在讨论特征值、特征向量 对角化
从这个例子可以看出，对于一些特殊的矩阵，它们的幂可以通过以下形式方便地计算出来： P D P − 1 {\displaystyle PDP^{-1}} P {\displaystyle P} D {\displaystyle D}
当然，给定一个矩阵，我们会想知道它是否可以表示为 P D P − 1 {\displaystyle PDP^{-1}} P {\displaystyle P} D {\displaystyle D}
鉴于动机部分，我们有以下定义。
练习。
以下是与对角化在某种程度上相关的重要的通用概念。
练习。
以下定理将可对角化矩阵与特征向量和特征值联系起来。
定理。 (对角化) 设 A {\displaystyle A} n × n {\displaystyle n\times n} A {\displaystyle A} 可对角化 A {\displaystyle A} n {\displaystyle n} 线性无关特征向量 v 1 , … , v n {\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}} A {\displaystyle A} 线性无关特征向量 特征值 λ 1 , … , λ n {\displaystyle \lambda _{1},\ldots ,\lambda _{n}} 定义 可逆 P {\displaystyle P} 列 v 1 , … , v n {\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}} 对角 D {\displaystyle D} 对角元素 λ 1 , … , λ n {\displaystyle \lambda _{1},\ldots ,\lambda _{n}} A = P D P − 1 . {\displaystyle A=PDP^{-1}.}
证明。 以下我们使用 ( v 1 ⋯ v n ) {\displaystyle {\begin{pmatrix}\mathbf {v} _{1}&\cdots &\mathbf {v} _{n}\end{pmatrix}}} v 1 , … , v n {\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}} A = P D P − 1 ⇔ A P = P D P P − 1 ⏟ I ⇔ A ( v 1 ⋯ v n ) = ( v 1 ⋯ v n ) ( λ 1 0 ⋯ 0 0 λ 2 ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ λ n ) ⇔ ( A v 1 ⋯ A v n ) = ( λ 1 v 1 ⋯ λ n v n ) ⇔ A v 1 = λ 1 v 1 , … , A v n = λ n v n . {\displaystyle {\begin{aligned}&&A&=PDP^{-1}\\&\Leftrightarrow &AP&=PD\underbrace {PP^{-1}} _{I}\\&\Leftrightarrow &A{\begin{pmatrix}\mathbf {v} _{1}&\cdots &\mathbf {v} _{n}\end{pmatrix}}&={\begin{pmatrix}\mathbf {v} _{1}&\cdots &\mathbf {v} _{n}\end{pmatrix}}{\begin{pmatrix}\lambda _{1}&0&\cdots &0\\0&\lambda _{2}&\cdots &0\\\vdots &\vdots &\ddots &\vdots \\0&0&\cdots &\lambda _{n}\end{pmatrix}}\\&\Leftrightarrow &{\begin{pmatrix}A\mathbf {v} _{1}&\cdots &A\mathbf {v} _{n}\end{pmatrix}}&={\begin{pmatrix}\lambda _{1}\mathbf {v} _{1}&\cdots &\lambda _{n}\mathbf {v} _{n}\end{pmatrix}}\\&\Leftrightarrow &A\mathbf {v} _{1}&=\lambda _{1}\mathbf {v} _{1},\ldots ,A\mathbf {v} _{n}=\lambda _{n}\mathbf {v} _{n}.\end{aligned}}} v 1 , … , v n {\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}} 特征向量 线性无关 线性无关 P {\displaystyle P} ◻ {\displaystyle \Box }
接下来，我们将介绍一种方便的方法来找到 特征值
示例. ( 1 2 4 4 5 2 0 0 9 ) {\displaystyle {\begin{pmatrix}1&2&4\\4&5&2\\0&0&9\\\end{pmatrix}}} 特征多项式 | 1 − t 2 4 4 5 − t 2 0 0 9 − t | = ( 9 − t ) | 1 − t 2 4 5 − t | = ( 9 − t ) [ ( 1 − t ) ( 5 − t ) − 2 ( 4 ) ] = − x 3 + 15 x 2 − 51 x − 27. {\displaystyle {\begin{vmatrix}1-t&2&4\\4&5-t&2\\0&0&9-t\\\end{vmatrix}}=(9-t){\begin{vmatrix}1-t&2\\4&5-t\\\end{vmatrix}}=(9-t)[(1-t)(5-t)-2(4)]=-x^{3}+15x^{2}-51x-27.}
证明. λ is an eigenvalue of A ⇔ A v = λ v for some v ≠ 0 ⇔ ( A − λ I n ) v = 0 for some v ≠ 0 ⇔ A − λ I n is non-invertible by simplified invertible matrix theorem ⇔ det ( A − λ I n ) = 0 {\displaystyle {\begin{aligned}&&\lambda {\text{ is an }}&{\text{eigenvalue of }}A\\&\Leftrightarrow &A\mathbf {v} &=\lambda \mathbf {v} \qquad {\text{for some }}\mathbf {v} \neq \mathbf {0} \\&\Leftrightarrow &(A-\lambda I_{n})\mathbf {v} &=\mathbf {0} \qquad {\text{for some }}\mathbf {v} \neq \mathbf {0} \\&\Leftrightarrow &A-\lambda I_{n}&{\text{ is non-invertible}}\qquad {\text{by simplified invertible matrix theorem}}\\&\Leftrightarrow &\det(A-\lambda I_{n})&=0\end{aligned}}}
◻ {\displaystyle \Box }
接下来，我们将介绍一个与特征向量
练习。
在介绍了这些术语和概念之后，我们有以下用于对 n × n {\displaystyle n\times n}
通过求解 det ( A − λ I ) = 0 {\displaystyle \det(A-\lambda I)=0} A {\displaystyle A} 特征值
对于矩阵 A {\displaystyle A} λ 1 , … , λ k {\displaystyle \lambda _{1},\ldots ,\lambda _{k}} 特征空间 E λ 1 , … , E λ k {\displaystyle E_{\lambda _{1}},\ldots ,E_{\lambda _{k}}} 基 β 1 , … , β k {\displaystyle \beta _{1},\ldots ,\beta _{k}}
如果 β 1 , … , β k {\displaystyle \beta _{1},\ldots ,\beta _{k}} n {\displaystyle n} v 1 , … , v n {\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}} A {\displaystyle A} P = ( v 1 ⋯ v n ) {\displaystyle P={\begin{pmatrix}\mathbf {v} _{1}&\cdots &\mathbf {v} _{n}\end{pmatrix}}}
我们有 A = P D P − 1 {\displaystyle A=PDP^{-1}} D {\displaystyle D} 对角 对角 v 1 , … , v n {\displaystyle \mathbf {v} _{1},\ldots ,\mathbf {v} _{n}} 特征值
例子。 （ 2 × 2 {\displaystyle 2\times 2} A = ( 19 − 8 48 − 21 ) {\displaystyle A={\begin{pmatrix}19&-8\\48&-21\\\end{pmatrix}}} P D P − 1 {\displaystyle PDP^{-1}}
首先， | 19 − λ − 8 48 − 21 − λ | = 0 ⟺ ( 19 − λ ) ( − 21 − λ ) + 8 ( 48 ) = 0 ⟹ λ 2 + 2 λ − 15 = 0 ⟺ λ = 3 or λ = − 5. {\displaystyle {\begin{vmatrix}19-\lambda &-8\\48&-21-\lambda \end{vmatrix}}=0\iff (19-\lambda )(-21-\lambda )+8(48)=0\implies \lambda ^{2}+2\lambda -15=0\iff \lambda =3{\text{ or }}\lambda =-5.} λ = 3 {\displaystyle \lambda =3} λ = − 5 {\displaystyle \lambda =-5}
对于特征值 λ = 3 {\displaystyle \lambda ={\color {green}3}} ( A − 3 I ) x = 0 ⟹ ( 16 − 8 48 − 24 ) x = 0 {\displaystyle (A-3I)\mathbf {x} =\mathbf {0} \implies {\begin{pmatrix}16&-8\\48&-24\\\end{pmatrix}}\mathbf {x} =\mathbf {0} } x = ( a , 2 a ) T {\displaystyle \mathbf {x} =(a,2a)^{T}} E 3 {\displaystyle E_{3}} { ( 1 , 2 ) T } {\displaystyle \{{\color {green}(1,2)^{T}}\}}
对于特征值 λ = − 5 {\displaystyle \lambda ={\color {blue}-5}} ( A + 5 I ) x = 0 ⟹ ( 24 − 8 48 − 16 ) x = 0 {\displaystyle (A+5I)\mathbf {x} =\mathbf {0} \implies {\begin{pmatrix}24&-8\\48&-16\\\end{pmatrix}}\mathbf {x} =\mathbf {0} } x = ( b , 3 b ) T {\displaystyle \mathbf {x} =(b,3b)^{T}} E − 5 {\displaystyle E_{-5}} { ( 1 , 3 ) T } {\displaystyle \{{\color {blue}(1,3)^{T}}\}}
然后，我们令 P = ( 1 1 2 3 ) {\displaystyle P={\begin{pmatrix}{\color {green}1}&{\color {blue}1}\\{\color {green}2}&{\color {blue}3}\\\end{pmatrix}}} D = ( 3 0 0 − 5 ) . {\displaystyle D={\begin{pmatrix}{\color {green}3}&0\\0&{\color {blue}-5}\end{pmatrix}}.} P − 1 = ( 3 − 1 − 2 1 ) . {\displaystyle P^{-1}={\begin{pmatrix}3&-1\\-2&1\\\end{pmatrix}}.} A = P D P − 1 = ( 1 1 2 3 ) ( 3 0 0 − 5 ) ( 3 − 1 − 2 1 ) , {\displaystyle A=PDP^{-1}={\begin{pmatrix}{\color {green}1}&{\color {blue}1}\\{\color {green}2}&{\color {blue}3}\\\end{pmatrix}}{\begin{pmatrix}{\color {green}3}&0\\0&{\color {blue}-5}\end{pmatrix}}{\begin{pmatrix}3&-1\\-2&1\\\end{pmatrix}},} A = P D P − 1 {\displaystyle A=PDP^{-1}} A n = P D n P − 1 , {\displaystyle A^{n}=PD^{n}P^{-1},} A n = ( 3 n + 1 − 2 ( − 5 ) n ( − 5 ) n − 3 n 6 ( 3 n ) − 6 ( − 5 ) n 3 ( − 5 ) n − 2 ( 3 n ) ) . {\displaystyle A^{n}={\begin{pmatrix}3^{n+1}-2(-5)^{n}&(-5)^{n}-3^{n}\\6(3^{n})-6(-5)^{n}&3(-5)^{n}-2(3^{n})\\\end{pmatrix}}.}
示例： （ 3 × 3 {\displaystyle 3\times 3} A = ( 0 0 1 0 1 0 1 0 0 ) {\displaystyle A={\begin{pmatrix}0&0&1\\0&1&0\\1&0&0\\\end{pmatrix}}} 不是 I 3 {\displaystyle I_{3}} A n {\displaystyle A^{n}} | − λ 0 1 0 1 − λ 0 1 0 − λ | = 0 ⟹ λ 2 ( 1 − λ ) − ( 1 − λ ) = 0 ⟹ ( 1 − λ ) ( λ 2 − 1 ) = 0 ⟹ λ = 1 (repeated) or λ = − 1. {\displaystyle {\begin{vmatrix}-\lambda &0&1\\0&1-\lambda &0\\1&0&-\lambda \end{vmatrix}}=0\implies \lambda ^{2}(1-\lambda )-(1-\lambda )=0\implies (1-\lambda )(\lambda ^{2}-1)=0\implies \lambda =1{\text{(repeated) or }}\lambda =-1.} λ = 1 {\displaystyle \lambda =1} λ = − 1 {\displaystyle \lambda =-1}
对于特征值 λ = 1 {\displaystyle \lambda ={\color {green}1}} ( A − I ) x = 0 ⟹ ( − 1 0 1 0 0 0 1 0 − 1 ) x = 0 ⟹ x = ( b , a , b ) T = b ( 1 , 0 , 1 ) T + a ( 0 , 1 , 0 ) T , {\displaystyle (A-I)\mathbf {x} =\mathbf {0} \implies {\begin{pmatrix}-1&0&1\\0&0&0\\1&0&-1\end{pmatrix}}\mathbf {x} =\mathbf {0} \implies \mathbf {x} =(b,a,b)^{T}=b(1,0,1)^{T}+a(0,1,0)^{T},} 2 {\displaystyle 2} E 1 {\displaystyle E_{1}} { ( 1 , 0 , 1 ) T , ( 0 , 1 , 0 ) T } {\displaystyle \{{\color {green}(1,0,1)^{T}},{\color {green}(0,1,0)^{T}}\}}
对于特征值 λ = − 1 {\displaystyle \lambda ={\color {blue}-1}} ( A + I ) x = 0 ⟹ ( 1 0 1 0 2 0 1 0 1 ) x = 0 ⟹ x = ( c , 0 , − c ) T , {\displaystyle (A+I)\mathbf {x} =\mathbf {0} \implies {\begin{pmatrix}1&0&1\\0&2&0\\1&0&1\end{pmatrix}}\mathbf {x} =\mathbf {0} \implies \mathbf {x} =(c,0,-c)^{T},} E − 1 {\displaystyle E_{-1}} { ( 1 , 0 , − 1 ) T } {\displaystyle \{{\color {blue}(1,0,-1)^{T}}\}}
Then, we let P = ( 1 0 1 0 1 0 1 0 − 1 ) {\displaystyle P={\begin{pmatrix}{\color {green}1}&{\color {green}0}&{\color {blue}1}\\{\color {green}0}&{\color {green}1}&{\color {blue}0}\\{\color {green}1}&{\color {green}0}&{\color {blue}-1}\end{pmatrix}}} three D = ( 1 0 0 0 1 0 0 0 − 1 ) {\displaystyle D={\begin{pmatrix}{\color {green}1}&0&0\\0&{\color {green}1}&0\\0&0&{\color {blue}-1}\end{pmatrix}}} λ = 1 {\displaystyle \lambda =1} P − 1 = ( 1 2 0 1 2 0 1 0 1 2 0 − 1 2 ) {\displaystyle P^{-1}={\begin{pmatrix}{\frac {1}{2}}&0&{\frac {1}{2}}\\0&1&0\\{\frac {1}{2}}&0&-{\frac {1}{2}}\end{pmatrix}}} A = P D P − 1 = ( 1 0 1 0 1 0 1 0 − 1 ) ( 1 0 0 0 1 0 0 0 − 1 ) ( 1 2 0 1 2 0 1 0 1 2 0 − 1 2 ) , {\displaystyle A=PDP^{-1}={\begin{pmatrix}{\color {green}1}&{\color {green}0}&{\color {blue}1}\\{\color {green}0}&{\color {green}1}&{\color {blue}0}\\{\color {green}1}&{\color {green}0}&{\color {blue}-1}\end{pmatrix}}{\begin{pmatrix}{\color {green}1}&0&0\\0&{\color {green}1}&0\\0&0&{\color {blue}-1}\end{pmatrix}}{\begin{pmatrix}{\frac {1}{2}}&0&{\frac {1}{2}}\\0&1&0\\{\frac {1}{2}}&0&-{\frac {1}{2}}\end{pmatrix}},} A n = P D n P − 1 = ( 1 0 1 0 1 0 1 0 − 1 ) ( 1 n 0 0 0 1 n 0 0 0 ( − 1 ) n ) ( 1 2 0 1 2 0 1 0 1 2 0 − 1 2 ) = ( 1 + ( − 1 ) n 2 0 1 − ( − 1 ) n 2 0 1 0 1 − ( − 1 ) n 2 0 1 + ( − 1 ) n 2 ) = { I 3 if n is even A if n is odd . {\displaystyle A^{n}=PD^{n}P^{-1}={\begin{pmatrix}{\color {green}1}&{\color {green}0}&{\color {blue}1}\\{\color {green}0}&{\color {green}1}&{\color {blue}0}\\{\color {green}1}&{\color {green}0}&{\color {blue}-1}\end{pmatrix}}{\begin{pmatrix}{\color {green}1}^{n}&0&0\\0&{\color {green}1}^{n}&0\\0&0&({\color {blue}-1})^{n}\end{pmatrix}}{\begin{pmatrix}{\frac {1}{2}}&0&{\frac {1}{2}}\\0&1&0\\{\frac {1}{2}}&0&-{\frac {1}{2}}\end{pmatrix}}={\begin{pmatrix}{\frac {1+(-1)^{n}}{2}}&0&{\frac {1-(-1)^{n}}{2}}\\0&1&0\\{\frac {1-(-1)^{n}}{2}}&0&{\frac {1+(-1)^{n}}{2}}\\\end{pmatrix}}={\begin{cases}I_{3}&{\text{if }}n{\text{ is even}}\\A&{\text{if }}n{\text{ is odd}}\end{cases}}.}
练习。
在接下来的内容中，我们将讨论对角化的一些数学应用，包括推导出序列公式以及求解 常微分方程
示例. (斐波那契数列) 考虑 斐波那契数列 F 0 , F 1 , … {\displaystyle F_{0},F_{1},\ldots } F 0 = 0 {\displaystyle F_{0}=0} F 1 = 1 {\displaystyle F_{1}=1} n {\displaystyle n} F n + 2 = F n + 1 + F n {\displaystyle F_{n+2}=F_{n+1}+F_{n}} n {\displaystyle n} ( F n + 2 F n + 1 ) = ( 1 1 1 0 ) ( F n + 1 F n ) = ( F n + 1 + F n F n ) . {\displaystyle {\begin{pmatrix}F_{n+2}\\F_{n+1}\end{pmatrix}}={\begin{pmatrix}1&1\\1&0\\\end{pmatrix}}{\begin{pmatrix}F_{n+1}\\F_{n}\end{pmatrix}}={\begin{pmatrix}F_{n+1}+F_{n}\\F_{n}\end{pmatrix}}.}
令 A = ( 1 1 1 0 ) {\displaystyle A={\begin{pmatrix}1&1\\1&0\\\end{pmatrix}}} ( F n + 1 F n ) = A ( F n F n − 1 ) = A 2 ( F n − 1 F n − 2 ) = ⋯ = A n ( F 1 F 0 ) = A n ( 1 0 ) . {\displaystyle {\begin{pmatrix}F_{n+1}\\F_{n}\end{pmatrix}}=A{\begin{pmatrix}F_{n}\\F_{n-1}\end{pmatrix}}=A^{2}{\begin{pmatrix}F_{n-1}\\F_{n-2}\\\end{pmatrix}}=\cdots =A^{n}{\begin{pmatrix}F_{1}\\F_{0}\end{pmatrix}}=A^{n}{\begin{pmatrix}1\\0\\\end{pmatrix}}.}
为了得到 F n {\displaystyle F_{n}} A n {\displaystyle A^{n}}
由于 | 1 − λ 1 1 − λ | = 0 ⟹ ( 1 − λ ) ( − λ ) − 1 = 0 ⟹ λ 2 − λ − 1 = 0 ⟹ λ = 1 + 5 2 or 1 − 5 2 . {\displaystyle {\begin{vmatrix}1-\lambda &1\\1&-\lambda \end{vmatrix}}=0\implies (1-\lambda )(-\lambda )-1=0\implies \lambda ^{2}-\lambda -1=0\implies \lambda ={\frac {1+{\sqrt {5}}}{2}}{\text{ or }}{\frac {1-{\sqrt {5}}}{2}}.} φ = 1 + 5 2 {\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}} 黄金分割 ψ = 1 − 5 2 {\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}} 黄金分割 共轭
对于特征值 λ = φ {\displaystyle \lambda ={\color {green}\varphi }} ( 1 − φ 1 1 − φ ) x = 0 , {\displaystyle {\begin{pmatrix}1-\varphi &1\\1&-\varphi \end{pmatrix}}\mathbf {x} =\mathbf {0} ,} ( 1 − φ 1 0 1 − φ 0 ) → r 1 ↔ r 2 ( 1 − φ 0 1 − φ 1 0 ) → − ( 1 − φ ) r 1 + r 2 → r 2 ( 1 − φ 0 0 1 + φ − φ 2 0 ) = ( 1 − φ 0 0 0 0 ) . {\displaystyle {\begin{pmatrix}1-\varphi &1&0\\1&-\varphi &0\\\end{pmatrix}}{\overset {\mathbf {r} _{1}\leftrightarrow \mathbf {r} _{2}}{\to }}{\begin{pmatrix}1&-\varphi &0\\1-\varphi &1&0\\\end{pmatrix}}{\overset {-(1-\varphi )\mathbf {r} _{1}+\mathbf {r} _{2}\to \mathbf {r} _{2}}{\to }}{\begin{pmatrix}1&-\varphi &0\\0&1+\varphi -\varphi ^{2}&0\\\end{pmatrix}}={\begin{pmatrix}1&-\varphi &0\\0&0&0\\\end{pmatrix}}.} [ 3] x = ( a φ , a ) T {\displaystyle \mathbf {x} =(a\varphi ,a)^{T}} E φ {\displaystyle E_{\varphi }} { ( φ , 1 ) T } {\displaystyle \{{\color {green}(\varphi ,1)^{T}}\}}
对于特征值 λ = ψ {\displaystyle \lambda ={\color {blue}\psi }} ( 1 − ψ 1 1 − ψ ) x = 0 , {\displaystyle {\begin{pmatrix}1-\psi &1\\1&-\psi \end{pmatrix}}\mathbf {x} =\mathbf {0} ,} ( 1 − ψ 0 0 0 0 ) {\displaystyle {\begin{pmatrix}1&-\psi &0\\0&0&0\\\end{pmatrix}}} [ 4] x = ( b ψ , b ) T {\displaystyle \mathbf {x} =(b\psi ,b)^{T}} E ψ {\displaystyle E_{\psi }} { ( ψ , 1 ) T } {\displaystyle \{{\color {blue}(\psi ,1)^{T}}\}}
Then, we let P = ( φ ψ 1 1 ) {\displaystyle P={\begin{pmatrix}{\color {green}\varphi }&{\color {blue}\psi }\\{\color {green}1}&{\color {blue}1}\end{pmatrix}}} D = ( φ 0 0 ψ ) {\displaystyle D={\begin{pmatrix}{\color {green}\varphi }&0\\0&{\color {blue}\psi }\end{pmatrix}}} P − 1 = 1 φ − ψ ( 1 − ψ − 1 φ ) {\displaystyle P^{-1}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}1&-\psi \\-1&\varphi \end{pmatrix}}} A = P D P − 1 {\displaystyle A=PDP^{-1}} A n = P D n P − 1 = 1 φ − ψ ( φ ψ 1 1 ) ( φ n 0 0 ψ n ) ( 1 − ψ − 1 φ ) = 1 φ − ψ ( φ n + 1 ψ n + 1 φ n ψ n ) ( 1 − ψ − 1 φ ) = 1 φ − ψ ( φ n + 1 − ψ n + 1 − ψ φ n + 1 + φ ψ n + 1 φ n − ψ n − ψ φ n + φ ψ n ) {\displaystyle A^{n}=PD^{n}P^{-1}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}{\color {green}\varphi }&{\color {blue}\psi }\\{\color {green}1}&{\color {blue}1}\end{pmatrix}}{\begin{pmatrix}{\color {green}\varphi }^{n}&0\\0&{\color {blue}\psi }^{n}\end{pmatrix}}{\begin{pmatrix}1&-\psi \\-1&\varphi \end{pmatrix}}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}\varphi ^{n+1}&\psi ^{n+1}\\\varphi ^{n}&\psi ^{n}\end{pmatrix}}{\begin{pmatrix}1&-\psi \\-1&\varphi \end{pmatrix}}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}\varphi ^{n+1}-\psi ^{n+1}&-\psi \varphi ^{n+1}+\varphi \psi ^{n+1}\\\varphi ^{n}-\psi ^{n}&-\psi \varphi ^{n}+\varphi \psi ^{n}\end{pmatrix}}} ( F n + 1 F n ) = 1 φ − ψ ( φ n + 1 − ψ n + 1 − ψ φ n + 1 + φ ψ n + 1 φ n − ψ n − ψ φ n + φ ψ n ) ( 1 0 ) = 1 φ − ψ ( φ n + 1 − ψ n + 1 φ n − ψ n ) . {\displaystyle {\begin{pmatrix}F_{n+1}\\F_{n}\end{pmatrix}}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}\varphi ^{n+1}-\psi ^{n+1}&-\psi \varphi ^{n+1}+\varphi \psi ^{n+1}\\\varphi ^{n}-\psi ^{n}&-\psi \varphi ^{n}+\varphi \psi ^{n}\end{pmatrix}}{\begin{pmatrix}1\\0\\\end{pmatrix}}={\frac {1}{\varphi -\psi }}{\begin{pmatrix}\varphi ^{n+1}-\psi ^{n+1}\\\varphi ^{n}-\psi ^{n}\end{pmatrix}}.} F n = φ n − ψ n φ − ψ {\displaystyle F_{n}={\frac {\varphi ^{n}-\psi ^{n}}{\varphi -\psi }}} φ = 1 + 5 2 {\displaystyle \varphi ={\frac {1+{\sqrt {5}}}{2}}} ψ = 1 − 5 2 {\displaystyle \psi ={\frac {1-{\sqrt {5}}}{2}}}
示例. (常微分方程组) 考虑常微分方程组 { d x d t = 2 x − 3 y d y d t = 4 x − 5 y {\displaystyle {\begin{cases}{\frac {dx}{dt}}&=2x-3y\\{\frac {dy}{dt}}&=4x-5y\\\end{cases}}} ( x , y ) = ( 1 , 2 ) {\displaystyle (x,y)=(1,2)} t = 0 {\displaystyle t=0}
Using the dot notation for differentiation, the system can be rewritten as ( x ˙ y ˙ ) = A ( x y ) {\displaystyle {\begin{pmatrix}{\dot {x}}\\{\dot {y}}\end{pmatrix}}=A{\begin{pmatrix}x\\y\end{pmatrix}}} A = ( 2 − 3 4 − 5 ) {\displaystyle A={\begin{pmatrix}2&-3\\4&-5\\\end{pmatrix}}} A = P D P − 1 ⟺ D = P − 1 A P {\displaystyle A=PDP^{-1}\iff D=P^{-1}AP} P {\displaystyle P} D {\displaystyle D} P − 1 = ( a b c d ) {\displaystyle P^{-1}={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}} a , b , c , d {\displaystyle a,b,c,d} ( u v ) = P − 1 ( x y ) = ( a b c d ) = ( a x + b y c x + d y ) {\displaystyle {\begin{pmatrix}u\\v\\\end{pmatrix}}=P^{-1}{\begin{pmatrix}x\\y\\\end{pmatrix}}={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}={\begin{pmatrix}ax+by\\cx+dy\\\end{pmatrix}}} u = a x + b y {\displaystyle u=ax+by} v = c x + d y {\displaystyle v=cx+dy} ( x y ) = P ( u v ) {\displaystyle {\begin{pmatrix}x\\y\\\end{pmatrix}}=P{\begin{pmatrix}u\\v\end{pmatrix}}} u ˙ = a x ˙ + b y ˙ {\displaystyle {\dot {u}}=a{\dot {x}}+b{\dot {y}}} v ˙ = c x ˙ + d y ˙ {\displaystyle {\dot {v}}=c{\dot {x}}+d{\dot {y}}} ( u ˙ v ˙ ) = ( a b c d ) ( x ˙ y ˙ ) = P − 1 ( x ˙ y ˙ ) = P − 1 A ( x y ) = P − 1 A P ( u v ) = D ( u v ) . {\displaystyle {\begin{pmatrix}{\dot {u}}\\{\dot {v}}\end{pmatrix}}={\begin{pmatrix}a&b\\c&d\\\end{pmatrix}}{\begin{pmatrix}{\dot {x}}\\{\dot {y}}\end{pmatrix}}=P^{-1}{\begin{pmatrix}{\dot {x}}\\{\dot {y}}\end{pmatrix}}=P^{-1}A{\begin{pmatrix}x\\y\\\end{pmatrix}}=P^{-1}AP{\begin{pmatrix}u\\v\\\end{pmatrix}}=D{\begin{pmatrix}u\\v\end{pmatrix}}.} D = ( d 1 0 0 d 2 ) {\displaystyle D={\begin{pmatrix}d_{1}&0\\0&d_{2}\end{pmatrix}}} { d u d t = d 1 u d v d t = d 2 v ⟹ { 1 u d u = d 1 d t 1 v d v = d 2 d t ⟹ { ∫ 1 u d u = ∫ d 1 d t ∫ 1 v d v = ∫ d 2 d t ⟹ { ln | u | = d 1 t + C 1 ln | v | = d 2 t + C 2 ⟹ { u = ± e d 1 t + C 1 v = ± e d 2 t + C 2 ⟹ { u = C 3 e d 1 t v = C 4 e d 2 t {\displaystyle {\begin{cases}{\frac {du}{dt}}&=d_{1}u\\{\frac {dv}{dt}}&=d_{2}v\end{cases}}\implies {\begin{cases}{\frac {1}{u}}\,du&=d_{1}\,dt\\{\frac {1}{v}}\,dv&=d_{2}\,dt\end{cases}}\implies {\begin{cases}\int {\frac {1}{u}}\,du&=\int d_{1}\,dt\\\int {\frac {1}{v}}\,dv&=\int d_{2}\,dt\end{cases}}\implies {\begin{cases}\ln |u|&=d_{1}t+C_{1}\\\ln |v|&=d_{2}t+C_{2}\end{cases}}\implies {\begin{cases}u&=\pm e^{d_{1}t+C_{1}}\\v&=\pm e^{d_{2}t+C_{2}}\end{cases}}\implies {\begin{cases}u&=C_{3}e^{d_{1}t}\\v&=C_{4}e^{d_{2}t}\end{cases}}} C 1 , C 2 {\displaystyle C_{1},C_{2}} C 3 = ± e C 1 , C 4 = ± e C 2 {\displaystyle C_{3}=\pm e^{C_{1}},C_{4}=\pm e^{C_{2}}}
然后，我们通过对角化 A {\displaystyle A} D {\displaystyle D} | 2 − λ − 3 4 − 5 − λ | = 0 ⟹ ( 2 − λ ) ( − 5 − λ ) + 12 = 0 ⟹ λ 2 + 3 λ + 2 = 0 ⟹ λ = − 1 or λ = − 2. {\displaystyle {\begin{vmatrix}2-\lambda &-3\\4&-5-\lambda \end{vmatrix}}=0\implies (2-\lambda )(-5-\lambda )+12=0\implies \lambda ^{2}+3\lambda +2=0\implies \lambda =-1{\text{ or }}\lambda =-2.} λ = − 1 {\displaystyle \lambda =-1} ( 3 − 3 4 − 4 ) x = 0 , {\displaystyle {\begin{pmatrix}3&-3\\4&-4\end{pmatrix}}\mathbf {x} =\mathbf {0} ,} x = ( s , s ) T {\displaystyle \mathbf {x} =(s,s)^{T}} E − 1 = { ( 1 , 1 ) T } {\displaystyle E_{-1}=\{(1,1)^{T}\}}
对于特征值 λ = − 2 {\displaystyle \lambda =-2} ( 4 − 3 4 − 3 ) x = 0 , {\displaystyle {\begin{pmatrix}4&-3\\4&-3\end{pmatrix}}\mathbf {x} =\mathbf {0} ,} x = ( t , 4 t / 3 ) T {\displaystyle \mathbf {x} =(t,4t/3)^{T}} E − 2 = { ( 1 , 4 / 3 ) T } {\displaystyle E_{-2}=\{(1,4/3)^{T}\}}
然后，令 P = ( 1 1 1 4 3 ) {\displaystyle P={\begin{pmatrix}1&1\\1&{\frac {4}{3}}\end{pmatrix}}} D = ( − 1 0 0 − 2 ) {\displaystyle D={\begin{pmatrix}-1&0\\0&-2\end{pmatrix}}} P − 1 = ( 4 − 3 − 3 3 ) {\displaystyle P^{-1}={\begin{pmatrix}4&-3\\-3&3\end{pmatrix}}} A = P D P − 1 {\displaystyle A=PDP^{-1}}
因此， d 1 = − 1 {\displaystyle d_{1}=-1} d 2 = − 2 {\displaystyle d_{2}=-2} u = C 3 e − t {\displaystyle u=C_{3}e^{-t}} v = C 4 e − 2 t {\displaystyle v=C_{4}e^{-2t}} ( x , y ) = ( 1 , 2 ) {\displaystyle (x,y)=(1,2)} t = 0 {\displaystyle t=0} ( u v ) = P − 1 ( 1 2 ) = ( 4 − 3 − 3 3 ) ( 1 3 ) = ( − 2 3 ) {\displaystyle {\begin{pmatrix}u\\v\\\end{pmatrix}}=P^{-1}{\begin{pmatrix}1\\2\\\end{pmatrix}}={\begin{pmatrix}4&-3\\-3&3\end{pmatrix}}{\begin{pmatrix}1\\3\\\end{pmatrix}}={\begin{pmatrix}-2\\3\end{pmatrix}}} t = 0 {\displaystyle t=0} u = − 2 e − t {\displaystyle u=-2e^{-t}} v = 3 e − 2 t {\displaystyle v=3e^{-2t}} ( x y ) = P ( u v ) = ( 1 1 1 4 3 ) ( − 2 e − t 3 e − 2 t ) = ( 3 e − 2 t − 2 e − t 4 e − 2 t − 2 e − t ) . {\displaystyle {\begin{pmatrix}x\\y\end{pmatrix}}=P{\begin{pmatrix}u\\v\end{pmatrix}}={\begin{pmatrix}1&1\\1&{\frac {4}{3}}\end{pmatrix}}{\begin{pmatrix}-2e^{-t}\\3e^{-2t}\end{pmatrix}}={\begin{pmatrix}3e^{-2t}-2e^{-t}\\4e^{-2t}-2e^{-t}\end{pmatrix}}.} ( x , y ) = ( 3 e − 2 t − 2 e − t , 4 e − 2 t − 2 e − t ) . {\displaystyle (x,y)=(3e^{-2t}-2e^{-t},4e^{-2t}-2e^{-t}).}
练习。
↑ 但即使 A {\displaystyle A} n {\displaystyle n} A {\displaystyle A} A {\displaystyle A} n {\displaystyle n} λ {\displaystyle \lambda } λ {\displaystyle \lambda } n {\displaystyle n} n {\displaystyle n} ↑ 它是复数 1 + 2 i {\displaystyle 1+2i} ↑ 1 + φ − φ 2 = 0 {\displaystyle 1+\varphi -\varphi ^{2}=0} 1 + φ − φ = − ( φ 2 − φ − 1 ⏟ 0 ) = 0 {\displaystyle 1+\varphi -\varphi =-(\underbrace {\varphi ^{2}-\varphi -1} _{0})=0} ↑ 特别地， 1 + φ − φ 2 = 1 + ψ − ψ 2 = 0. {\displaystyle 1+\varphi -\varphi ^{2}=1+\psi -\psi ^{2}=0.} φ , ψ {\displaystyle \varphi ,\psi } λ 2 − λ − 1 = 0 {\displaystyle \lambda ^{2}-\lambda -1=0} 
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