I do simulations, when t->infinity, X->[1/n, 1/n, ..., 1/n]
but I don't know how to prove it.
If X=[x1,x2,...,xn]' where x1+x2+...+xn=1.
If A1,A2,...,An are nXn square matrices, all elements are greater than 0 and less than 1.
The sum of all elements of A1 is n. It is also true for A2,...,An.
A1+A2+...+An=E, where E is a matrix of all 1s.
Now we update X by doing followings:
x1(t+1)=X(t)'*A1*X(t)
x2(t+1)=X(t)'*A2*X(t)
...
xn(t+1)=X(t)'*An*X(t)
when t->infinite, what is X?
Thanks.
I think you can use Banach's fixed point theorem here, although I don't have all the details worked out.
Let the map be defined by
The assumptions on the imply that satisfies , this is, is a fixed point of If you can show that is a contraction mapping, then the fixed point is unique by Banach's theorem and is the limit of the iterative sequence