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Thread: Intersection between two sets

  1. #1
    Senior Member Spec's Avatar
    Aug 2007

    Intersection between two sets

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

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


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

    EDIT: Another one:

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

    Prove that $\displaystyle dim(U_1) + dim(U_2) \leq dim(V)$
    Last edited by Spec; Jan 27th 2009 at 08:03 AM.
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  2. #2
    Jul 2008
    I assume you mean U and V are the span of those vectors. Then for the first part, let $\displaystyle x\in U\cap V$. 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 $\displaystyle x=a(1,1,1)+b(1,0,-1)=c(2,1,1)+d(1,0,1)$. Then comparing coordinates of these representations of x, we get the following equations:
    $\displaystyle a+b=2c+d$,$\displaystyle a=c$, $\displaystyle a-b=c+d$

    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?
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