I’ve got these vectors

`vecs= {{0,1,0,0,0,0,0,-1,0}, {1,-1,1,0,0,0,-1,1,-1}, {1,0,-1,1,0,-1,1,0,-1}, {1,0,-1,1,0,0,-1,0,1}, {1,0,0,-1,0,1,0,0,-1}, {1,0,0,-1,1,-1,1,-1,1}, {1,0,0,0,-1,0,0,1,0}, {-1,0,1,0,0,-1,1,0,-1}, {-1,0,1,0,0,0,-1,0,1}, {-1,1,-1,1,-1,1,0,0,-1}, {-1,1,-1,1,0,-1,1,-1,1}, {-1,1,0,-1,0,1,0,-1,1}, {-1,1,0,-1,1,-1,0,1,0}, {0,-1,0,0,1,0,0,0,-1}, {0,-1,0,1,-1,1,0,-1,1}, {0,-1,0,1,0,-1,0,1,0}, {0,-1,1,-1,0,1,-1,1,0}, {0,0,-1,0,0,0,1,0,0}} `

And I would like to find all linear dependencies with positive coefficients between them. I started with

`ns = NullSpace[Transpose[vecs]] `

which gave me

`{{2,2,-1,0,-1,0,0,0,0,0,0,0,0,0,0,0,0,3}, {2,-1,2,0,-1,0,0,0,0,0,0,0,0,0,0,0,3,0}, {2,-1,-1,0,2,0,0,0,0,0,0,0,0,0,0,3,0,0}, {1,1,1,0,1,0,0,0,0,0,0,0,3,0,3,0,0,0}, {2,-1,-1,0,-1,0,3,0,0,0,0,0,0,3,0,0,0,0}, {-1,2,2,0,-1,0,0,0,0,0,0,3,0,0,0,0,0,0}, {-1,2,-1,0,2,0,0,0,0,0,3,0,0,0,0,0,0,0}, {-1,2,-1,0,-1,3,0,0,0,3,0,0,0,0,0,0,0,0}, {-1,-1,2,0,2,0,0,0,3,0,0,0,0,0,0,0,0,0}, {-1,-1,-1,3,2,0,0,3,0,0,0,0,0,0,0,0,0,0}} `

so there is one linear dependence with nonnegative coefficients (the fourth one). To check whether there are others, I made a system of inequalities with

`ineqs = Simplify[Union[Map[# >= 0 &, Table[x[k], {k, Length[ns]}].ns]]] `

which returns

`{x[1]>=0,x[2]>=0,x[3]>=0,x[4]>=0,x[5]>=0,x[6]>=0,x[7]>=0,x[8]>=0,x[9]>=0,x[10]>=0, 2 x[1]+2 x[2]+2 x[3]+x[4]+2 x[5] >= x[6]+x[7]+x[8]+x[9]+x[10], 2 x[1]+x[4]+2 (x[6]+x[7]+x[8]) >= x[2]+x[3]+x[5]+x[9]+x[10], 2 x[2]+x[4]+2 (x[6]+x[9]) >= x[1]+x[3]+x[5]+x[7]+x[8]+x[10], 2 x[3]+x[4]+2 (x[7]+x[9]+x[10]) >= x[1]+x[2]+x[5]+x[6]+x[8]} `

but my notebook runs out of memory on both `Solve[ineqs]`

and `Reduce[ineqs]`

.

What is the proper way?