京都大学 情報学研究科 通信情報システム専攻 2024年2月実施 専門基礎A [A-1]

标签：

• Kyoto-University
• Limit
• Double-Integral
• Eigenvalues-and-Eigenvectors
• Matrix-Power

Author

祭音Myyura

Description

(1)

Find the limit

$$\lim_{x \to 0} \frac{\sinh x}{x}$$

(2)

Evaluate the integral

$$D = \{(x,y) \mid 0 \leq x+y \leq 1, 0 \leq x-y \leq 1\}$$

$$I = \iint_D x^2 \ dxdy$$

(3)

Given the matrix:

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

(i) Find the eigenvalues and corresponding eigenvectors of $A$, assuming it has distinct eigenvalues.

(ii) Using the eigenvalues and eigenvectors from (i), find $A^n$, where $n$ is a positive integer.

(iii) If the eigenvalues are the same, find $A^n$ for a positive integer $n$.

题目描述

1. 求极限

   $$\lim_{x\to0}\frac{\sinh x}{x}.$$

2. 对区域

   $$D=\{(x,y)\mid0\le x+y\le1,\ 0\le x-y\le1\},$$

   计算

   $$I=\iint_D x^2\,dx\,dy.$$

3. 给定上三角矩阵

   $$A=\begin{pmatrix}a&b\\0&d\end{pmatrix}.$$
   1. 假设 $A$ 有两个不同特征值，求特征值及相应特征向量。
   2. 利用第 1 小问的特征值、特征向量，求正整数 $n$ 时的 $A^n$。
   3. 若两个特征值相同，求正整数 $n$ 时的 $A^n$。

Kai

(1)

We know the Maclaurin series for $\sinh x$ is:

$$\sinh x = x + \frac{x^3}{6} + O(x^5)$$

Thus,

$$\frac{\sinh x}{x} = 1 + \frac{x^2}{6} + O(x^4)$$

As $x \to 0$, the limit is:

$$\lim_{x \to 0} \frac{\sinh x}{x} = 1$$

(2)

Put $u=x+y$ and $v=x-y$. Then $0\le u,v\le1$, $x=\frac{u+v}{2}$, $y=\frac{u-v}{2}$, and $\left|\partial(x,y)/\partial(u,v)\right|=\frac12$.

The integral in the $uv$-coordinate system is given by:

$$\iint _{D}x^{2}\,dA=\int _{0}^{1}\int _{0}^{1}\frac{(u+v)^{2}}{4}\cdot \frac{1}{2}\,du\,dv=\int _{0}^{1}\int _{0}^{1}\frac{(u+v)^{2}}{8}\,du\,dv$$

We integrate with respect to $u$ first, then $v$:

$$\int _{0}^{1}\frac{1}{8}\left[\frac{(u+v)^{3}}{3}\right]_{0}^{1}\,dv=\int _{0}^{1}\frac{1}{24}[(1+v)^{3}-v^{3}]\,dv=\int _{0}^{1}\frac{1+3v+3v^{2}}{24}\,dv$$

Evaluating the final integral:

$$\frac{1}{24}\left[v+\frac{3v^{2}}{2}+v^{3}\right]_{0}^{1}=\frac{1}{24}\left(1+\frac{3}{2}+1\right)=\frac{1}{24}\left(\frac{7}{2}\right)=\frac{7}{48}$$

(3)

(i)

The characteristic equation is:

$$\text{det}(A - \lambda I) = 0$$

$$\begin{pmatrix} a - \lambda & b \\ 0 & d - \lambda \end{pmatrix}$$

The determinant is:

$$(a - \lambda)(d - \lambda) = 0$$

Thus, the eigenvalues are $\lambda_1 = a$ and $\lambda_2 = d$.

The corresponding eigenvectors are:

• For $\lambda_1 = a$, the eigenvector is $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$.
• For $\lambda_2 = d$, the eigenvector is $\begin{pmatrix} \frac{b}{d-a} \\ 1 \end{pmatrix}$.

(ii)

If $A$ has distinct eigenvalues, it is diagonalizable. We can write $A=PDP^{-1}$, where $D$ is the diagonal matrix of eigenvalues and $P$ is the matrix whose columns are the corresponding eigenvectors:

$$P=\left(\begin{matrix}1&\frac{b}{d-a}\\ 0&1\end{matrix}\right),\quad D=\left(\begin{matrix}a&0\\ 0&d\end{matrix}\right)$$

The inverse of $P$ is $P^{-1}=\left(\begin{matrix}1&-\frac{b}{d-a}\\ 0&1\end{matrix}\right)$.Then $A^{n}=PD^{n}P^{-1}$:

$$A^{n}=\left(\begin{matrix}1&\frac{b}{d-a}\\ 0&1\end{matrix}\right)\left(\begin{matrix}a^{n}&0\\ 0&d^{n}\end{matrix}\right)\left(\begin{matrix}1&-\frac{b}{d-a}\\ 0&1\end{matrix}\right)$$

$$A^{n}=\left(\begin{matrix}a^{n}&\frac{bd^{n}}{d-a}\\ 0&d^{n}\end{matrix}\right)\left(\begin{matrix}1&-\frac{b}{d-a}\\ 0&1\end{matrix}\right)=\left(\begin{matrix}a^{n}&-\frac{ba^{n}}{d-a}+\frac{bd^{n}}{d-a}\\ 0&d^{n}\end{matrix}\right)$$

$$A^{n}=\left(\begin{matrix}a^{n}&b\frac{d^{n}-a^{n}}{d-a}\\ 0&d^{n}\end{matrix}\right)$$

(iii)

If $a=d$, write

$$A=aI+N,\qquad N=\begin{pmatrix}0&b\\0&0\end{pmatrix},\qquad N^2=0.$$

The binomial theorem gives

$$A^n = \begin{pmatrix} a & b \\ 0 & a \end{pmatrix}^n = \begin{pmatrix} a^n & nba^{n-1} \\ 0 & a^n \end{pmatrix}$$

for every positive integer $n$, with the upper-right entry interpreted as $b$ when $n=1$. (When $b=0$, $A=aI$ is diagonalizable, and the same formula still applies.)