手动推导如下公式。
证明：
- 首先将如下矩阵对角化：
{ 1 − a a b 1 − b } \begin {Bmatrix} 1-a & a \\ b & 1-b \end {Bmatrix} {1−aba1−b}
(1)求如下矩阵的特征值：
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\begin {Bmatrix} 1-a & a \\ b & 1-b \end {Bmatrix} \begin {Bmatrix} x_1 \\x_2 \end {Bmatrix} = \lambda \begin {Bmatrix} x_1 \\x_2 \end {Bmatrix} == >
{1−aba1−b}{x1x2}=λ{x1x2}==>
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\begin {vmatrix} 1-a - \lambda& a \\ b & 1-b - \lambda \end {vmatrix} = 0 ==>
1−a−λba1−b−λ
=0==>
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(1-a- \lambda)(1-b - \lambda) - ab = 0 ==>
(1−a−λ)(1−b−λ)−ab=0==>
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\lambda^2 +(a+b-2)\lambda + (1-a-b) = 0 ==> \\ \lambda = \frac{(2-a-b) +- \sqrt{(a+b-2)^2-4(1-a-b)}}{2} = \\ \frac{(2-a-b) +- (a+b)}{2} = (1) or (1-a-b)
λ2+(a+b−2)λ+(1−a−b)=0==>λ=2(2−a−b)+−(a+b−2)2−4(1−a−b)=2(2−a−b)+−(a+b)=(1)or(1−a−b)
(2)求得正交特征向量
∣ − a a b − b ∣ ∣ x 1 x 2 ∣ = 0 = = > x 1 = 1 , x 2 = 1 \begin {vmatrix} -a & a \\ b &-b \end {vmatrix} \begin {vmatrix} x_1 \\x_2 \end {vmatrix} = 0 ==> x_1 = 1,x_2 = 1 −aba−b x1x2 =0==>x1=1,x2=1
∣ b a b a ∣ ∣ x 1 x 2 ∣ = 0 = = > x 1 = a , x 2 = − b \begin {vmatrix} b & a \\ b &a \end {vmatrix} \begin {vmatrix} x_1 \\x_2 \end {vmatrix} = 0 ==> x_1 = a,x_2 = -b bbaa x1x2 =0==>x1=a,x2=−b
也即：
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A = P^{-1} \Lambda P = \begin {Bmatrix} \frac{1}{\sqrt{2}} & \frac{a}{\sqrt{a^2+b^2}} \\\\ \frac{1}{\sqrt{2}} & \frac{-b}{\sqrt{a^2+b^2}} \end {Bmatrix} \begin {Bmatrix} 1 & 0\\\\ 0& 1 - a - b \end {Bmatrix} \begin {Bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \\ \frac{a} {\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}} \end {Bmatrix}
A=P−1ΛP=⎩
⎨
⎧2121a2+b2aa2+b2−b⎭
⎬
⎫⎩
⎨
⎧1001−a−b⎭
⎬
⎫⎩
⎨
⎧21a2+b2a21a2+b2−b⎭
⎬
⎫文章来源：https://www.toymoban.com/news/detail-636887.html
- 求矩阵的相似矩阵。上式中的 Λ \Lambda Λ为特征值在对角线上的对角矩阵。而 P − 1 P^{-1} P−1和P即为特征向量组成的矩阵和转置矩阵(正交矩阵的转置矩阵跟逆矩阵相同)。
A n = P − 1 Λ n P = { 1 2 a a 2 + b 2 1 2 − b a 2 + b 2 } { 1 0 0 ( 1 − a − b ) n } { 1 2 1 2 a a 2 + b 2 − b a 2 + b 2 } = { 1 2 + a 2 ( 1 − a − b ) 2 a 2 + b 2 1 2 + − a b ( 1 − a − b ) 2 a 2 + b 2 1 2 + − a b ( 1 − a − b ) 2 a 2 + b 2 1 2 + b 2 ( 1 − a − b ) 2 a 2 + b 2 } A^n = P^{-1} \Lambda^n P = \begin {Bmatrix} \frac{1}{\sqrt{2}} & \frac{a}{\sqrt{a^2+b^2}} \\\\ \frac{1}{\sqrt{2}} & \frac{-b}{\sqrt{a^2+b^2}} \end {Bmatrix} \begin {Bmatrix} 1 & 0\\\\ 0& (1 - a - b)^n \end {Bmatrix} \begin {Bmatrix} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \\ \frac{a} {\sqrt{a^2+b^2}} & \frac{-b}{\sqrt{a^2+b^2}} \end {Bmatrix} =\\ \\ \begin {Bmatrix} \frac{1}{2} + \frac{a^2 (1-a-b)^2}{a^2+b^2} & \frac{1}{2} + \frac{-ab (1-a-b)^2}{a^2+b^2} \\\\ \frac{1}{2} + \frac{-ab (1-a-b)^2}{a^2+b^2} & \frac{1}{2} + \frac{b^2 (1-a-b)^2}{a^2+b^2} \end {Bmatrix} An=P−1ΛnP=⎩ ⎨ ⎧2121a2+b2aa2+b2−b⎭ ⎬ ⎫⎩ ⎨ ⎧100(1−a−b)n⎭ ⎬ ⎫⎩ ⎨ ⎧21a2+b2a21a2+b2−b⎭ ⎬ ⎫=⎩ ⎨ ⎧21+a2+b2a2(1−a−b)221+a2+b2−ab(1−a−b)221+a2+b2−ab(1−a−b)221+a2+b2b2(1−a−b)2⎭ ⎬ ⎫文章来源地址https://www.toymoban.com/news/detail-636887.html
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