Linear system associated to a translation of a subspace Let $ X \subset R^n$ such that $ X = x_o + S$, where S is a subspace of $ R^n$ with dimension k. If $ m = n-k$, show that $\exists A \in R^{m \times n} $ and $b \in R^m$ such that 
$$X = \{ x \in R^n | Ax=b\}$$
My attempt
I first tried to consider the case that X is a subspace (i.e, X=S). In this case
$$X=\left\{x_1\begin{bmatrix}
s_{11}\\ 
s_{12}\\ 
.\\ 
.\\ 
.\\ 
s_{1n}
\end{bmatrix} +...+x_k\begin{bmatrix}
s_{1k}\\ 
s_{2k}\\ 
.\\ 
.\\ 
.\\ 
s_{nk}
\end{bmatrix}; x_i \in R \right\}$$
My idea is to get the last m lines of this vector and create the variables $x_{k+1},...,x_{n} such that they equal to the last (or first) m lines. In this case, X would be the system formed by this equalities. But I am not sure about it, I don't know how good or rigorous it is. Trully, I am having difficults to organize the steps.
A help would be great! I hope my idea is clear.
Tahnks in advance!
 A: Lemma: $\exists A\in\mathbb{R}^{m\times n}: S=\{x|Ax=0\}$
Proof:
Since $S$ is a subspace of $\mathbb{R}^{n}$ of dimension $k$, we can find a basis $\{v_i\}_{i=1}^k$ for $S$. 
$$
\therefore S=\text{span }\{v_i\}_{i=1}^k=\left\{x=\sum_{i=1}^k a^iv_i|a\in\mathbb{R}^{k}\right\}
$$
Note that for each $a\in\mathbb{R}^{k}$, the corresponding $x$ has coordinates $(a^i)$ in basis $\{v_i\}_{i=1}^k$. To find its coordinates in the standard basis, we change basis.
For each $i$, there is a unique $\{v_i^j\}_{j=1}^n$ such that
$$
v_i=\sum_{i=1}^n v_i^je_j
$$
The change of basis matrix $B\in\mathbb{R}^{n\times k}$ is then given by $B_{ij}=v_i^j$.
Now $S$ becomes the image of $B$. By Fundamental Theorem of Linear Algebra (unfortunate naming. sigh), the image of $B$ is the orthogonal component of the kernal of $B^T$. ie. $S=\left(\ker{B^T}\right)^\perp$. Note that $B^T$ has rank $k$ as well.
The explicit basis of $\ker{B^T}$ can be obtained by finding the RREF of $B^T$. Since $B^T$ has rank $k$, its kernal has rank $m$. $\ker{B^T}$ therefore can be considered as the image of a $n\times m$ matrix name it $C$. 
($C$ is the last $k\times m$ block of the RREF of $B$ extended to $n\times m$ by a $m\times m$ identity matrix.)
Repeat the use of Fundamental Theorem of Linear Algebra, we get that $$
\left(\ker{B^T}\right)^\perp=\left(\text{im }C\right)^\perp=\ker{C^T}$$
with $C^T\in\mathbb{R}^{m\times n}$. $C^T$ is the $A$ we want.

Proof of original statement:
Note that $X=\{x+x_0|Ax=0, x\in\mathbb{R}^n\}$
$\therefore X=\{x\in\mathbb{R}^n|A(x-x_0)=0\}=\{x|Ax=Ax_0\}$
Let $b=Ax_0$ and we are done.

Comment: The Lemma can be proven in many different ways. I just tried to re-cycle as many theorems as possible to make it easy for me to think. It is very likely that you have a different (and probably more elegant) proof that is easier for you. So try proving the Lemma yourself!
