1. Intersection between two sets

$U=[(1,1,1), (1,0,-1)]$
$V=[(2,1,1), (1,0,1)]$

Find a basis for the set $U\cap V$

SOLVED
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While we're at it, how do I mathematically prove that $dim(M)=3$ if $M=\{(x_1, x_2, x_3, x_4)\in R^4: x_1+x_2+x_3=0\}$

EDIT: Another one:

$U_1 \subseteq V$, $U_2 \subseteq V$ and $U_1\oplus U_2 = \{\}$

Prove that $dim(U_1) + dim(U_2) \leq dim(V)$

2. I assume you mean U and V are the span of those vectors. Then for the first part, let $x\in U\cap V$. Then because x is in U(respectively, V), it is in the span of those vectors, so in particular there exist a,b (c,d) such that $x=a(1,1,1)+b(1,0,-1)=c(2,1,1)+d(1,0,1)$. Then comparing coordinates of these representations of x, we get the following equations:
$a+b=2c+d$, $a=c$, $a-b=c+d$

Solve the system of equations to figure out the form of such x. By my calculation, the intersection is of dimension one.

As for the other problem, you need to show that there are 3 linearly independent vectors, but not 4. Clearly (0,0,0,1) is in the set. We can take for example (1,0,-1,0) and (0,1,-1,0) as two more vectors, and so far our set is linearly independent so the dimension is at least 3. Can you find a,b,c,d such that a(w,x,y,z)+b(1,0,-1,0) + c(0,1,-1,0)+d(0,0,0,1)=0 where w+x+y=0?