令：

P = [ ε 1 , ε 2 , ε 3 ] = [ 1 1 1 0 2 − 2 − 1 1 1 ] P =[\varepsilon_1,\varepsilon_2,\varepsilon_3]= \begin{bmatrix} 1 & 1 &1\\ 0 & \sqrt{2} &-\sqrt{2}\\ -1 & 1 & 1 \end{bmatrix} P=[ε1​,ε2​,ε3​]=⎣⎡​10−1​12 ​1​1−2 ​1​⎦⎤​

根据前文提到的定理

P − 1 A P = [ − 2 0 0 0 2 2 0 0 0 0 ] P^{-1}AP = \begin{bmatrix} -\sqrt{2}&0&0\\ 0&2\sqrt{2}&0\\ 0&0&0 \end{bmatrix} P−1AP=⎣⎡​−2 ​00​022 ​0​000​⎦⎤​

A n = P [ − 2 0 0 0 2 2 0 0 0 0 ] n P − 1 = [ 1 1 1 0 2 − 2 − 1 1 1 ] [ − 2 0 0 0 2 2 0 0 0 0 ] n [ 1 1 1 0 2 − 2 − 1 1 1 ] − 1 = 1 4 [ 1 1 1 0 2 − 2 − 1 1 1 ] [ ( − 2 ) n 0 0 0 ( 2 2 ) n 0 0 0 0 ] [ 2 0 − 2 1 2 1 1 − 2 1 ] = 2 n 4 [ 2 ( − 1 ) n + 2 n 2 ⋅ 2 n ( − 1 ) n + 1 2 + 2 n 2 n 2 2 n + 1 2 n 2 ( − 1 ) n + 1 2 + 2 n 2 ⋅ 2 n ( − 1 ) n 2 + 2 n ] \begin{aligned}A^n &= P\begin{bmatrix} -\sqrt{2}&0&0\\ 0&2\sqrt{2}&0\\ 0&0&0 \end{bmatrix}^nP^{-1}\\ &=\begin{bmatrix} 1&1&1\\ 0&\sqrt{2}&-\sqrt{2}\\ -1&1&1 \end{bmatrix} \begin{bmatrix} -\sqrt{2}&0&0\\ 0&2\sqrt{2}&0\\ 0&0&0 \end{bmatrix}^{n} \begin{bmatrix} 1&1&1\\ 0&\sqrt{2}&-\sqrt{2}\\ -1&1&1 \end{bmatrix}^{-1}\\ &=\frac{1}{4} \begin{bmatrix} 1&1&1\\ 0&\sqrt{2}&-\sqrt{2}\\ -1&1&1 \end{bmatrix} \begin{bmatrix} (-\sqrt{2})^n&0&0\\ 0&(2\sqrt{2})^n&0\\ 0&0&0 \end{bmatrix} \begin{bmatrix} 2&0&-2\\ 1&\sqrt{2}&1\\ 1&-\sqrt{2}&1 \end{bmatrix}\\ & = \frac{\sqrt{2}^n}{4} \begin{bmatrix} 2(-1)^n+2^n&\sqrt{2}\cdot2^n&(-1)^{n+1}2+2^n\\ 2^n\sqrt{2}&2^{n+1}&2^n\sqrt{2}\\ (-1)^{n+1}2+2^n&\sqrt{2}\cdot 2^n&(-1)^n2+2^n \end{bmatrix} \end{aligned} An​=P⎣⎡​−2 ​00​022 ​0​000​⎦⎤​nP−1=⎣⎡​10−1​12 ​1​1−2 ​1​⎦⎤​⎣⎡​−2 ​00​022 ​0​000​⎦⎤​n⎣⎡​10−1​12 ​1​1−2 ​1​⎦⎤​−1=41​⎣⎡​10−1​12 ​1​1−2 ​1​⎦⎤​⎣⎡​(−2 ​)n00​0(22 ​)n0​000​⎦⎤​⎣⎡​211​02 ​−2 ​​−211​⎦⎤​=42 ​n​⎣⎡​2(−1)n+2n2n2 ​(−1)n+12+2n​2 ​⋅2n2n+12 ​⋅2n​(−1)n+12+2n2n2 ​(−1)n2+2n​⎦⎤​​