$H$ is an $n\times m$ matrix with non-negative coefficients and $n < m$. $H'$ is the transpose of $H$.


Are the following statements true?


1. If $\det(HH’) > 0$, the rows of $H$ define the edges of an $n$-dimensional conic polyhedron. No row of $H$ is a linear combination of the other rows of $H$ using non-negative coefficients.
2. If $\det(HH’) = 0$, a subset of the rows of $H$ define the edges of an $n$-dimensional conic polyhedron, but at least one row of $H$ either duplicates another row up to some scalar, lies on a face of the conic polyhedron or lies inside of the conic polyhedron. That is, at least one row is a linear combination of the other rows using non negative coefficients.


If these are true, how can I find one row (or enumerate each row) of $H$ which is a linear combination of the other rows of $H$ with non-negative coefficients when $\det(HH’) = 0$?



$\det(HH’) = 0$ for nonnegative $H$
1. Yes, because $H$ must be of rank $n$ (for $n=\operatorname{rk}(HH^\top)\leq\operatorname{rk}(H)$), so none of its rows is a linear combination of the others.
2. No, the matrix $$H = \begin{pmatrix} 1 & 1 & 1 & 0 & 1 \\ 0 & 1 & 1 & 1 & 0 \\ 1 & 1 & 1 & 1 & 0 \\ 0 & 1 & 1 & 0 & 1 \end{pmatrix}$$ provides a counter-example. The simplest linear relation between its rows with non-negative coefficients is $H\_{1,\*}+H\_{2,\*}=H\_{3,\*}+H\_{4,\*}$.


