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?

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.
