column space for this vector space

Problem is find a basis and its dimension.

Let W={(a,b,c):a-3b+c=0, b-2c=0, 2b-c=0}. I know that W=Nul A and it can be set up as

[1 -3 1][a]=[0]

[0 1 -2][b]=[0]

[0 2 -1][c]=[0]

when row reduced it turns out to be

[1 0 0][a]=[0]

[0 1 0][b]=[0]

[0 0 1][c]=[0]

The Nul A={**0**} and there exists no basis for it so the dimension is 0 but what about a basis for the column space. Wouldn't its dimension be 3 and a basis be

{(1 0 0),(-3 1 2),(0 2 -1)}? I can see it might be a basis since the only vector in W is (0 0 0).

Re: column space for this vector space

A Basis for a vector space is a set of linearly independent vectors which span the vector space. Now, the vector space W only contains $\displaystyle 0_w $ . So does not have a basis and $\displaystyle Dim(W) = 0 $ .