The kernel to the left of the matrix

    Nm = {{
        -1, 1, -1, 1, 1, 0},
       {2, -2, 0, 0, -1, 1 },
       {0, 0, -1, 1, 1, 0 },
       {0, 0, 1, -1, 0, -1},
       {0, 0, 0, 0, -1, 1}};

may be generated by {0, 0, 1, 1, 1},{2, 1, 0, 2, 1}, as easily checked

    {0, 0, 1, 1, 1} . Nm
    {2, 1, 0, 2, 1} . Nm
I need a script which finds a positive (nonnegative) basis  for arbitrary integer valued matrix. For now, I am working with 

    tx = Table[x[i], {i, 5}];Solve[Thread[tx . Nm == 0], tx]
and then searching with pencil for an  answer. I tried things like adding positivity conditions 

    Solve[Join[Thread[tx . Nm == 0], Thread[tx >= 0]], tx]
or 

    FindInstance[Join[Thread[tx . Nm == 0], Thread[tx > 0]], tx, 2]

but they don't seem to advance me, so I'm asking for help.  

Here's a second example, where all the methods offered are in difficulty

    ST = {{-1, 0}, {0, 1}, {1, -1}, {-2, 1}, {1, -1}}
    nu=NullSpace[ST // Transpose]

The null-space contains two positive vectors out of three
(*{{1, 1, 0, 0, 1}, {-2, -1, 0, 1, 0}, {1, 1, 1, 0, 0}}*)

The

    HermiteDecomposition[nu][[2]]
    (*{{1, 0, 0, -1, -1}, {0, 1, 0, 1, 2}, {0, 0, 1, 0, -1}}*)

discovers a third vector. The script I offered has 5 arbitrary constants instead of three, and playing around with substituting values yields another answer {0,1,2,1,0}. The problem of offering a script which works always remains open :)