矩阵对角化给计算矩阵的幂提供了简便方法
我们将矩阵进行分解，然后在计算矩阵的幂
假设我们求解
u
k
=
A
k
u
0
boldsymbol{u}_k=A^kboldsymbol{u}_0
uk=Aku0
由于
X
X
−
1
=
I
XX^{-1}=I
XX−1=I，所以下式可以简化
图中矩阵 X X X 为矩阵 A A A 的特征向量构成的矩阵、矩阵 Λ Lambda Λ 为矩阵 A A A 的特征值构成的对角矩阵
Step 1：先求解出特征向量矩阵 X X X、特征值矩阵 Λ Lambda Λ、矩阵 X − 1 X^{-1} X−1
Step 2：求解
X
−
1
u
0
X^{-1}boldsymbol{u}_0
X−1u0，结果记作
c
boldsymbol{c}
c，即
c
=
X
−
1
u
0
boldsymbol{c}=X^{-1}boldsymbol{u}_0
c=X−1u0
实际上是将向量
u
0
boldsymbol{u}_0
u0 写为了特征向量的线性组合
Step 3：求解
Λ
k
X
−
1
u
0
Lambda^kX^{-1}boldsymbol{u}_0
ΛkX−1u0，即
Λ
k
c
Lambda^kboldsymbol{c}
Λkc
Step 3：求解
X
Λ
k
X
−
1
u
0
XLambda^kX^{-1}boldsymbol{u}_0
XΛkX−1u0，即
X
Λ
k
c
XLambda^kboldsymbol{c}
XΛkc
Step4：写出结果
例子：
求解特征向量矩阵、特征值矩阵的详细步骤请参考本人博客：特征值、特征向量、迹
Step1：
X
=
[
x
⃗
1
x
⃗
2
]
=
[
2
1
1
−
1
]
X
−
1
=
1
d
e
t
X
[
−
1
−
1
−
1
2
]
=
[
1
/
3
1
/
3
1
/
3
−
2
/
3
]
Λ
=
[
λ
1
0
0
λ
2
]
=
[
2
0
0
−
1
]
X=begin{bmatrix}vec{x}_1 & vec{x}_2end{bmatrix}=begin{bmatrix}2 & 1\ 1 & -1end{bmatrix}\ ~\ X^{-1}=frac{1}{det X}begin{bmatrix}-1 & -1\ -1 & 2end{bmatrix}= begin{bmatrix}1/3 & 1/3\ 1/3 & -2/3end{bmatrix}\ ~\ Lambda=begin{bmatrix}lambda_1 & 0\ 0 & lambda_2end{bmatrix}= begin{bmatrix}2 & 0\ 0 & -1end{bmatrix}
X=[x
1x
2]=[211−1] X−1=det X1[−1−1−12]=[1/31/31/3−2/3] Λ=[λ100λ2]=[200−1]
Step2：
c
=
X
−
1
u
0
=
[
1
/
3
1
/
3
1
/
3
−
2
/
3
]
[
1
0
]
=
[
1
/
3
1
/
3
]
=
[
c
1
c
2
]
boldsymbol{c}=X^{-1}boldsymbol{u}_0= begin{bmatrix}1/3 & 1/3\ 1/3 & -2/3end{bmatrix} begin{bmatrix}1\ 0end{bmatrix}= begin{bmatrix}1/3\ 1/3end{bmatrix}= begin{bmatrix}c_1\ c_2end{bmatrix}
c=X−1u0=[1/31/31/3−2/3][10]=[1/31/3]=[c1c2]
Step3：
Λ
k
=
[
λ
1
k
0
0
λ
2
k
]
=
[
2
k
0
0
(
−
1
)
k
]
Λ
k
c
=
[
λ
1
k
0
0
λ
2
k
]
[
c
1
c
2
]
=
[
2
k
0
0
(
−
1
)
k
]
[
1
/
3
1
/
3
]
=
[
1
3
(
2
k
)
1
3
(
−
1
)
k
]
=
1
3
[
2
k
(
−
1
)
k
]
Lambda^k=begin{bmatrix}lambda_1^k & 0\ 0 & lambda_2^kend{bmatrix}= begin{bmatrix}2^k & 0\ 0 & (-1)^kend{bmatrix}\ ~\ Lambda^kboldsymbol{c}= begin{bmatrix}lambda_1^k & 0\ 0 & lambda_2^kend{bmatrix} begin{bmatrix}c_1\ c_2end{bmatrix}= begin{bmatrix}2^k & 0\ 0 & (-1)^kend{bmatrix} begin{bmatrix}1/3\ 1/3end{bmatrix}= begin{bmatrix}frac{1}{3}(2^k)\ frac{1}{3}(-1)^kend{bmatrix}= frac{1}{3}begin{bmatrix}2^k\ (-1)^kend{bmatrix}
Λk=[λ1k00λ2k]=[2k00(−1)k] Λkc=[λ1k00λ2k][c1c2]=[2k00(−1)k][1/31/3]=[31(2k)31(−1)k]=31[2k(−1)k]
Step4：
u
k
=
X
Λ
k
c
=
[
2
1
1
−
1
]
1
3
[
2
k
(
−
1
)
k
]
=
1
3
2
k
[
2
1
]
+
1
3
(
−
1
)
k
[
1
−
1
]
=
[
F
k
+
1
F
k
]
boldsymbol{u}_k=XLambda^kboldsymbol{c}=begin{bmatrix}2 & 1\ 1 & -1end{bmatrix}frac{1}{3}begin{bmatrix}2^k\ (-1)^kend{bmatrix}= frac{1}{3}2^kbegin{bmatrix}2\ 1end{bmatrix}+ frac{1}{3}(-1)^kbegin{bmatrix}1\ -1end{bmatrix}= begin{bmatrix}F_{k+1}\ F_{k}end{bmatrix}
uk=XΛkc=[211−1]31[2k(−1)k]=312k[21]+31(−1)k[1−1]=[Fk+1Fk]
F
k
F_k
Fk通项公式如下：
F
k
=
1
3
2
k
⋅
1
+
1
3
(
−
1
)
k
⋅
(
−
1
)
=
2
k
−
(
−
1
)
k
3
F_k=frac{1}{3}2^kcdot1+frac{1}{3}(-1)^kcdot(-1)=frac{2^k-(-1)^k}{3}
Fk=312k⋅1+31(−1)k⋅(−1)=32k−(−1)k
基本思想：Follow the eigenvectors
