Solution for A*X = 0

Hello,

I want to find a solution for the following equation:

 A*X = 0

Obviously X = 0 is a solution, but I want to find a non trivial one, so I restrict X on being orthonormal (the question is not restricted to this constraint). Then I can use null(A) to get the kernel of A. However this matrix varies in size (depending on the rank of A). I would rather find a square matrix X and just minimize A*X. Any ideas on how to do this?

Normaly I have:

 A*X = B

Then I can compute

 X = (A' * A) \ (A' * B);

But that does not work for B = 0.

[Titus Edelhofer]

Sascha,

I don't yet understand why null is not what you are looking for. Yes, the size will change with the rank of A (it's size will be equal to the rank deficiency). But why is this a problem? What "other" solution do you look for?

Titus

[Sascha]

Basically this is not a math problem, but a practical one. I want to transform my vector space so that the following equation is satisfied as close as possible: S=U. So I define A = S-U. When I transform everything to zero, it will not help me at all, so I thought I would restrict it to an orthogonal transformation.