I am looking at a particular full-dimensional pointed cone $C \subset R^{11}$ with $14$ generators. In matrix form, with each column being a generator, I have the matrix
\begin{pmatrix} 1 & 1 & 1 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1 & 1 & 1 & 0 & 0 & 0 &0\\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 1 & 1 & 1\\ \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & -\frac{1}{2} & 0 & 0 & 0 & 0 & \frac{1}{2} & 0 & 0 & 0 & 0\\ -\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & 0 & 0 & \frac{1}{2} & -\frac{1}{2}& 0 & 0 & 0 & 0 & 0 & 0\\ -\frac{1}{2} & \frac{1}{2} & -\frac{1}{2} & \frac{1}{2} & 0 & \frac{1}{2} & 0 & 0 & -\frac{1}{2}& 0 & 0 & 0 & 0 & 0\\ \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 & -\frac{1}{4} & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4}\\ -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 & -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4}\\ -\frac{1}{4} & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & 0 & 0 & 0 & 0 & 0 & 0 & -\frac{1}{4} & \frac{3}{4} & -\frac{1}{4} & -\frac{1}{4}\\ 0 & 0 & 0 & 0 & \frac{1}{3} & -\frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} & -\frac{1}{6} & \frac{1}{3} & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & -\frac{1}{6} & -\frac{1}{6} & \frac{1}{3} & \frac{1}{3} & -\frac{1}{6} & -\frac{1}{6} & 0 & 0 & 0 & 0\\ \end{pmatrix}
I'd like to get the inequality representation of $C$. Towards this end I am trying to determine the cone's facets. It is easy (at least I think it is easy) to see two of them: The first $10$ columns and the last $10$ columns. There have to be others because there is a result of Eggleston, Grunbaum and Klee that says that every $k$-face of a pointed cone has a complementary $k$-face. The two facets I mentioned are complementary to the extreme rays in the first four and last four columns, because the first ten columns contain all zeros in the third row and the last ten all zeros in the first row, from which it's easy to infer that they form a face. However, I can't see how to get a face out of any other set of columns that span a ten-dimensional space; i.e. I can't seem to figure out how to get complementary facets for the extreme rays in the middle six columns.
Does anyone recognize this kind of matrix? It's easy to show that these columns are a frame for $C$, hence each one (if my understanding is correct), is an extreme ray of $C$, so the result I referred to above seems applicable. I have a sneaking suspicion I'm missing something significant, but I can't see what.
Thanks in advance for any references to relevant work in this area.