Basis of a subspace of $\mathbb{R}^5$ spanned by vectors $(1,3,-1,2,3)$, $(2,7,-2,5,2),$ and $(1,4,-1,3,-1)$. 
Let $W$ be a subspace of $\mathbb{R}^5$ spanned by vectors $(1,3,-1,2,3)$, $(2,7,-2,5,2),$ and $(1,4,-1,3,-1)$.
How can I find a basis for $W$ and what is the dimension of $W$?

I know I will create a matrix by using vectors as columns of matrix and reduce it to row echelon form but I don't know what to do after than.
And also, how can I find a basis for complement of $W$ and what will be the relationship between basis of $W$ and its complement?
 A: You may first check if the three vectors are linearly independent, if this is the case then they're the basis themselves, if one of them is made up from two others then remove that one, the other two will still span the same space as the removed one is a linear combination of these two, you can remove one by one as long as you can to reach a linearly independent set which spans the whole subspace, this is the basis, and the dimension of the subspace is the number of elements of basis. In this case the third is really a linear combination of two others:
$-1 (1,3,-1,2,3) + (2,7,-2,5,2) = (1,4,-1,3,-1)$
but the first and second one are linearly independent, So the basis is
$\{ (1,3,-1,2,3), (2,7,-2,5,2)\}$
and the dimension is $2$. note that complement of $W$ is not a vector space as it does not contain the $0$, adding the $0$ also may not make it a subspace. Let's think of $\mathbb{R}^2$ for easier imagination, assume $W$ be the $x=y$ line, is complement of $W$ an interesting set? Not at all. So speaking about a vector subspace complement usually doesn't make sense, instead it's better to think about Orthogonal complement, you may find more details here.
