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I would like to solve the following infinite linear system subject to $x_i \ge 0$ that minimizes $x_3$.

enter image description here

The third column contains no additional nonzero values beyond what is shown. Though the first 7 rows are jumbled, the pattern of entries is more clear for rows 8 and beyond.

One solution to the system is

$$\mathbf{x}= \left( 1,0,1,0,1,0,1,0,1,0,1,0,\dots \right)$$

But I would like to show that there is no solution where $x_3<1$. I know that infinite matrices are generally a devilish topic, but I'm hoping that standard linear programming and the simplex method (which I am not too familiar with) would get the job done for this matrix.

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  • $\begingroup$ Are you sure that there is more than one solution to the equality? $\endgroup$
    – Dirk
    Jun 7, 2017 at 18:57
  • $\begingroup$ I believe there are an infinite number of solutions, if we don't include the inequality constraints. The matrix is in row echelon form, but with top-bottom and left-right orientation reversed. So $x_1,x_5,x_9,x_{13}...$ are all free variables. $\endgroup$ Jun 8, 2017 at 1:59

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