Linear Algebra finding a basis of the intersection of 2 vector spaces and its dimensions Let 
$$U = \mathrm{span} \{ (1,2,0,1) , (-1,1,1,1)\}$$
and
$$V = \mathrm{span} \{ (0,0,1,1) , (2,2,2,2)\}.$$
Find $U \cap V$ and compute its dimension. Determine a basis of $U \cap V$.
 A: Let $x\in U\cap V$. Then $x=a(1,2,0,1)+b(-1,1,1,1)=c(0,0,1,1)+d(2,2,2,2)$ for some $a,b,c,d\in R$. Hence $x=(a-b,2a+b,b,a+b)=(2d,2d,c+2d,c+2d)$ so that we have $a-b=2d,2a+b=2d,b=c+2d,a+b=c+2d$. It follows that $a=b=0$, so that $U\cap V=\varnothing$ with basis $\varnothing$.
A: Clearly $U \cap V \subseteq U$ and $U \cap V \subseteq V$ hence we should consider which vectors spanning $U$ are linear combinations of the spanning set of $V$ and vice-versa. Therefore, calculate $rref[A|B]$ and $rref[B|A]$ where $U  = span(A)$ and $V = span(B)$.
$$ rref[A|B] = I \qquad and \qquad rref[B|A] = I $$
therefore, by the column correspondence property no vector in either set is a linear combination of the other. It follows that the intersection is trivial; $U \cap V=\{ 0\}$.
A: Let $U=span\{u_1,u_2\}$, $V=span\{v_1,v_2\}$, then for any $w \in U \cap V$, we have $w=a_1u_1+a_2u_2=b_1v_1+b_2v_2$ for some $a_1, a_2, b_1, b_2 \in \mathbb{R}$. 
So we have $Ac=0$, where $A=[u_1,u_2,v_1,v_2]$ and $c=[a_1, a_2, -b_1, -b_2]^T$. It follows that $\dim(U \cap V)=\dim(\ker(A))=0$ since $\det(A) \neq 0$.
