# how it remove redundancy about finding extremal rays?

Let $$A = \left( \begin{array}{ccc} -3 & -1 & 3 & 0 & 1 \\ 4 & 2 & 0 & 1 & -3 \\ 1 & 1 & 1 & 0 & -1 \end{array} \right)\\ S1 = \left( \begin{array}{ccc} 3 & 1 & 0 & 0 & 0 \\ 0 & 0 & 3 & 1 & 0 \\ 3 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 3 & 0 & 1 & 0 \end{array} \right).$$ Then $$A*S1 = \left( \begin{array}{ccc} 0 & 0 & 0 & 0 & 0 \\ 12 & -5 & 6 & -1 & 1 \\ 6 & -2 & 4 & 0 & 0 \end{array} \right).$$ Futher $$S2 = \left( \begin{array}{ccc} 5 & 1 & 0 & 0 & 0 & 0\\ 12 & 0 & 6 & 1 & 0 & 0\\ 0 & 0 & 5 & 0 & 1 & 0 \\ 0 & 12 & 0 & 0 & 6 & 1 \\ 0 & 0 & 0 & 5 & 0 & 1 \end{array} \right)$$

i do not understand an example using the following theorem about how it remove redundancy of above $S2$ to below $S2$ Ajacency, A necessary and sufficient condition for two extremal rays $x$ and $y$ of the polyhedral solution cone of $A*x = 0$ $x \geq 0$ to be adjacent, is that there exist no other exremal solutions with zeros at the same positions as the common zeros of $x$ and $y$. Call $Ixy = \{k | Xk = 0 \text{ and } Yk = 0\}$, the set of indices of the common zeros of $x$ and $y$ then $x$ and $y$ are adjacent if and only if there exist no other extreme solutions z with $Ixy \subset Iz$

$$S2 = \left( \begin{array}{ccc} 5 & 0 & 0 & 0\\ 12 & 1 & 0 & 0\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 6 & 1 \\ 0 & 5 & 0 & 1 \end{array} \right)$$

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 Are you aware of the preview and edit functionalities? There's a preview under the edit window, so you can catch serious formatting problems like the visible $\TeX$ code all over your question before you post. If you do somehow end up with such problems in your post, you can always edit it using the edit link under the question. In its current garbled form the question is likely to get less attention than it would otherwise. – joriki Jul 31 '12 at 4:41