Every subspace is the kernel of a linear map I know that every kernel of a linear map from $\mathbb R^n$ to $\mathbb R^m$ is a subspace of $\mathbb R^n$.
I am wondering if the converse is true, i.e. every subspace of $\mathbb R^n$ is the kernel of a certain linear map fromm $\mathbb R^n$ to $\mathbb R^m$.
It was for me easy to see that this is true for  $n=m=2$ and I am pretty sure that this is true in general. But I dont know how to prove this. 
 A: take a subspace $F$ of $\mathbb{R^n}$ then they exist $E$ subspace of $\mathbb{R^n}$ such that :
$$
\mathbb{R^n}=E\oplus F
$$
Take $P:\mathbb{R^n} \to \mathbb{R^n}$ the projection into $E$ then $P$ is linear and 
$$
Ker (P)=E
$$
A: MathLearner, to get such a map you need to make a choice of basis (directly or indirectly) at some point.
In general, every subspace of a (finite dimensional) vector space naturally a kernel. Let $W\subset V$. Then $W$ is the kernel of the natural projection $\pi:V\to V/W$ (the quotient space).
This is kind of 'the answer', since every map $\phi:V\to U$ ($U$ a third vector space) with $\ker \phi\subset W$ factors through $\pi$, this is not hard to see.
If $V=\mathbb{R}^n$, then you can compose $\pi$ with an isomorphism from  $V/W$ to some $\mathbb{R}^m$, giving you the desired map. (Again, every map as you wish just differs from the others by such isomorphism.) This isomorphism is, in some sense, a choice of basis.
A: Yes, this is true.
Let $W$ be a subspace of $\mathbb{R}^n$, and let $B_W=\{w_1, \ldots, w_k \}$ be a basis of $W$. Complete $B_W$ to a basis of $V$, $B_V = \{w_1, \ldots, w_k ,v_{k+1},\ldots v_n\}$.
Now define $T:\mathbb{R}^n \rightarrow \mathbb{R}^m$ as such: $\forall i, T(w_i) = 0$, and send $v_i$ to non-zero vectors. This uniquely defines a linear map which has $W$ as its kernel. 
