When does a system of inequalities have only the solution x = 0 ?Let A be a n × n real valued matrix, n ≥ 2. For every column n-vector x from A ⋅ x = 0 we get n equations a 1 1 x 1 + a 1 2 x 2 + ⋯ + a 1 n x n = 0 a 2 1 x 1 + a 2 2 x 2 + ⋯ + a 2 n x n = 0 … a n 1 x 1 + a n 2 x 2 + ⋯ + a n n x n = 0I know that if d e t ( A ) ≠ 0then the only solution to A ⋅ x = 0 is x = 0.Now replace all the previous equalities with inequalities in order to obtain the following system: a 1 1 x 1 + a 1 2 x 2 + ⋯ + a 1 n x n ≥ 0 a 2 1 x 1 + a 2 2 x 2 + ⋯ + a 2 n x n ≥ 0 … a n 1 x 1 + a n 2 x 2 + ⋯ + a n n x n ≥ 0Which conditions on the matrix A will guarantee that the only solution of this system of inequalities is x = 0?

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When does a system of inequalities have only the solution    x  =  0 ?Let   A be a   n  ×  n real valued matrix,   n  ≥  2. For every column n-vector   x from   A  ⋅  x  =  0 we get   n equations      a    1    1        x    1    +      a    1    2        x    2    +  ⋯  +      a    1    n        x    n    =  0        a    2    1        x    1    +      a    2    2        x    2    +  ⋯  +      a    2    n        x    n    =  0    …        a    n    1        x    1    +      a    n    2        x    2    +  ⋯  +      a    n    n        x    n    =  0I know that if   d  e  t  (  A  )  ≠  0then the only solution to   A  ⋅  x  =  0 is   x  =  0.Now replace all the previous equalities with inequalities in order to obtain the following system:      a    1    1        x    1    +      a    1    2        x    2    +  ⋯  +      a    1    n        x    n    ≥  0        a    2    1        x    1    +      a    2    2        x    2    +  ⋯  +      a    2    n        x    n    ≥  0    …        a    n    1        x    1    +      a    n    2        x    2    +  ⋯  +      a    n    n        x    n    ≥  0Which conditions on the matrix A will guarantee that the only solution of this system of inequalities is   x  =  0?

Reagan Madden · Accepted Answer

As you have said, the left hand sides can be viewed as the action of a matrix on a vector   A      x    →  .If   A has full rank, then it is invertible, then your requirement cannot be met.If   A does not have full rank, then its kernel is not trivial, then there would be other vectors being mapped to the zero vector, so your reuqirement cannot be met.This means that it is impossible.

When does a system of inequalities have only the solution x = 0 ? Let A be a n &#

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