広島大学 先進理工系科学研究科 情報科学プログラム 2017年8月実施 専門科目I 問題1

标签：

• Hiroshima-University
• Eigenvalues-and-Eigenvectors

Author

samparker, 祭音Myyura

Description

$A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 2 & -4 \\ -1 & 0 & 1 \end{bmatrix}$ とする。

(1) $A$ を対称行列 $S \ (S^T = S)$ と交代行列 $T \ (T^T = -T)$ の和 $(A = S + T)$ に分解せよ。ただし、$A^T$ は、行列 $A$ の転置を表す。

(2) $S$ のすべての固有値と対応する固有空間を求めよ。

(3) $T$ のすべての固有値と対応する固有空間を求めよ。

(4) 一般に、実交代行列の固有値は $0$ または純虚数であることを示せ。

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Let $A = \begin{bmatrix} 1 & 3 & 3 \\ 1 & 2 & -4 \\ -1 & 0 & 1 \end{bmatrix}$.

(1) Decompose $A$ into the sum $A = S + T$ of the symmetric matrix $S$ ($S^T = S$) and the alternative matrix $T$ ($T^T = -T$). Here $A^T$ denotes the transpose of the matrix $A$.

(2) Find all the eigenvalues of the symmetric matrix $S$ and a basis of the corresponding eigenspaces.

(3) Find all the eigenvalues of the alternative matrix $T$ and a basis of the corresponding eigenspaces.

(4) Show that the eigenvalues of the real alternative matrix are $0$ or purely imaginary numbers.

题目描述

给定

$$A=\begin{bmatrix}1&3&3\\1&2&-4\\-1&0&1\end{bmatrix}.$$

1. 将 $A$ 分解为对称矩阵 $S$（$S^T=S$）与实反对称矩阵 $T$（$T^T=-T$）之和 $A=S+T$，其中 $A^T$ 表示 $A$ 的转置。
2. 求 $S$ 的全部特征值及各特征值对应的特征空间的一组基。
3. 求 $T$ 的全部特征值及各特征值对应的特征空间的一组基。
4. 一般地，证明实反对称矩阵的特征值只能是 $0$ 或纯虚数。

考点

• 特征值与特征向量：先由矩阵的对称与反对称分解得到 $S,T$，再分别求其谱与特征空间，并利用反对称性证明实反对称矩阵的特征值类型。

Kai

(1)

Let $S = \begin{bmatrix} a & b & c \\ b & d & e \\ c & e & f \end{bmatrix}$ and $T = \begin{bmatrix} 0 & g & h \\ -g & 0 & i \\ -h & -i & 0 \end{bmatrix}$. Then we have

$$\begin{bmatrix} 1 & 3 & 3 \\ 1 & 2 & -4 \\ -1 & 0 & 1 \end{bmatrix} = \begin{bmatrix} a & b+g & c+h \\ b-g & d & e+i \\ c-h & e-i & f \end{bmatrix}$$

Solving the equations we have

$$a = 1, d=2, f = 1, h=2, c=1, g=1, b=2, i=-2, e=-2$$

Hence

$$S = \begin{bmatrix} 1 & 2 & 1 \\ 2 & 2 & -2 \\ 1 & -2 & 1 \end{bmatrix}, T = \begin{bmatrix} 0 & 1 & 2 \\ -1 & 0 & -2 \\ -2 & 2 & 0 \end{bmatrix}$$

(2)

Eigenvalues

$$\lambda_1 = 2, \lambda_2 = 4, \lambda_3 = -2$$

A basis of the corresponding eigenspaces

$$v_1=(1,0,1), v_2=(-1,-2,1), v_3=(-1,1,1)$$

(3)

Eigenvalues

$$\lambda_1 = 3i, \lambda_2 = -3i, \lambda_3 = 0$$

A basis of the corresponding eigenspaces

$$v_1=(1-3i,1+3i,4), v_2=(1+3i,1-3i,4), v_3=(-2,-2,1)$$

(4)

Let $\lambda$ be an eigenvalue of $A$ and let $\mathbf{x}$ be an eigenvector corresponding to the eigenvalue $\lambda$. That is, we have

$$A\mathbf{x}=\lambda \mathbf{x}.$$

Multiplying by $\bar{\mathbf{x}}^{T}$ from the left, we have

$$\begin{align} \bar{\mathbf{x}}^{T}A\mathbf{x}=\lambda \bar{\mathbf{x}}^{T} \mathbf{x}=\lambda ||\mathbf{x}||^2. \tag{*} \end{align}$$

Note that the left hand side $\bar{\mathbf{x}}^{T}A\mathbf{x}$ is the dot (inner) product of $\bar{\mathbf{x}}$ and $A\mathbf{x}$. Since the dot product is commutative, we have

$$\begin{aligned} &\text{The left hand side of (*)}\\ &=\bar{\mathbf{x}}^{T}A\mathbf{x}=(A\mathbf{x})^{T}\bar{\mathbf{x}}\\ &=x^{T}A^{T}\bar{\mathbf{x}}. \end{aligned}$$

Since $A$ is skew-symmetric, we have $A^{T}=-A$. Substituting this into the above equality, we have

$$\begin{aligned} &\text{The left hand side of (*)}\\ &=x^{T}A^{T}\bar{\mathbf{x}}=-\mathbf{x}^{T}A\bar{\mathbf{x}} \end{aligned}$$

Taking conjugate of $A\mathbf{x}=\lambda\mathbf{x}$ and use the fact that $A$ is real, we have

$$A\bar{\mathbf{x}}=\bar{\lambda}\bar{\mathbf{x}}.$$

Thus, we have

$$\begin{aligned} &\text{The left hand side of (*)}\\ &=-\mathbf{x}^{T}A\bar{\mathbf{x}}\\ &=-\mathbf{x}^{T}\bar{\lambda}\bar{\mathbf{x}}=-\bar{\lambda}||\mathbf{x}||^2. \end{aligned}$$

Therefore comparing the left and right hand sides of (*) yields

$$-\bar{\lambda}||\mathbf{x}||^2=\lambda ||\mathbf{x}||^2.$$

Since $\mathbf{x}$ is an eigenvector, it is nonzero by definition. Thus $||\mathbf{x}||\neq 0$.

Hence we have

$$-\bar{\lambda}=\lambda,$$

and this implies that $\lambda$ is either $0$ or purely imaginary number.