Dino
Posts:
2
Registered:
5/16/13


Linear System in Mathematica  Solve returns only the zero vector (though there are more solutions)
Posted:
May 16, 2013 5:46 AM


Hey guys, say I want to find the null space of this matrix (which I define as 'Tbl'): {{c, 1/2, 0}, {0, 1 + c, 1}, {0, 1/2, 3 + c}}
I try this: Solve[Tbl.{x, y, z} == {0, 0, 0}, {x, y, z}] and I get this: {{x > 0, y > 0, z > 0}}
But the zero vector is not the only solution. For example if I do this: c = Sqrt[3/2]  2 then Solve[Tbl.{x, y, z} == {0, 0, 0}, {x, y, z}] returns {{x > (((2  Sqrt[6]) z)/(4 + Sqrt[6])), y > (2 + Sqrt[6]) z}} which is closer to what I want. I actually want something like that with 'c' in it. (And then I have some other equation that I'll use to find the value of c. Btw, the matrix above is only an example. I actually care to find a solution to a whole range of matrices of larger and larger sizes.)
Can anyone help? I guess I want to tell Mathematica somehow that c is some arbitrary constant and there may be more solutions, depending on the value of c.
Thanx in advance, Dino

