Is the following set a polyhedron, where $a_1, a_2 \in \mathbb{R}^{n}$? \begin{align} U = \{ X \in \mathbb{R}^{n \times n}: a^T_1Xa_1 \leq a^T_2 X a_2 \} \end{align}
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Depends on what you mean by a polyhedron. $U$ is a linear halfspace orthogonal to the vector whose $i, j$-th coordinate is $v_{ij} = (a_1)_i (a_1)_j - (a_2)_i (a_2)_j.$
Igor Rivin
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The usual definition for polyhedron in combinatorial optimization is: a polyhedron is the intersection of finitely many halfspaces of the form $P = {x \in \mathbb{R}^n : Ax \leq b }$ – AlexGuevara Apr 16 '17 at 20:45
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@AlexGuevara Wel, $1$ is finitely many... – Igor Rivin Apr 16 '17 at 20:47
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are there any other common definitions of polyhedron which may change the fact whether the expression is one or not? – AlexGuevara Apr 16 '17 at 20:48
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I also do not directly see why from the orthogonality property the $Ax \leq b$ condition follows – AlexGuevara Apr 16 '17 at 20:55
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@AlexGuevara polyhedra are sometimes assumed to be compact. As for the last comment, think about it. – Igor Rivin Apr 16 '17 at 21:03