solution to infinite dimensional matrix

Consider the equations Ax=0 where both the number of columns and rows of A are countably infinite and all entries are either 1, 0 or -1.

Is the following statement true or false?

Ax=0 has a nonnegative bounded solution (i.e., Ax=0 for some x=(x1,x2,...) with xi>=0 for all i and sum_i(xi)<infinity)

iff

Ax=0 has a nonnegative bounded solution with at most finitely many nonzeros (i.e., Ax=0 for some x=(x1,x2,...) with xi>=0 for all i and sum_i(xi)<infinity AND xi=0 for all but finitely many i).

I guess the answer is no.

I greatly appreciate any reference. I have checked a few books but can't find the answer.

Re: solution to infinite dimensional matrix

Quote:

Originally Posted by

**vivian6606** Consider the equations Ax=0 where both the number of columns and rows of A are countably infinite and all entries are either 1, 0 or -1.

Is the following statement true or false?

Ax=0 has a nonnegative bounded solution (i.e., Ax=0 for some x=(x1,x2,...) with xi>=0 for all i and sum_i(xi)<infinity)

iff

Ax=0 has a nonnegative bounded solution with at most finitely many nonzeros (i.e., Ax=0 for some x=(x1,x2,...) with xi>=0 for all i and sum_i(xi)<infinity AND xi=0 for all but finitely many i).

I guess the answer is no.

I greatly appreciate any reference. I have checked a few books but can't find the answer.

Suppose that A looks like this:

$\displaystyle A = \begin{bmatrix}1&-1&-1&-1&\ldots\\ 0&1&-1&-1&\ldots\\ 0&0&1&-1&\ldots\\ \vdots&\vdots&&\ddots&\ddots\end{bmatrix}$,

with 1s on the main diagonal, –1s everywhere above it and 0s everywhere below it. Then Ax=0, where $\displaystyle x = \bigl(\tfrac12,\tfrac14,\tfrac18,\ldots,2^{-n},\ldots\bigr).$

But if Ty = 0 with $\displaystyle y_n=0$ whenever n≥N, then by looking at the (N–1)th coordinate of Ay you can see that $\displaystyle y_{N-1}=0$, and by "backwards induction" $\displaystyle y_n=0$ for all n.

Re: solution to infinite dimensional matrix

THANK YOU SO MUCH!!!

That helps a lot.