# Linear algebra problem

1. Jul 15, 2014

### c00ter

Let V be a 3-dimensional vector space with a chosen basis α = {e1,e2,e3}. Consider the subspace W = span(u,v,w). Represent W as span(α, [v]α, [w]α), and then write
W as a solution space {X : AX = 0} for some matrix A.

2. u = e1e2 + e3
v = e1 + e3
and w = e1 + e2 + e3

3. so i got w= span([1 -1 1],[1 0 1],[1 1 1])
and through row reduction I found that the vectors in the span are not linearly independent so removing one will make it linearly independent. I removed the last one so we have
w= span([1 -1 1],[1 0 1]) which is linearly indep.

From here I am not sure how to proceed. I was told that at this point I need to "find the number of parameters needed for the solution space (i.e. W) of a hypothetical system of equations" which will give me the number of equations( I am unsure of how the #of free parameters relates to the number of equations). And I think since the space is 3-dimensional this implies I will have 3 unknowns. Knowing # of equations = # of rows and #of unknowns = # of columns for the matrix A. Then find the possible RREF's of such a matrix and see which one has the correct # of free parameters and equations then use this to find the matrix A.

Any help would be greatly appreciated.

2. Jul 16, 2014

### vela

Staff Emeritus

You're saying [1 1 1] = span([1 -1 1],[1 0 1],[1 1 1]). w and W aren't the same thing.

When you say a vector $\vec{x}$ is in W, what exactly does that mean?