Let $V$ be a subspace of $\mathbb{R}^4$, spanned by $v$ and $u$. Find a linear transformation whose kernel is $V$. And the vectors given are $v = (1,0,3,-2)$ and $u = (0,1,4,1)$.
It asks me to find the linear transformation from $\mathbb{R}^4$ to $\mathbb{R}^2$, where the kernel of that transformation is $V$. 
So what I know is that: the transformation I'm trying to find, applied to every vector in the span of $(1,0,3,-2)$ and $(0,1,4,1)$, will give the zero vector. 
Please let me know if that interpretation is incorrect. 
I've really no idea how to get started on this question. I have the equation $Av = 0$ where $A$ is the matrix of the transformation in question, and v is any vector of the subspace V, but...I don't think that gets me anywhere. Any help is greatly appreciated.
 A: We are searching for a matrix $A$ such that
$$ A = \left[\begin{matrix} a \ &b \ &c \ &d \\ x \ & y \ &z \ &t \end{matrix} \right] \left[\begin{matrix} 1\\ 0\\ 3 \\ -2 \end{matrix} \right] =\left[\begin{matrix} 0 \\ 0 \end{matrix} \right] $$
And we do the same with the other vector $u$.
So we get the equation 
$$\left\{
\begin{aligned}
    a + 3c -2d  &= 0\\
    b +4c +d    &= 0
\end{aligned}
\right.$$
We choose arbitrary $a=1$ and $b=5$ and we get $c=-1$ and $d=-1$.
And from the other 2 equations,
$$\left\{
\begin{aligned}
    x + 3y -2t  &= 0\\
    y +4z +t    &= 0
\end{aligned}
\right.$$
We also choose arbitrary $x=-7$ and $y=-2$ we get that $z=1$ and $t=-2$
So, $$ A = \left[\begin{matrix} 1 \ &5 \ &-1 \ &-1 \\ -7 \ & -2 \ &1 \ &-2 \end{matrix} \right]$$
and $Av=0$ and $Au=0$
And by the rank theorem we have that $$ n= nullity + rank$$
 So, in this case we get that $nullity = 2$ and so $$Ker(A)=span\{u,v\}$$
A: You have to find a $2$ by $4$ matrix  whose rows are linearly independent and orthogonal to the given vectors $u$ and $v$. 
One such matrix is $$ A = \left[\begin{matrix} 2 \ &-1 \ &0 \ &1 \\ -3 \ & -4 \ &1 \ &0 \end{matrix} \right] $$
The desired linear transformation is defined by $ T(v)=Av$ for $v\in R^4$ 
Note that according to the rank theorem $$ n= nullity + rank$$ which in this case we have $$4=2+2$$ therefore the kernel of your transformation is a two dimensional subspace generated by the given vectors $u$ and $v$ 
