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.

  • $\begingroup$ is the solution unique? $\endgroup$ – Yolbarsop Nov 25 '20 at 14:58
  • 1
    $\begingroup$ yes, it is, please, see the reference $\endgroup$ – Asatur Khurshudyan Nov 26 '20 at 15:02
  • $\begingroup$ Thank you so much for your useful information! $\endgroup$ – Yolbarsop Nov 26 '20 at 17:44
  • $\begingroup$ sure thing @Yolbarsop $\endgroup$ – Asatur Khurshudyan Nov 27 '20 at 18:12

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).

  • $\begingroup$ So the Cramer's rule is true also for infinite systems of LAE? $\endgroup$ – Asatur Khurshudyan Apr 24 '18 at 3:05
  • $\begingroup$ @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 $\endgroup$ – gt6989b Apr 24 '18 at 3:10

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