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Math Help - Basic Solutions and Extreme Points

  1. #1
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    Basic Solutions and Extreme Points

    Let \hspace{1 mm} S = \{x:x^Tx \hspace{1 mm} $\leq$1\}.\hspace{1 mm}Prove\hspace{1 mm}that\hspace{1 mm}the\hspace{1 mm}extreme\hspace{1 mm}points\hspace{1 mm}of\hspace{1 mm}S\hspace{1 mm}are\hspace{1 mm}the\hspace{1 mm}points\hspace{1 mm}on\hspace{1 mm}its\hspace{1 mm}boundary.

    So x is an extreme point of S iff it is a basic feasible solution.

    Would proof by contradiction be appropriate? Assume that the extreme points of S are not on the boundary?
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    If I have understood correctly your question, for any linear function f:S\to \mathbb{R} , f(x)=\sum_{j=1}^nc_jx_j (not all c_j\neq 0 ) we have no critical points in x^tx<1 . On the other hand, S is compact so, being f continuous on S we have absolute maximum and minimum on S. Necessarily this must occur on the boundary of S : x^tx=1 .
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