Show that every subspace of $\mathbb{R}^n$ is a kernel of a linear map. Let $S$ be a subspace of $\mathbb R^n$. Show that there is an $n \times n$ matrix $A$ such that 
$$S= \{x \in \mathbb R^n : Ax=0\}.$$
How to proceed?
 A: Let $(e_1,\ldots,e_p)$ a basis for $S$ and we complete this to a basis $(e_1,\ldots,e_p,e_{p+1},\ldots,e_n)$ for $\mathbb{R}^n$.
We define the endomorphism $f$ of $\mathbb{R}^n$ by
$$f(e_i)=0 \quad i=1,\ldots,p\quad \mathrm{and}\quad f(e_i)=e_i\quad i=p+1,\ldots,n$$
then the matrix of $A$ on this basis verify the desired result.
A: Imagine a matrix whose rows form a basis for $S$. You probably know that you can find a basis $B$ for the nullspace of this matrix. Now form a matrix whose rows are the elements of $B$. Fill this matrix out to $n\times n$ by adding rows of all zeros, if necessary. Voila! you have your matrix $A$. 
A: Find an orthnormal basis for $S$, call it $v_1,..., v_m$ assume $1<m<n$, else take the 0 matrix or the identity marix. (assume also that $v_i$ are colum vectors
Then extend this to a basis for $\mathbb{R}^n$, say $v_1,...,v_m,u_1,...u_{n-m}$
Then the matrix given by $(u_1^T;... u_{n-m}^T; u_{n-m}^T...; u_{n-m}^T)$
; means stating a new row. so this is $n$ row vectors.
where $u_{n-m}^T$ apppears $m+1$ times. Do you see why this matrix must have kernel $S$?
A: You can define a translation as $T$ from basis of $R^n$ to $S$ such that consider same base for $S$ and $R^n$ then if you define $T$ as it goes every same base to 0  then you will find $A$ matrix.
you are arbitrary  that your $T$ maps basis of $(R^{n}-S ) to (R^n-s)$ how you like it!
then you will have alternative $T$
A: Do you know much about linear transformations? If so, you may know that $T : \mathbb{R}^n \to \mathbb{R}^n$, where $T(u)$ is the component of $u$ orthogonal to $S$, is a linear transformation. Then $\ker T = S$, so by taking $A$ to be the standard matrix of $T$, we have $$S = \ker T = \{x \in \mathbb{R}^n : T(x) = 0\} = \{x \in \mathbb{R}^n : Ax = 0\}.$$
A: We have that $S\oplus S^\perp= \mathbb{R}^n$. Here $S^\perp=\{x\in\mathbb{R}^n : \langle s,x\rangle=0 \,\forall s\in S  \}$ is a linear subspace of $\mathbb{R}^n$.  Set the linear operator 
$
T:\mathbb{R}^n\to \mathbb{R}^n 
$
whit fellow $T(s)=0\;\forall s\in S$ and $T(x)=x\;\forall x\in S^\perp$. Then $A$ is the matrix of $T$ in canonical base of $\mathbb{R}^n$. 
Update.
To fix ideas we write 
$$
e_1=
\left[
\begin{array}{c}
1 \\
0\\
\vdots\\
0\\
\end{array}
\right]
\quad 
e_2=
\left[
\begin{array}{c}
0 \\
1\\
\vdots\\
0\\
\end{array}
\right]
\; \ldots \;
e_n=
\left[
\begin{array}{c}
0 \\
0\\
\vdots\\
1\\
\end{array}
\right]
$$
For a construtive answer let $d=\dim(S)$ and  $e_{i_1},\ldots e_{i_d}$ the vectors of canonical basis $\{e_1,\ldots e_n\}$ such that $S=\mbox{Span}\{e_{k_1},\ldots e_{k_d}\}$.
Analogously, we have  $n-d=\dim(S)$ and  $e_{l_1},\ldots e_{l_{(n-d)}}$ the vectors of canonical basis $\{e_1,\ldots e_n\}$ such that $S=\mbox{Span}\{e_{l_1},\ldots e_{l_{(n-d)}}\}$. Define,
$$
\lambda_i=
\left\{
\begin{array}{ll}
0 & \mbox{ if } i\in\{k_1,\ldots,k_d\}\\
1 & \mbox{ if } i\in\{l_1,\ldots,l_{(n-d)}\}
\end{array}
\right.
$$
Then $x\in S$ implies $x= (1-\lambda_1)\cdot x_1\cdot e_1+\ldots +(1-\lambda_n)\cdot x_n\cdot e_n $ and 
$$
A=
\left[
\begin{array}{ccccc}
\lambda_1&\cdots&0&\cdots &0 \\
\vdots & \ddots &\vdots &\quad &\vdots\\
0&\cdots&\lambda_i&\cdots &0 \\
\vdots & \quad &\vdots &\ddots &\vdots\\
0&\cdots&0&\cdots &\lambda_n \\
\end{array}
\right]
$$
