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In general polyhedron(as a feasible region of a lpp) is nothing but intersection of finite number of half spaces. Now polyhedron in standard form is represented by intersection of hyperplanes(Ax=b) and halfspaces(x>=0). So, it should also match the 1st definition. So can a hyperplane be considered as a halfspace too?

Ani
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    No, one hyperplane defines one half-space (which side of the constraint are you on) – Johan Löfberg Sep 28 '15 at 10:29
  • Think of a plane as target space. A hyperplane is just a straight line through the origin. This is different of a halfspace. – mfl Sep 28 '15 at 11:01

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Provided that $V$ is vector space of dimension $n$ and $A \colon V \to V$ linear operator of rank $1$, the set $$P_b = \{v \in V \mid Av=b\}$$ is called a hyperplane. It divides the space $V$ into two half-spaces $$H_b = \{v \in V \mid Av=\lambda b, \lambda > 0\}$$ $$H_b' = \{v \in V \mid Av=\lambda b, \lambda < 0\}$$ in the sense that the whole space $V=P_b \cup H_b \cup H_b'$ is a disjoint union of these three sets. Half-space is not a hyperplane: the latter is affine subspace of $V$, while the former isn't.

Zoran Loncarevic
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