Date: May 16, 2013 5:35 PM
Author: RGVickson@shaw.ca
Subject: Re: Linear System in Mathematica -- Solve returns only the zero<br> vector (though there are more solutions)
On Thursday, May 16, 2013 2:46:02 AM UTC-7, Dino wrote:

> 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

The same thing happens in Maple. The problem is that both Maple and Mathematica do not test the determinant, which is D = c*(2c^2 + 8c + 5)/2 and this is nonzero except for c = 0 and c = -2 +- sqrt(6)/2. When the determinant is nonzero there is only the one solution [0,0,0]. When the determinant is zero the matrix has rank 2.