How do I get a solution set equals to a sub space? I've four vectors that spans the $\mathbb{R}^4$ sub-space $W_1$:
$\alpha_1 = \{-1,0,1,2 \}$,
  $\alpha_2 = \{3,4,-2,5 \}$,
  $\alpha_3 = \{0,4,1,11 \}$,
  $\alpha_4 = \{1,4,0,9 \}$
And I'm asked to find an homogeneous system of linear equations such that the solution set equals $W_1$.
I found that $\alpha_1$, $\alpha_2$ and $\alpha_4$ are linearly independent, but from that point, I'm not quite sure how to continue.
Any hint is appreciated.
 A: There is an algorithm in Wikipedia, but here are some helpful ways of thinking about the problem.
The set of solutions to a system of homogenous linear equations is the null space of the corresponding matrix $A$.  Another way of thinking about $A\mathbf{x}=\mathbf{0}$ is: the dot product of any row in $A$ with $\mathbf{x}$ is zero, i.e. the rows of $A$ are a bunch of vectors, each one of which is perpendicular to every solution $\mathbf{x}$.
$W_1$ is a $3$-dimensional subspace of $\mathbb{R}^4$; the orthogonal complement of $W_1$ is the subspace of vectors perpendicular to all the vectors in $W_1$.  The rows of [the matrix you want for your system of equations whose null space is $W_1$] are a set of basis vectors for the orthogonal complement of $W_1$. Since $W_1$ is is a $3$-dim subspace, its orthogonal complement in $\mathbb{R}^4$ will be a $1$-dim subspace. So if you can find a single non-zero vector $\mathbf{v}$ perpendicular to each of the vectors in your basis for $W_1$, then you can use $\mathbf{v}$ as the row in your matrix.
In fact, what happens if you try translate "a single non-zero vector $\mathbf{v}=(v_1,v_2,v_3,v_4)$ perpendicular to each of the vectors in your basis for $W_1$" into equations?
