\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}
次に\boldsymbol{v}_1を求める．
\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}
最後に\boldsymbol{v}_{-1}を求める．
\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}
