Thread: Quadratic forms

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 do simulations, when t->infinity, X->[1/n, 1/n, ..., 1/n]

but I don't know how to prove it.

Originally Posted by hasami2

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.

Originally Posted by hasami2

I do simulations, when t->infinity, X->[1/n, 1/n, ..., 1/n]

but I don't know how to prove it.

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

Originally Posted by hasami2

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.

Originally Posted by hasami2

I do simulations, when t->infinity, X->[1/n, 1/n, ..., 1/n]

but I don't know how to prove it.

Originally Posted by JakeD

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

In trying to prove that is a contraction map, I found that it is not and does not have a unique fixed point.
Take



Although is a fixed point of , the sequence using the starting vector converges to not