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 $w \in W$?

I figured something out myself, so I'll post it. Let $\{w_1,w_2,...,w_k\}$ be a basis of $W$ and let's extend this set to a basis of $\mathbb{F}^n$, to obtain $\{w_1,w_2,...,w_n\}$.
Now, if we define a linear map  $T: \mathbb{F^n} \to \mathbb{F^n}$, such that $T(w_i) = 0$ for $1 \leq i \leq k$ and $T(w_j) = w_j$ for $k+1 \leq j \leq n$. As a side-note, we can see that $\rm{dim}(\rm{null}(T)) = k$ & $\rm{dim}(\rm{range}(T)) = n-k$. Consider the matrix $A$ corresponding to this linear map $T$. Clearly, $Ax = 0$ is the desired system of homogeneous equations!
It remains to verify that this construction of $A$ actually works, i.e. the solution space of $Ax = 0$ is $W$ and only $W$ - but I'll not include that here for brevity.
