# When the solution of an infinite system of linear algebraic equations exist and is bounded?

I face an infinite system of linear algebraic equations. What conditions must the matrix $\bf A$ and the vector $\bf b$ satisfy for the existence of the bounded solution of the linear system $${\bf A} {\bf x} = {\bf b}?$$

Discussions Cramer's rule and Physics forum do not help.

The answer seems to be given in the well-known monograph "Approximate methods of higher analysis" by Kantorovich and Krylov (pp. 26-27).

The infinite system $${\bf A}{\bf x} = {\bf b}$$ is called

• regular if $$\sum_{j = 1}^\infty |a_{i j}| < 1,$$
• fully regular if for a positive constant $\theta$, $$\sum_{j = 1}^\infty |a_{i j}| \leq 1 - \theta < 1.$$ Here $a_{i j}$ are the elements of the matrix ${\bf A}$.

Denote by $$c_i = 1 - \sum_{j = 1}^\infty |a_{i j}|.$$ Then, the following theorem holds.

Theorem. If the infinite system of linear equations $${\bf A}{\bf x} = {\bf b}$$ is regular, and the inequality $$|b_i| \leq K c_i$$ holds for a constant $K > 0$, then the bounded solution $$|x_i| \leq K$$ to the infinite system exists.

Remark If the infinite system is fully regular, and $$|b_i| \leq K \theta$$ holds, then the bounded solution $$|x_i| \leq K$$ to the infinite system exists.

There are only 3 possibilities when you solve linear systems: 0 solutions, 1 solution or infinite amount of solutions. This can be shown by proving that if there are 2 solutions, there must be an infinite amount of solutions -- how do you do that?

Once that is proven, the only possibility for the unique solution to exist is that $b$ is in the column space of $A$ (so at least one solution exists) and $A$ has full rank (so at most one solution exists).

• So the Cramer's rule is true also for infinite systems of LAE? – Asatur Khurshudyan Apr 24 '18 at 3:05
• @AsaturKhurshudyan I am not sure you can extend Cramer's rule to infinite systems, but you can possible define the determinant of a linear operator as a product of its eigenvalues, although I am not sure where that will lead you exactly – gt6989b Apr 24 '18 at 3:10