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

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.

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

\[ \begin{aligned} \bar{\mathbf{x}}^{T}A\mathbf{x}=\lambda \bar{\mathbf{x}}^{T} \mathbf{x}=\lambda ||\mathbf{x}||^2. \tag{*} \end{aligned} \]

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.