I have the following question: In this article
it is stated that when considering a bounded convex polytope $P=\{x \mid Ax\le b\}$ with the matrix $A$ and the vector $b$ having rational entries, then it must be that the volume of $P$ is also rational. Can anybody give me an explanation for this? I tried to establish a link to determinants, but failed.
Best regards Eukloidamos
Here is a simple explanation if you know some linear algebra and polyhedral geometry: Given $d$ vectors in $R^d$, that form a $d \times d$ matrix $A$, the volume of the parallelopiped $\{ \sum_i c_i v_i \, | \, 0 \leq c_i \leq 1\}$ spanned by the vectors is the absolute value of the determinant of $A$. The volume of the simplex $\{ \sum_i c_i v_i \, | \, c_i \geq 0, \sum_i c_i \leq 1\}$ is the absolute value of the determinant of $A$, divided by $d!$. Any convex polytope can be triangulated into simplexes (full dimensional polytopes with $d+1$ vertices, analagous to triangles in 2d) where the vertices of the simplexes are vertices of the polytope, and the union of the simplexes is the polytope, and the interiors of the simplexes are disjoint. Thus the volume of the polytope is the sum of the volume of the simplexes in the triangulation. For a given simplex, translate it so that one vertex is the origin. Then the vectors that define $A$ for the volume of the simplex are all rational, so the determinant of $A$ is rational, provided that the vertices are rational, so the volume of the simplex is rational. So your overall polytope volume is a sum of rationals and hence rational.
To see that the vertices of the polytope are all rational, note that every vertex of $P$ has the form $x = M^{-1}c$ where $M$ and $c$ are submatrices of $A$ and $b$, and the inverse of a matrix with rational entries has rational entries, which can be seen by the adjoint formula for the inverse.