京都大学 情報学研究科 知能情報学専攻 2022年2月実施 基礎科目 F1-1

标签：

• Kyoto-University
• Matrix-Inverse
• Eigenvalues-and-Eigenvectors
• Matrix-Power
• Sparse-Matrix

Author

祭音Myyura (co-authored with GPT 5.6 SOL)

Description

Q.1

Consider the matrix

$$A= \begin{pmatrix} 0&1&0\\ 2&1&0\\ 0&-1&1 \end{pmatrix}.$$

(1) Derive the inverse matrix of $A$.

(2) Derive all eigenvalues of $A$.

(3) Derive $A^{10}$.

Q.2

Let

$$\boldsymbol{x}= \begin{pmatrix} 1&2&3&4&5&6&7&8&9&10 \end{pmatrix}^{\mathsf T},$$

and let $B$ be a $10\times10$ matrix. All entries of $B$ are zero except

$$\begin{gathered} B(1,6)=8,\qquad B(3,7)=2,\qquad B(4,8)=\frac18,\\ B(6,4)=5,\qquad B(7,1)=\frac14,\qquad B(8,10)=4,\qquad B(10,3)=\frac1{10}. \end{gathered}$$

Derive $B^{50}\boldsymbol{x}$.

题目描述

1. 对矩阵

   $$A=\begin{pmatrix}0&1&0\\2&1&0\\0&-1&1\end{pmatrix},$$

   求 $A^{-1}$、全部特征值及 $A^{10}$。

2. 给定 $\boldsymbol{x}=(1,2,\ldots,10)^\mathsf T$ 和题设七个非零元素所定义的 $10\times10$ 稀疏矩阵 $B$，求 $B^{50}\boldsymbol{x}$。

Kai

Q.1

(1)

Gaussian elimination gives

$$\boxed{ A^{-1}= \begin{pmatrix} -\frac12&\frac12&0\\ 1&0&0\\ 1&0&1 \end{pmatrix}.}$$

Indeed, direct multiplication gives $AA^{-1}=I$.

(2)

The characteristic polynomial is

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

Thus

$$\boxed{\lambda=-1,\ 1,\ 2}.$$

(3)

The eigenvalues are distinct, so the characteristic polynomial is also the minimal polynomial. The quadratic polynomial

$$r(t)=341t^2-340$$

agrees with $t^{10}$ at $t=-1,1,2$. Therefore $t^{10}-r(t)$ is divisible by the characteristic polynomial, and Cayley--Hamilton gives

$$A^{10}=341A^2-340I.$$

Since

$$A^2= \begin{pmatrix} 2&1&0\\ 2&3&0\\ -2&-2&1 \end{pmatrix},$$

we obtain

$$\boxed{ A^{10}= \begin{pmatrix} 342&341&0\\ 682&683&0\\ -682&-682&1 \end{pmatrix}.}$$

Q.2

For a nonzero entry $B(i,j)$, view multiplication by $B$ as sending coordinate $j$ to coordinate $i$ with multiplier $B(i,j)$. The seven active coordinates form the cycle

$$1\xrightarrow{1/4}7\xrightarrow{2}3\xrightarrow{1/10}10 \xrightarrow{4}8\xrightarrow{1/8}4\xrightarrow{5}6\xrightarrow{8}1.$$

The product of its multipliers is

$$\frac14\cdot2\cdot\frac1{10}\cdot4\cdot\frac18\cdot5\cdot8=1.$$

Hence $B^7$ is the identity on coordinates $\{1,3,4,6,7,8,10\}$ and zero on coordinates $\{2,5,9\}$. Moreover, $B\boldsymbol{x}$ is supported on the seven active coordinates. Since $50=7\cdot7+1$,

$$B^{50}\boldsymbol{x} =(B^7)^7B\boldsymbol{x} =B\boldsymbol{x}.$$

Therefore,

$$\boxed{ B^{50}\boldsymbol{x}= \begin{pmatrix} 48\\ 0\\ 14\\ 1\\ 0\\ 20\\ \frac14\\ 40\\ 0\\ \frac3{10} \end{pmatrix}.}$$