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Linear Algebra - SubSpaces

  1. Sep 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Let V = V1 + V2, where V1 and V2 are vector spaces. Define M ={(x1, 0vector2): x1 in V1}

    and N = {(0vector1, x2) : x2 in V2
    0vector 1 is the 0v of V1 and 0vector is the 0v of V2 and 0v is 0 vector of V

    a) prove hat both M and N are subspace of V
    b) show that M n N = {0v}
    c) show that M+N=V

    2. Relevant equations



    3. The attempt at a solution
    I am not clear about what M intersection N is
    is it that the intersection of M and N is the 0 vector? If so what are the first steps to show this?

    as for a)

    do uprove using cx1 + x2, where Yi = (Yi, 0v2) Yj = (Yj,0v2)
    and so on...?
     
  2. jcsd
  3. Sep 25, 2013 #2

    Office_Shredder

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    For the intersection, (x,y) is in M and N if x=? and y=?.

    For doing a, your notation is terrible so it's impossible to say whether you are making a reasonable statement. I say your notation is terrible because you start with vectors x1 and x2, and then you have vectors Yi and Yj coming from nowhere whose definitions depends on themselves.
     
  4. Sep 25, 2013 #3
    let y1 = (m1, 0v2) be in M and let y2 = (0v1, m2) also be in M

    so prove that cy1 + y2 is also in M?
     
  5. Sep 25, 2013 #4

    Office_Shredder

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    Yes, except you should prove it for y2 = (m2, ov2)!
     
  6. Sep 25, 2013 #5
    Sorry about that typo.

    So for the intersection, is when the two vectors intersect right? And this wants me to prove that their intersection is at the 0v. So can I do something like (x,y) - (a,b) = (0,0) <--(where x,y and a,b are two vectors, so the intersection is when they are at the same point, so their difference is 0?)
     
  7. Sep 25, 2013 #6

    Office_Shredder

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    What are x,y, and and b, how do they relate to M and N, and why is the intersection of M and N at all related to subtraction?

    A vector (x,y) is contained in the intersection of M and N if (x,y) is contained in M and (x,y) is contained in N. If (x,y) is contained in M, what do we know about x and y? If (x,y) is contained in N, what do we know about x and y?
     
  8. Sep 26, 2013 #7
    So M and N intersect when (x,y) is contained in both M and N...

    if M = (x1, 0v2) and N = (0v1, x2) shouldnt the intersection calculated by when the difference of those is = 0?
    So you can get the point where both M and N contains?
     
  9. Sep 26, 2013 #8

    Office_Shredder

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    M is not a vector, and neither is N so those equalities you wrote don't make sense.
     
  10. Sep 26, 2013 #9
    can we just assume there is a non-zero vector a1, a2 that's in both m and n for it to be in m, we know a1 = x 1 and a2 = 0 component vector thingie and since it's a non-zero vector, you know x1 does not equal the zero component vector now, apply a1,a2 to N since a1 is not a zero-component vector, a1 does not equal to the zero component, which is a requirement for the vector in N so a1 a2 is not in both sets which is a contradiction
    so by proof of contradiction, there is no vector outside the zero vector in both m and n hence m intersection n is just the zero vector
     
  11. Sep 26, 2013 #10

    Office_Shredder

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    Yes, that sounds good to me. Technically you also should prove that the zero vector is contained in the intersection.
     
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