Find a matrix X∈V such that U∩W=span{X} 
Here is my problem.  I've tried reading other people's related questions, but they're always just slightly different, I can't find one like mine and don't really know how to approach this problem.  Any help would be appreciated.
 A: The standard method for this kind of problem (I would think) is to combine the those spanning vectors and look for a complete set of relations between them.
Concretely you have $\def\sp{\operatorname{span}}U=\sp(u_1,u_2)$ and $V=\sp(v_1,v_2)$, and you are looking for common linear combinations $au_1+bu_2=cv_1+dv_2$, for any scalars $a,b,c,d$. That equation can also be written $au_1+bu_2-cv_1-dv_2=0$, which is a linear relation between $u_1,u_2,v_1,v_2$ with coefficients $a,b,-c,-d$ (the minus signs for $c,d$ are annoying, but you can consider $-c,-d$ as new unknown scalars). You can find all linear relations by solving a linear system $A\cdot(a,b,-c,-d)^T=(0,0,0)^T$ where $A$ has $u_1,u_2,v_1,v_2$ as columns. Here concretely
$$
  \begin{pmatrix}-3&-2&-5&-4\\-1&2&-1&2\\2&-1&0&-3\end{pmatrix}
\cdot\begin{pmatrix}a\\b\\-c\\-d\end{pmatrix}
=\begin{pmatrix}0\\0\\0\end{pmatrix}.
$$
Solving this leaves one free parameter in the general solution, which I choose to be$~c$, and the general solution can be given as $(a,b,-c,-d)=c(1,-1,-1,1)$. This means that $u_1-u_2-v_1+v_2=0$ and this is essentially the only linear relation between those vectors (any other relation is a scalar multiple of it). The $u_1-u_2=v_1-v_2$ should be a common linear combination of $u_1,u_2$ and of $v_1,v_2$; indeed both sides check out to be $[-1,-3,3]$. Moreover all other common linear combination of $u_1,u_2$ and of $v_1,v_2$ are multiples of this vector, so $\{[-1,-3,3]\}$ is a basis of $U\cap V$.
More generally, the general solution of the homogeneous linear system $A\cdot\vec x=0$ can be written as a linear combination of certain specific solutions (with the coefficients being freely chosen parameters); those solutions (in the example there was only one) form a basis of $\ker(A)$. For each of those specific solutions one can take the first few coordinates and form the corresponding linear combination of the basis vectors of$~U$; this will also be minus the linear combination with the remaining coordinates of the basis vectors of$~U$, hence give a vector of $U\cap V$. Moreover, after running through the basis of $\ker(A)$, one is guaranteed to have obtained a basis of $U\cap V$ (since all vectors in $U\cap V$ must correspond to some common linear combination).
