Eigenvectors...

Q1.
Ae(i)=λi*e(i), i=1,2,3,...,n
v=a(1)e(1)+a(2)e(2)+...+a(n)e(n)=Σa(i)e(i)
Av=AΣa(i)e(i)=Σa(i)Ae(i)=Σa(i)λi*e(i)=Σλi*a(i)e(i)
A^2 v=A(Av)=Σλi^2*a(i)e(i)
...,A^k v=Σλi^k *a(i)e(i)=(λ1^k)*Σ[(λi/λi)^k*a(i)e(i)]
lim(k->inf) [(λi/λ1)^k]=0 (i=2,3,4,...), so
lim(k->inf) [A^k v/λ1^k]=a(1)e(1)
Q2.
A=[1 -1 0 // -1 2 -1//0 0 1]
eigenvalues of A are 1, (3+/- √5)/2
Let λ1=(3+√5)/2, λ2=1, λ3=(3-√5)/2, then |λ1|>|λ2|>|λ3|
and the coorespoinding eigenvectors are
e1=(2, -1-√5, 0)^t (transpose)
e2=(1, 0, -1)^t
e3=(2, -1+√5, 0)^t
x(0)=(50,0,10)^t=a e(1)+b e(2)+ c e(3),
(a, b, c)=(15-3√5, -10, 15+3√5)
x(r+1)=Ax(r), then x(r)=A^r x(0)
By Q1, A^12 x(0)/λ1^12 approaches to a*e(1),
thus A^12 x(0) approaches to [(3+√5)/2]^12 *(15-3√5)*(2, -1-√5, 0)^t=u
If fact, (by Excel) A^12 x(0)=(1719420, -2782080, 10)^t and
u=[(3+√5)/2]^12 *(15-3√5)* (2, -1-√5, 0)^t
=(1719419.9996, -2782080.0003,0)^t

2010-06-30 14:17:15 補充：
更正:If fact, (by Excel) A^12 x(0)=(1719410, -2782080, 10)^t

2010-07-02 22:56:27 補充：
x(r+1)=A*x(r), so, x(1)=A*x(0)
x(2)=A*x(1)=A*A*x(0)=A^2 x(0)
...
x(r)=A^r x(0)

1)Ae(i)=λi*e(i), i=1,2,3,...,n
v=a(1)e(1)+a(2)e(2)+...+a(n)e(n)=Σa(i)e(i)
Av=AΣa(i)e(i)=Σa(i)Ae(i)=Σa(i)λi*e(i)=Σλi*a(i)e(i)
A^2 v=A(Av)=Σλi^2*a(i)e(i)
...,A^k v=Σλi^k *a(i)e(i)=(λ1^k)*Σ[(λi/λi)^k*a(i)e(i)]
lim(k->inf) [(λi/λ1)^k]=0 (i=2,3,4,...), so
lim(k->inf) [A^k v/λ1^k]=a(1)e(1)
2)A=[1 -1 0 // -1 2 -1//0 0 1]
eigenvalues of A are 1, (3+/- √5)/2
Let λ1=(3+√5)/2, λ2=1, λ3=(3-√5)/2, then |λ1|>|λ2|>|λ3|
and the coorespoinding eigenvectors are
e1=(2, -1-√5, 0)^t (transpose)
e2=(1, 0, -1)^t
e3=(2, -1+√5, 0)^t
x(0)=(50,0,10)^t=a e(1)+b e(2)+ c e(3),
(a, b, c)=(15-3√5, -10, 15+3√5)
x(r+1)=Ax(r), then x(r)=A^r x(0)
By Q1, A^12 x(0)/λ1^12 approaches to a*e(1),
thus A^12 x(0) approaches to [(3+√5)/2]^12 *(15-3√5)*(2, -1-√5, 0)^t=u
If fact, (by Excel) A^12 x(0)=(1719420, -2782080, 10)^t and
u=[(3+√5)/2]^12 *(15-3√5)* (2, -1-√5, 0)^t
=(1719419.9996, -2782080.0003,0)^t

更正:If fact, (by Excel) A^12 x(0)=(1719410, -2782080, 10)^t

當然是要答問題!!

這個是特徵向量
每個特徵向量一定有一個對應的值
叫做特徵值

但是你這是要翻譯還是想要了解原因??