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let S={x:Ax$\le$b} and $x_o$ satisfies A$x_0$<0. Show that $x_0$ cannot be an extreme point of S. The question is posted by the professor when we were learning about the chapter 2 of NONLINEAR PROGRAMMING Theory and Algorithms

DAN
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    Is there more information about this question? As it stands, $x_0$ could well be an extreme point of $S$. For example, if $A$ is the $3 \times 3$ identity matrix, and both $x_0$ and $b$ are the column vector in $\Bbb{R}^3$ consisting of all $-1$s, then $x_0$ is an extreme point in $S$, but $Ax_0 = b < 0$. – Theo Bendit Sep 28 '22 at 06:58
  • Thank you very much. But that's all about the question. – DAN Sep 28 '22 at 07:06
  • It might be a typo then. Or maybe there are hidden assumptions in whatever text provided this question. Either way, this isn't a question we can answer here (beyond my comment above). I would ask whomever set this question if there is a typo. – Theo Bendit Sep 28 '22 at 07:10
  • “That’s all about the question”. I’m guessing the question didn’t come to you from a floating piece of paper out of the sky. Where does this question come from? On a chapter in a book about matrices? Or is this just real analysis or what? Please provide more context and background as to what book this question comes from. Furthermore, you need to tell us what you have tried to solve the problem and where you got stuck, otherwise we don’t know where to help you. – Adam Rubinson Sep 28 '22 at 07:11
  • The question is posted by the professor when we were learning about the chapter 2 of NONLINEAR PROGRAMMING Theory and Algorithms . – DAN Sep 28 '22 at 07:22
  • @DAN That is information pertinent to the question, so please add it to the body of the question - not as a comment. In fact, Please add all relevant knowledge and details about the question in the body of the question... – Adam Rubinson Sep 28 '22 at 08:16
  • Thanks for your suggestion. This is the first time I post the question in Stack. – DAN Sep 28 '22 at 08:34

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$\exists x_1\in S$, $Ax_1=b$, and $x_2\in S$, $Ax_2<b$ (maybe $x_2=x_0$), then $\forall \lambda \in(0,1)$, $A(\lambda x_1+(1-\lambda)x_2)= \lambda Ax_1+(1-\lambda)Ax_2<0$, hence $x_0$ can be represented as $x_0=\lambda x_1+(1-\lambda)x_2$, and $x_1\ne x_2$, which contradicts to the definition of extreme points, i.e. $x_0$ cannot be an extreme point of S.

DAN
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