Subspace of V is also a null space of T Prove that any subspace of vector space $V$ is a null space over some linear transformation $V \rightarrow V$.
So far I have:
Let $W$ be the subspace of $V$, let $(e_1, e_2, \ldots, e_r)$ be the basis of $W$, where $r \leq \dim(V)$.
It seems I need to find a linear transformation $T: V \rightarrow V$, so that:.
$T(s) = 0$ if $s \in W$
$T(s) \neq 0$ if $s \notin W$.
So $T(x) = 0$, if $x$ is a linear combination of $(e_1, e_2, \ldots, e_r)$ and $T(x) \neq 0$ if it's not.
How do I construct the matrix of this linear transform?
 A: You are doing well: first find a basis $(e_1,\ldots,e_r)$ for $W$.
Then, remember that every linearly independent subset of $V$ can be extended to a basis. Since $(e_1,\ldots,e_r)$ is linearly independent, we can extend it to a basis:
$(e_1,\ldots,e_r,f_1,\ldots,f_{n-r})$, where $n=\dim(V)$, $r=\dim(W)$, $r\leq n$.
Then remember that a linear transformation may be specified by saying what it does to a basis: proceeding as you did before, send each and every one of $e_1,\ldots,e_r$ to $0$. 
How do we ensure nothing else is mapped to $0$? How about mapping each $f_i$ to itself?
Suppose $a_1e_1+\cdots+a_re_r + b_1f_1+\cdots+b_{n-r}f_{n-r}$ is mapped to $0$. That means that
$$\begin{align*}
0 &= T(a_1e_1+\cdots+a_re_r+b_1f_1+\cdots+b_{n-r}f_{n-r})\\
&= a_1T(e_1)+\cdots+a_rT(e_r) + b_1T(f_1) + \cdots + b_{n-r}T(f_{n-r})\\
&= a_10 + \cdots +a_r0 + b_1f_1 + \cdots + b_{n-r}f_{n-r}\\
&= b_1f_1+\cdots+b_{n-r}f_{n-r}.
\end{align*}$$
What can we conclude about $b_1,\ldots,b_{n-r}$?
Now, it is easy to find the matrix of $T$ with respect to the basis $(e_1,\ldots,e_r,f_{1},\ldots,f_{n-r})$. What is it? Express the image of each basis vector in terms of the basis vectors, and those are the columns of the matrix of $T$.
