京都大学 情報学研究科 知能情報学専攻 2023年2月実施 基礎科目 F1-1

标签：

• 京都大学
• 特征多项式
• 特征值与特征向量
• 格拉姆–施密特正交化
• 投影算子

Author

祭音Myyura (co-authored with GPT 5.6 SOL)

Description

Let

$$A=\begin{pmatrix} 1&a&0\\ b&3&0\\ -1&1&2 \end{pmatrix},\qquad a,b\in\mathbb R.$$

1. Find the condition on $a,b$ under which all eigenvalues of $A$ are distinct real numbers.
2. Let $\lambda_1$ and $\lambda_2$ be the largest and smallest eigenvalues, and let $v_1,v_2$ be corresponding eigenvectors. For $a=6,b=4$:
   1. find $\lambda_1,\lambda_2,v_1,v_2$;
   2. find an orthonormal basis of $W=\operatorname{span}\{v_1,v_2\}$;
   3. find the orthogonal projection of $r=(1,4,-6)^\top$ onto $W$.

题目描述

设

$$A=\begin{pmatrix} 1&a&0\\ b&3&0\\ -1&1&2 \end{pmatrix},\qquad a,b\in\mathbb R.$$

1. 求使 $A$ 的特征值均为两两不同实数的 $a,b$ 条件。
2. 记最大、最小特征值分别为 $\lambda_1,\lambda_2$，对应特征向量为 $v_1,v_2$。当 $a=6,b=4$ 时：
   1. 求 $\lambda_1,\lambda_2,v_1,v_2$；
   2. 求 $W=\operatorname{span}\{v_1,v_2\}$ 的一组标准正交基；
   3. 求 $r=(1,4,-6)^\top$ 在 $W$ 上的正交投影。

Kai

Q.1

The characteristic polynomial factors as

$$\det(\lambda I-A) =(\lambda-2)\left((\lambda-1)(\lambda-3)-ab\right) =(\lambda-2)\left((\lambda-2)^2-(1+ab)\right).$$

Hence the eigenvalues are

$$2,\qquad 2+\sqrt{1+ab},\qquad 2-\sqrt{1+ab}.$$

They are real and pairwise distinct if and only if

$$\boxed{ab>-1}.$$

Q.2

For $a=6,b=4$, the eigenvalues are $7,2,-3$. Therefore

$$\lambda_1=7,\qquad \lambda_2=-3.$$

(1)

Solving $(A-7I)v_1=0$ and $(A+3I)v_2=0$, we may take

$$\boxed{v_1=\begin{pmatrix}1\\1\\0\end{pmatrix}}, \qquad \boxed{v_2=\begin{pmatrix}-3\\2\\-1\end{pmatrix}}.$$

(2)

Normalize $v_1$ and apply Gram-Schmidt to $v_2$:

$$e_1=\frac1{\sqrt2}\begin{pmatrix}1\\1\\0\end{pmatrix},$$

$$v_2-(v_2^\top e_1)e_1 =\frac12\begin{pmatrix}-5\\5\\-2\end{pmatrix}.$$

Thus an orthonormal basis is

$$\boxed{ e_1=\frac1{\sqrt2}\begin{pmatrix}1\\1\\0\end{pmatrix}, \qquad e_2=\frac1{3\sqrt6}\begin{pmatrix}-5\\5\\-2\end{pmatrix} }.$$

(3)

Using this basis,

$$\begin{aligned} \operatorname{proj}_W r &=(r^\top e_1)e_1+(r^\top e_2)e_2\\ &=\begin{pmatrix}5/2\\5/2\\0\end{pmatrix} +\begin{pmatrix}-5/2\\5/2\\-1\end{pmatrix} =\boxed{\begin{pmatrix}0\\5\\-1\end{pmatrix}}. \end{aligned}$$