For $n \in \mathbb{N}$ and $W \leq \mathbb{F}^n$, there exists a homogeneous system of linear equations whose solution space is $W$ 
For $n \in \mathbb{N}$ and $W \leq \mathbb{F}^n$, show that there exists a homogeneous system of linear equations whose solution space is $W$.

Here's my work:
Since $W \leq \mathbb{F}^n$, $k = dim(W) \leq dim(\mathbb{F}^n)$. Let's say that $\{w_1,w_2,...,w_k\}$ is a basis of $W$. Now, construct a matrix $A$ (of size $k \times n$) such that its rows are elements from the basis of $W$, stacked together. The row space of $A$ is $W$, so the row space of its row-echelon form is $W$ too. At this point, I'm stuck! I'm trying to come up with a homogeneous system with the help of $A$, though there may exist other easier ways of approaching this problem.
Could someone show me the light?
P.S. $W \leq \mathbb{F}^n$ stands for $W$ is a subspace of $\mathbb{F}^n$.
P.P.S. Isn't this equivalent to saying that $W$ is the null-space of some matrix? Can we go ahead along these lines, and construct a matrix $P$ such that $Pw = 0$ for all $w \in W$?

Extend $\{w_1,\dots,w_k\}$ to a basis $\beta=\{w_1,\dots,w_n\}$ for $\Bbb F^n$.
Now define the matrix $A$ whose first $n-k$ columns are $\{w_{k+1},\dots,w_n\}$. Fill the rest in with zeros.
Use the basis $\beta$ for the domain, and the standard basis for the range.  So, we need to multiply $A$ by the (inverse of the) transition matrix, whose columns are the elements of $\beta$.  Call that matrix $B$.
The homogeneous system corresponding to $AB^{-1}$ has solution space equal to $W$.
