pmatrix}.$$

Then $Au=u$ , $Aw=w$ , $u\cdot e=0$ , $u\cdot w=0$ , and $||u||=||w||=1$ .