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 $w \in W$?
 A: Consider a basis of $W$ to be the vectors $\{w_1, \ldots, w_k\}$. Now take the homogeneous system with matrix the row vectors $(w_1, \ldots, w_k)$. If you take the space of solutions of this system and find a basis for it then you got your desired matrix.
We can make this a bit more explicit. The matrix with the row vectors $(w_1, \ldots, w_k)$ has size $k \times n$.  Say it is in block form $(A,B)$ where $A$ is non-singular $k\times k$, and $B$ is $k \times n-k$.  Solving the homogeneous system given by the matrix $(A,B)$ expresses the first $k$ components in terms of the the last $n-k$ components ( invert $A$, ...). To find the basis for the space of solutions, just check that
$$[A,B]\cdot \begin{bmatrix}-A^{-1}B \\ I_{n-k} \end{bmatrix} =0_{k,n-k}$$
Therefore, we can take the transpose of the matrix $\begin{bmatrix}-A^{-1}B \\ I_{n-k} \end{bmatrix} $ and find a desired homogenous system.
A: 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.
A: 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 $W$.
A: Consider $W$ as normal subgroup of $\mathbb F^n$ (it's normal because $\mathbb F^n$ as a group is abelian). Then we can define the quotient group $\mathbb F^n/W$ with the equivalence classes $x+W$, with $x\in\mathbb F^n$, as its elements.
First, I assert that $\Bbb F^n/W$ works as a vector space over $\Bbb F$ (you can skip this part if you already know it). Given $x,y\in\Bbb F^n$ and $\alpha, \beta\in\Bbb F$:

*

*It's an abelian group for the sum defined as $(x+W)+(y+W)=(x+y)+W$ (we know this from group theory).


*We can define the scalar product $\alpha(x+W)=\alpha x+W$. For this to be well defined we have to show that if $x+W=y+W$ then $\alpha x+W=\alpha y+W$. Indeed, we have that $x-y\in W$, so there is some $w\in W$ such that $x-y=w$, and thus $\alpha(x-y)=\alpha x-\alpha y=\alpha w$. Since $W$ is a vector subspace, $\alpha w\in W$, so $\alpha x+W=\alpha y+W$.


*$\alpha(\beta(x+W))=\alpha(\beta x+W)=\alpha\beta x+W=(\alpha\beta)x+W=\alpha\beta(x+W)$.


*If we name $1$ the identity element for the product in $\Bbb F$, $1(x+W)=1x+W=x+W$.


*$\alpha((x+W)+(y+W))=\alpha((x+y)+W)=\alpha(x+y)+W=(\alpha x+W)+(\alpha y+W)=\alpha(x+W)+\alpha(y+W)$.


*$(\alpha+\beta)(x+W)=(\alpha+\beta)x+W=(\alpha x+\beta x)+W=(\alpha x+W)+(\beta x+W)$.

Now let's consider the canonical projection $\pi:\Bbb F^n\to\Bbb F^n/W$ given by $\pi(x)=x+W$. This is a linear map, since $\pi(\alpha x+\beta y)=(\alpha x +\beta y)+W=(\alpha x+W)+(\beta y+W)=\alpha(x+W)+\beta(x+W)=\alpha\pi(x)+\beta\pi(y)$.
We can use the fact that any linear transformation between two finite dimensional vector spaces can be represented by a matrix (the proof in the link uses $\Bbb R$ as field, but it can be easily generalized). In our case we know $\Bbb F^n/W$ is finite dimensional because $\pi$ is a surjective linear map (and linear maps preserve linear dependence).
Finally, we know that $\ker(\pi)=W$, so taking the matrix representation you can represent the linear map's kernel as a system of homogeneous linear equations whose solution is the nullspace of the matrix, this is, $W$.
Note: Now we know $\ker(\pi)=W$ we can use the rank-nullity theorem to get the dimension of $\mathbb F^n/W$, since $\text{Im}(\pi)=W$ (remember $\pi$ is onto): $\dim(\mathbb F^n)=\dim(\text{Im}(\pi))+\dim(\ker(\pi))=\dim(\mathbb F^n/W)+\dim(W)\Rightarrow\mbox{$\dim(\mathbb F^n/W)=\dim(\mathbb F^n)-\dim(W)$}.$
