Hi, I had a basic linear algebra question Question #1 1. The problem statement, all variables and given/known data Find a basis for the subspace of R3 for which the components in all of the vectors sum to zero. 2. Relevant equations If u and v are in w and w is a subspace, then a*u + b*v is in w. 3. The attempt at a solution w = {v in R3 : v1 + v2 + v3 = 0} Okay, so let's say you have Ax = b, where the column space of A is the basis B, and b is a vector which is in w. I really don't know how to work with this problem beyond that. I can imagine a basis looking something like: [1, 0, 0], [0, -1/2, 0], [0, 0, 1/2] Because if you add those vectors together, all of the components sum to 0. And those are indeed linearly independent. But I don't know if those are the right basis vectors. Thanks, Al.
Call the subspace described in the problem W. If v = (x, y, z) is in W, then x+y+z=0. One equation, three unknowns => 2 parameters, so let y=s, z=t. Then we have v = (-y - z, y, z) = (-1, 1, 0)s + (-1, 0, 1)t. (Note that the condition that x+y+z=0 for each v=(x,y,z) in W is equivalent to saying that W is the perpendicular subspace of span(1, 1, 1).)
Or, slightly different approach, since v1+ v2+ v3= 0, v3= -v1- v2. Let v1= 1, v2= 0 so v3= -1. We have (1, 0, -1). Let v1= 0, v2= 1 so v3= -1. We have (0, 1, -1). Those are basis vectors. That's not the same two vectors as Unco got but there are an infinite number of different bases for this subspace.
Oh, so when he says "Find a basis", he doesn't mean find all of the bases, he just means find a single vector in the basis? So if I had another question "Find a basis for a subspace of R3 in which all vectors satisfy: (1 1 0) v = 0 Then I could just give a vector like: (-1, 1, 0) and then say that I found a basis?