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(linear algebra) union of subspaces

  1. Feb 8, 2010 #1
    Eh, kind of stuck on this question. I need some suggestions on how to tackle the problem..

    1. The problem statement, all variables and given/known data
    Let U and V be the subspaces of R_3 defined by:

    U = {x: aT * x = 0} and V = {x: bT * x = 0} (T means transpose)

    where

    a = [1; 1; 0] and b = [0; 1; -1]

    Demonstrate that the union of U and V is not a subspace of R_3..

    2. Relevant equations
    See Above


    3. The attempt at a solution
    Should I just combine U and V, into something like UuV = {x: aT * x = b*T * x = 0}, since both equations equal to 0, just kind of combine them together..

    any tips? am I on the right track?
     
  2. jcsd
  3. Feb 8, 2010 #2
    describe / visualize what the subspace U is first. Then describe/visualize what V is. finally, what is U union V? It will be obvious after you do this.
     
  4. Feb 8, 2010 #3
    Describing U:
    U: aT *x = [1 1 0]*[x1;x2;x3] = x1+x2 = 0

    Describing V:
    V: bT *x = [0 1 -1]*[x1;x2;x3] = x2-x3 = 0

    union of U and V {x: x1+x2=x2-x3=0}

    after moving around x's i get
    union of U and V {x: x1+x3 = 0}

    But as i test this subspace (zero vector, sum of two arbitrary vectors and scalar multiplication) , the union of U and V passes all of the tests...

    I have a feeling that Im doing something wrong when Im adding two arbitrary vectors:
    u = [-1; 1; 1] and v = [-2; 2; 2] (both vectors are in subspace of U and V).
    when i add them i get u+v = [-3; 3; 3], which satisfies the x1+x3=0 equation, so it passed that test..

    am I wrong? :( tired of banging my head against the table!
     
  5. Feb 8, 2010 #4
    coupla things wrong here:

    -you're confused as to what 'union' means. union means 'in V OR U'. not ' V AND U'.

    -sum of 2 arbitrary vectors
    its supposed to be the sum of any 2 vectors is also inside. not just the random 2 that you happen to pick.


    this is what i meant by 'describing' earlier:

    U is the set of all vectors perpendicular to a. (dot product=0)
    V is the set of all vectors perpendicular to b.


    U union V is the set of all vectors perpendicular to either a or b.


    now, can you find 2 vectors in [U union V] whose sum is not in [U union V]?
     
  6. Feb 8, 2010 #5
    Eh, i still don't get it..

    So the sum of two arbitrary vectors will need to satisfy either U or V in order to have the new "U union V" to be considered subspace??
     
    Last edited: Feb 8, 2010
  7. Feb 8, 2010 #6

    vela

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    What geometric shapes do the equations x1+x2=0 and x2-x3=0 describe?
     
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