Find a basis for the subspace of $\Bbb{R}^3$ that is spanned by the vectors:
$$v_1=(1,0,0), \space v_2=(1,0,1), \space v_3=(2,0,1), \space v_4=(0,0,-1)$$
I am not sure how to solve this problem. I know that if these vectors span $\Bbb{R}^3$ then we can express them as: $$(a,b,c)= k_1(1,0,0)+k_2(1,0,1)+k_3(2,0,1)+k_4(0,0,-1)$$
How would I attempt to find a basis?
Best Answer
Notice that $v_{1} = v_{3} - v_{2}$. So $\{v_{1}, v_{4}\}$ is your basis, as you can form $v_{2}$ and $v_{3}$ from linear combinations of $v_{1}, v_{4}$.
Edit: I'll add a bit more on a basis. Note that a basis is a maximally independent set of vectors that spans the space. In $\mathbb{R}^{3}$, we would have three basis vectors. For subspaces of any vector space, we are permitted fewer vectors. A subspace is a vector space itself, so the number of basis vectors describes the dimension of the subspace.
Now in general vector spaces, we can use the determinant test to see if a set of vectors spans. The determinant test is nice, because (for a matrix $M$) $det(M) = 0$ if and only if the set of vectors in $M$ are linearly dependent. So if $det(M) \neq 0$ and $dim(M) = dim(V)$, for $V$ your vector space, then the vectors in $M$ form a basis.
In a subspace, you generally have fewer than $dim(V)$ vectors. I say generally, as a vector space is trivially a subspace of itself. So you can row-reduce a matrix to find the number of independent vectors. Or you can solve systems of linear equations yourself. If you are given vectors, these are your two options. If you are choosing your own basis without such constraints, selecting a subset of the standard basis is the way to go.