Find a matrix such that $Ax=0$ Let
$$W = span\left\{ {\left( {\matrix{
   1  \cr 
   0  \cr 
   0  \cr 
   1  \cr 
 } } \right),\left( {\matrix{
   0  \cr 
   2  \cr 
   1  \cr 
   { - 1}  \cr 
 } } \right)} \right\}$$
I was asked to find a matrix, $A$ which satisify the following:
$$W = \left\{ {x \in {\mathbb{R}^4}|Ax = 0} \right\}$$
$A$ must be $n\times 4$. $W$ is actually $null(A)$, so every linear combination of the vectors is a solution for $Ax = 0$.
Other then those observations, I am not sure how to find $A$.
Looking for guidance. Thanks. 
 A: Here's one method.  You know that $(w,x,y,z)^T \in \ker(A)$ iff
$$
\begin{pmatrix}
w\\x\\y\\z
\end{pmatrix} = 
s \begin{pmatrix} 1 \\0 \\ 0 \\1 \end{pmatrix} + t \begin{pmatrix} 0 \\2 \\ 1 \\-1 \end{pmatrix}
= \begin{pmatrix} s \\ 2t \\ t \\ s-t \end{pmatrix}
$$
for some $s, t \in \mathbb{R}$.  There are two free variables and two linear relations:
\begin{array}{cccccc}
w & & - y & - z &=& 0\\
& x & -2y & & = &0 & \, .
\end{array}
Using these coefficients as the rows of our matrix, we find that
$$
A =
\begin{pmatrix}
1 & 0 & -1 & -1\\
0 & 1 & -2 & 0
\end{pmatrix}
$$
satisfies the criteria.
A: Suppose $\;A=(a_{ij})\;$, then we want
$$\underline x\in W\iff \underline0=A\underline x=\begin{pmatrix}\sum_{j=1}^4a_{1j}x_j\\{}\ldots\\\sum_{j=1}^4a_{4j}x_j\end{pmatrix}$$
But the above means
$$\sum_{i=1}^4 a_{ij}x_k=0\;,\;\;\forall\,j=1,2,3,4\iff r_i\perp\underline x\;\;\;\forall\,i=1,2,3,4$$
with $\;r_i=$ the i-th row of $\;A\;$, and thus
$$\forall\,i=1,2,3,4\;,\;\;r_i\perp\underline X\;,\;\;\forall\,\underline x\in W\iff r_i\perp \alpha_k\;,\;\;$$
where $\;\{\alpha_k\}\;$ is any basis of $\;W\;$ , and since the two given vectors are clearly a basis of $\;W\;$ (why?), you can form your matrix's rows with elements from $\;W^\perp\;$ . Note that
$$W^\perp:=\left\{\;\begin{pmatrix}x\\y\\z\\w\end{pmatrix}\in\Bbb R^4\;:\;\begin{cases}x+w=0\\2y+z-w=0\end{cases}\;\right\}$$
and , of course, $\;\dim W=2\iff \dim W^\perp=2\;$
