Prove One but not both of these systems is consistent.

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SUMMARY

This discussion focuses on proving the consistency of two systems related to an m x n matrix A: (Sy 1) defined as Ax = 0, x ≥ 0, x ≠ 0, and (Sy 2) defined as A^Ty > 0. The user successfully demonstrated that if (Sy 1) is inconsistent, then (Sy 2) must be inconsistent as well, using the Separation Theorem. However, they are seeking assistance in proving the converse, that if (Sy 1) is consistent, then (Sy 2) must also be consistent.

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  • Understanding of linear algebra concepts, particularly matrix equations.
  • Familiarity with the Separation Theorem in convex analysis.
  • Knowledge of the properties of the kernel of a matrix.
  • Basic proficiency in mathematical proofs and logical reasoning.
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  • Study the implications of the Separation Theorem in greater detail.
  • Research the properties of the kernel of a matrix and its relationship to system consistency.
  • Explore techniques for proving implications in mathematical logic, particularly contrapositive proofs.
  • Examine examples of consistent and inconsistent systems in linear algebra to solidify understanding.
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Mathematicians, students of linear algebra, and anyone involved in theoretical computer science or optimization who seeks to understand the consistency of linear systems.

Kevin_H
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Given the two systems below for an ##m \times n## matrix ##A##:

  • (Sy 1): ##Ax = 0, x \geq 0, x \neq 0##
  • (Sy 2): ##A^Ty > 0 ##

I seek to prove: (Sy 1) is consistent ##\Leftrightarrow## (Sy 2) is inconsistent.

I figured out how to prove Q ##\Rightarrow ## P by proving the contrapositive (##\neg##P ##\Rightarrow## ##\neg## Q). Sketch Proof for ##\neg##P ##\Rightarrow## ##\neg## Q: Let (Sy 1) be inconsistent, then ##0 \notin K = \{Ax: x \geq 0, x \neq 0\}##.
>Separation Theorem 1 states: Let ##X## be a closed convex set in ##\mathbb{R}^n##. If ##p \notin X##, then for some ##p \neq 0## we have: ##\langle v, p \rangle < \inf_{x \in X}\langle v, x \rangle##. Since ##K## is a closed convex set, then by Separation Theorem 1, ##\exists y## such that
\begin{eqnarray}
\langle y, 0 \rangle = 0 & < & \inf_{x \geq 0, x \neq 0}\langle y, Ax \rangle
\end{eqnarray}

Since ##x \geq 0, x\neq 0##, then ##\inf \langle y, Ax, \rangle = y^TAx = (A^Ty)^Tx > 0##. We must now verify that ##A^Ty > 0 ##.

If not true, then for some index ##i##, ##(A^Ty)_i \leq 0##. Pick ##x \in \mathbb{R}^n## to be the vector with coordinate ##x_j = \lambda \delta_{ij}##. Then:
\begin{eqnarray}
\langle y, Ax\rangle = \sum_{j = 1}^{n}(A^Ty)_jx_j = \lambda(A^Ty)_{i}
\end{eqnarray}
As ##\lambda \rightarrow \infty##, then ##\lambda(A^Ty)_{i} \rightarrow -\infty## if ##(A^Ty)_i < 0## or ##\lambda(A^Ty)_{i} = 0## if ##(A^Ty)_i = 0##.

Thus ##\langle y, Ax \rangle \leq 0 = \langle y, 0\rangle##. This is a contradiction, thus ##A^Ty > 0##. Q.E.D.Now I seek to prove P ##\Rightarrow ## Q. However, I am stuck. I tried to mimic something similar to my proof for ##\neg##P ##\Rightarrow## ##\neg##Q.

I let (Sy 1) be consistent, which meant ##0 \notin S(A) = \{x: x\in \text{Ker}(A), x\geq 0, x \neq 0\}##. However, I did not seem to get very far. Any suggestions on how I should go about it?

Thank You for taking the time to read this. I appreciate any suggestions or feedback you give me. Take care and have a wonderful day.
 
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