Geometric series

The condition for this series converge is
lim(k→∞) A^k=0
The limit of I + A + A^2 + A^3 + A^4 + ... is
(I-A)^-1
Proof
We prove this in two parts
Part A
We show I-A is nonsingular. Suppose that I-A is singular. Then there is a non zero solution to (I-A)x=0
let call it x1
Thus Ax1=x1
Multiplying both sides by A
A^2(x1)=A(Ax1)=Ax1=x1
and in general
A^kx1=x1
Taking the limit as k→∞, we have
lim(k→∞) A^kx1=x1
which is a contradiction. Part a follows
Part b
We show I + A + A^2 + A^3 + A^4 + ... = (I-A)^-1
To do this, note that
(I-A)(I + A + A^2 + A^3 + A^4 + ...+A^k-1) =I-A^k
So
I + A + A^2 + A^3 + A^4 + ...+A^k-1=(I-A)^-1(I-A^k)
Taking the limit on k→∞, We see that the partial sums converge, and
I + A + A^2 + A^3 + A^4 + ... = (I-A)^-1


2007-03-12 13:41:58 補充：
其實我想說若lim(k→∞) A^k=0 這個條件成立則 I + A + A^2 + A^3 + A^4 + ... 一定收歛part (a) 只是想證明若果lim(k→∞) A^k=0成立則I-A 一定是 nonsingular

2007-03-14 13:10:04 補充：
因為lim(k→∞) I + A + A^2 + A^3 + A^4 + ...+A^k-1=lim(k→∞) (I-A)^-1(I-A^k)lim(k→∞) I + A + A^2 + A^3 + A^4 + ...+A^k-1=(I-A)^-1lim(k→∞)(I-A^k)lim(k→∞) I + A + A^2 + A^3 + A^4 + ...+A^k-1=(I-A)^-1SoI + A + A^2 + A^3 + A^4 + ... = (I-A)^-1

Matrix Analysis 由 Roger A. Horn, Charles R. Johnson 著作


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