Chapter 02無料公開

# (1) 解答

2021.01.17に更新

\boldsymbol{A}の固有方程式は，固有値を\lambdaとすると

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

である．ここで

\begin{aligned} \det(\lambda\boldsymbol{I}-\boldsymbol{A})&=\begin{vmatrix} \lambda-1 & -\sqrt{2} & 0\\ -\sqrt{2} & \lambda-1 & -\sqrt{2}\\ 0 & -\sqrt{2} & \lambda-1 \end{vmatrix}\\ &=(\lambda-1)^3-4(\lambda-1)\\ &=(\lambda-1)\left[(\lambda-1)^2-4\right]\\ &=(\lambda-1)(\lambda-1-2)(\lambda-1+2)\\ &=(\lambda-3)(\lambda-1)(\lambda+1) \end{aligned}

であるから，\boldsymbol{A}の全ての固有値は

(\lambda-3)(\lambda-1)(\lambda+1)=0
\therefore \lambda=3,1,-1\tag{答}

である．それぞれに対応する，ノルムが1かつ第一要素が非負実数である固有ベクトルを\boldsymbol{v}_3,\boldsymbol{v}_1,\boldsymbol{v}_{-1}とする．
まず\boldsymbol{v}_3を求める．

\begin{aligned} 3\boldsymbol{I}-\boldsymbol{A}&=\begin{pmatrix} 2 & -\sqrt{2} & 0\\ -\sqrt{2} & 2 & -\sqrt{2}\\ 0 & -\sqrt{2} & 2 \end{pmatrix}\\ &\xrightarrow{\text{基本変形}}\begin{pmatrix} \sqrt{2} & -1 & 0\\ 0 & 1 & -\sqrt{2}\\ 0 & 0 & 0 \end{pmatrix}\\ &\left\{\begin{aligned} \sqrt{2}x-y&=0\\ y-\sqrt{2}z&=0 \end{aligned}\right. \end{aligned}
\therefore \boldsymbol{v}_3=\frac{1}{2}\begin{pmatrix}1\\\sqrt{2}\\1\end{pmatrix}\tag{答}

\begin{aligned} \boldsymbol{I}-\boldsymbol{A}&=\begin{pmatrix} 0 & -\sqrt{2} & 0\\ -\sqrt{2} & 0 & -\sqrt{2}\\ 0 & -\sqrt{2} & 0 \end{pmatrix}\\ &\xrightarrow{\text{基本変形}}\begin{pmatrix} 1 & 0 & 1\\ 0 & 1 & 0\\ 0 & 0 & 0 \end{pmatrix}\\ &\left\{\begin{aligned} x+z&=0\\ y&=0 \end{aligned}\right. \end{aligned}
\therefore \boldsymbol{v}_1=\frac{1}{\sqrt{2}}\begin{pmatrix}1\\0\\-1\end{pmatrix}\tag{答}

\begin{aligned} -\boldsymbol{I}-\boldsymbol{A}&=\begin{pmatrix} -2 & -\sqrt{2} & 0\\ -\sqrt{2} & -2 & -\sqrt{2}\\ 0 & -\sqrt{2} & -2 \end{pmatrix}\\ &\xrightarrow{\text{基本変形}}\begin{pmatrix} 1 & 0 & -1\\ 0 & 1 & \sqrt{2}\\ 0 & 0 & 0 \end{pmatrix}\\ &\left\{\begin{aligned} x-z&=0\\ y+\sqrt{2}z&=0 \end{aligned}\right. \end{aligned}
\therefore \boldsymbol{v}_{-1}=\frac{1}{2}\begin{pmatrix}1\\-\sqrt{2}\\1\end{pmatrix}\tag{答}