京都大学 情報学研究科 知能情報学専攻 2021年2月実施 基礎科目 F1-1

标签：

• 京都大学
• 线性方程组
• 特征值与特征向量
• 矩阵对角化

Author

祭音Myyura (co-authored with GPT 5.6 SOL)

Description

Q.1

Consider the system

$$\begin{cases} x+2y+z=3,\\ ay+4z=10,\\ 2x+6y+az=b. \end{cases}$$

(1) Find those values of $a$ for which the system has a unique solution.

(2) Find those pairs of values $(a,b)$ for which the system has more than one solution.

Q.2

Consider the matrix

$$A= \begin{pmatrix} 3&1&-1\\ 1&3&-1\\ 1&1&1 \end{pmatrix}.$$

(1) Find all eigenvalues of $A$.

(2) Find a maximum set $S$ of linearly independent eigenvectors of $A$.

(3) Determine whether $A$ is diagonalizable. If it is, find $P$ such that $D=P^{-1}AP$ is diagonal; otherwise, prove that it is not diagonalizable.

题目描述

1. 对方程组

   $$\begin{cases} x+2y+z=3,\\ ay+4z=10,\\ 2x+6y+az=b \end{cases}$$

   回答：

   1. 求使方程组有唯一解的 $a$；
   2. 求使方程组有多于一个解的 $(a,b)$。

2. 对矩阵

   $$A=\begin{pmatrix}3&1&-1\\1&3&-1\\1&1&1\end{pmatrix},$$

   求全部特征值、线性无关特征向量的最大集合，并判断是否可对角化；若可，求 $P$ 使 $P^{-1}AP$ 为对角矩阵。

Kai

Q.1

The coefficient determinant is

$$\begin{aligned} \det \begin{pmatrix} 1&2&1\\ 0&a&4\\ 2&6&a \end{pmatrix} &=a^2-2a-8\\ &=(a-4)(a+2). \end{aligned}$$

Hence the system has a unique solution exactly when

$$\boxed{a\ne4,-2}.$$

For more than one solution, the coefficient matrix must be singular and the augmented system must be consistent.

• If $a=4$, the third coefficient row satisfies $R_3=2R_1+\frac12R_2$. Thus consistency requires $b=2\cdot3+\frac12\cdot10=11$.
• If $a=-2$, the third coefficient row satisfies $R_3=2R_1-R_2$. Thus consistency requires $b=2\cdot3-10=-4$.

Therefore,

$$\boxed{(a,b)=(4,11)\ \text{or}\ (-2,-4)}.$$

In either case the consistent singular system has infinitely many solutions.

Q.2

(1)

The characteristic polynomial is

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

Thus the eigenvalues are

$$\boxed{\lambda=2\ \text{(multiplicity 2)},\qquad \lambda=3}.$$

(2)

For $\lambda=2$,

$$(A-2I)\boldsymbol{x}=0 \iff x+y-z=0,$$

so its eigenspace has dimension $2$. For $\lambda=3$, the eigenspace is $\operatorname{span}\{(1,1,1)^\mathsf{T}\}$. One maximum linearly independent set is therefore

$$\boxed{ S=\left\{ \begin{pmatrix}1\\-1\\0\end{pmatrix}, \begin{pmatrix}1\\0\\1\end{pmatrix}, \begin{pmatrix}1\\1\\1\end{pmatrix} \right\}.}$$

(3)

The three vectors in $S$ are linearly independent. Taking them as columns gives

$$P= \begin{pmatrix} 1&1&1\\ -1&0&1\\ 0&1&1 \end{pmatrix}, \qquad \det P=-1\ne0.$$

Consequently,

$$\boxed{P^{-1}AP= \begin{pmatrix} 2&0&0\\ 0&2&0\\ 0&0&3 \end{pmatrix}.}$$

Thus $A$ is diagonalizable.