Given two vectors and Nul A, find Matrix A I am trying to solve this problem of finding matrix A, given null space.
Let u = $\begin{bmatrix}1 \\1 \\ 2\end{bmatrix}$ and v = $\begin{bmatrix}1 \\0 \\ -1\end{bmatrix}$. Find matrix A such that Nul A = Span {u, v}
I know that if you want to find A given only one vector (e.g. w) in Nul A, you just have to find all solutions Aw = 0. But what about two vectors? Should I find sum of them?
 A: If the size doesn't matter, then let's look for a matrix $\begin{bmatrix}x & y &z\end{bmatrix}$ such that:
$$\begin{bmatrix}x & y &z\end{bmatrix}\begin{bmatrix}1 \\1 \\ 2\end{bmatrix}=0$$
$$\begin{bmatrix}x & y &z\end{bmatrix}\begin{bmatrix}1 \\0 \\ -1\end{bmatrix}=0$$
We get:
$$\begin{cases} x+y+2z=0 \\ x-z=0 \end{cases}$$
The solution of that system is $\text{sp}\{(1,-3,1)\}$.
So we can choose $A=\begin{bmatrix}1 & -3 &1\end{bmatrix}$
A: A solution is 
$$P=\begin{pmatrix}\ \ 1&-3&\ \ 1\\
    -3& \ \ \ 9&-3\\
    \ \ 1&-3&\ \ 1\end{pmatrix}.$$
I obtained it as (a multiple of) the matrix of orthogonal projection on the line defined by vector $w=u \times v$ using the following (apparently complicated) formula:
$$P=I_3-M(M^TM)^{-1}M^T$$
(where $M:=(u|v)$ is the $3 \times 2$ matrix with columns $u$ and $v$).
Proof: We are going to prove that $Pu=0$ and $Pv=0$ at the same time by proving that $P(u|v)=0$ i.e. by proving that $PM=0$. The proof is straightforward:
$$PM=(I_3-M(M^TM)^{-1}M^T)M=I_3M-M(M^TM)^{-1}(M^TM)=M-M=0$$
A: The vectors given are in the 3d-space, and they are linearly independent. So their span is some plane passing through the origin.  Geometrically there is a unique line passing through the origin that is perpendicular to the plane. Take a point $(a,b,c)$ in that line. So the equation is $ax+by+cz=0$ which will be satisfied precisely by the span of $u$ and $v$. To find the vector $(a,b,c)$ perpendicular to given vectors $u$ and $v$ take their cross-product $u\times v$ as in JeanMarie's answer.
