It would be helpful if you quoted the original question along with my response.
The solutions to z' * x = 0 form a linear subspace of C^n.
________________________________ Eric J. Wingler (firstname.lastname@example.org) Dept. of Mathematics and Statistics Youngstown State University One University Plaza Youngstown, OH 44555-0001 330-941-1817
"junoexpress" <email@example.com> wrote in message news:firstname.lastname@example.org... > I'm not sure there is much I can make out of this. > > The problem I am having using the fact you provided is that it is not > obvious to me what it implies in Cn. > Essentially, z has 2n free parameters and the eqns > (i.a) z' * x = 1 > (i.b) z' * x = 0 i > gives me constraints on 2 of the params. > > Doing the subtraction you suggested essentially tells me that the 2n > params in z1 - z2 solve two homogeneous linear equations, which still > means that there are 2n-2 free params. > > So it seems that an algebraic approach is not going to get me anywhere. > Geometrically, I am not sure what z' * x = 0 means. If the vectors were > in Rn, I would know that z'*x=0 means that they are orthogonal, but I > am not sure that this is even what z'*x=0 means in Cn. > > Juno >