I have to solve this:
Let $W$ denote the subspace of $R^5$ consisting of all the vectors having coordinates that sum to zero. The vectors
$$u_1=(2,-3,4,-5,2), \quad u_2=(-6,9,-12,15,-6), $$
$$u_3=(3,-2,7,-9,1), \quad u_4=(2,-8,2,-2,6), $$
$$u_5=(-1,1,2,1,-3), \quad u_6=(0,-3,-18,9,12), $$
$$u_7=(1,0,-2,3,-2), \quad u_8=(2,-1,1,-9,7) $$
generate $W$. Find a subset of $\{u_1,u_2,….,u_8\}$ that is a basis for $W$
What I have done is try to find which of the vectors are linear combinations of others. I put the vectors on a matrix and found that four rows became filled with zeroes thus the basis would have only four elements i.e., $dim(W=4)$ and I don't understand why?
My second question is if I can be sure that the vectors that did not become filled with zeroes are lineraly independent.
Best Answer
The dimension is 4 because you effectively 4 free variables. You can pick any 4 of the positions, that forces the fifth to be the additive inverse of the sum of those. In other words, a generic vector in the space looks like $$(x,y,z,w,-x-y-z-w)$$ The "simplest" basis is $$(1,0,0,0,-1),(0,1,0,0,-1),(0,0,1,0,-1),(0,0,0,1,-1)$$
If you row reduce and get 4 non-zero rows, yes, that guarantees those 4 are linearly independent. Any dependent vectors turn into all 0s.