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Linear algebra - subspaces

  1. Aug 21, 2010 #1
    Suppose V is a vector space over R and [tex]U_1, U_2, W [/tex] are subspaces of V.

    Prove or contradict:

    1. [tex] W \cap (U_1+U_2) = (W \cap U_1)+(W \cap U_2)[/tex]
    2. [tex] If\ U_1 \oplus W = U_2 \oplus W\ then\ U_1=U_2 [/tex]

    I'm not sure how to approach this problem, and will appreciate any guidance.
    Thanks!

    (For the first one I had an idea to use theorem which links between dimension and subspaces equivalence, but both sides of the equation are to complex)
     
    Last edited: Aug 21, 2010
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  3. Aug 21, 2010 #2

    Office_Shredder

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    If you want to prove two subspaces are equal to each other, it's usually done the same as any time you want to prove two sets are equal: Show that each subspace is contained inside the other. For example, if [tex]v\in W\cap(U_1 + U_2)[/tex] can you prove that [tex]v\in (W\cap U_1) + (W\cap U_2)[/tex]? That and showing the reverse would give you that 1 is true.

    I'm not claiming this is true though, I haven't checked whether 1 is true. Whenever you have a problem that asks to 'prove or contradict', it's usually a good idea to try to make a couple of examples first; either you'll get a counterexample and you'll be done or you'll be able to get some intuition for why the statement is true that might help when making the general proof
     
  4. Aug 21, 2010 #3
    Thanks for the quick response!
    I actually tried to show bi-directional contain, but the fact that V is not necessarily finite dimensional made me uncomfortable. Don't I have to worry about that?
     
  5. Aug 21, 2010 #4

    Office_Shredder

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    You should only worry about it if you're going to pick a basis, or apply a theorem that only holds for finite dimensional vector spaces (like rank nullity). You shouldn't have to do that though.

    Have you tried coming up with some examples for these? For 1), just pick a vector space like R2 and a couple of one dimensional vector subspaces and see what happens
     
  6. Aug 21, 2010 #5
    I think I have contra-example for 1:

    [tex] W=Sp\{(1,0),(0,1)\}\ ,U_1=Sp\{(1,0)\}\ ,U_2=Sp\{(1,1)\} [/tex]

    Then: [tex] R^2=W\cap(U_1+U_2)=(W \cap U_1) + (W \cap U_2)=<not\ sure\ how\ to\ prove\ this\, for\ now\only\ based\ on\ intuition>=Sp(\{1,0\})[/tex]

    What do you think?
     
  7. Aug 22, 2010 #6

    Office_Shredder

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    In your example [tex]W\cap U_1=U_1[/tex] and [tex]W \cap U_2=U_2[/tex] so the right hand side is also the plane. Having [tex]U_1[/tex] and [tex]U_2[/tex] span the plane is a good idea, but W needs to be refined
     
  8. Aug 22, 2010 #7
    [tex] W=Sp\{(1,0)\}, U_1= Sp\{(1,1)\}, U_2= Sp\{(3,1)\} [/tex]

    Then: [tex] R^2=W\cap(U_1+U_2)=(W \cap U_1) + (W \cap U_2)=Sp(\{0,0\})=Sp\{0\}[/tex]

    It seems to contradict 2 also, am I right this time? (Set operations confuse me when applied on vector spaces)
     
    Last edited: Aug 22, 2010
  9. Aug 22, 2010 #8

    Office_Shredder

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    [tex]W\cap(U_1+U_2)=W[/tex], not R2, but other than that it's a good example. You can use what you have there to give a counterexample for 2, but you haven't said yet how it does so
     
  10. Aug 22, 2010 #9
    [tex]W=Sp\{(1,0)\}, U_1= Sp\{(1,1)\}, U_2= Sp\{(3,1)\}[/tex]

    [tex] (*1)\ U_1+W = U_2+W [/tex]
    [tex] (*2)\ U_1 \cap W= (0,0,0) [/tex] <Should I prove such things, or it's ok to leave it in this way?>
    [tex] (*3)\ U_2 \cap W= (0,0,0) [/tex]

    From (*1) and (*2) and (*3) [tex]U_1 \oplus W = U_2 \oplus W[/tex] is valid, but [tex] U_1 \neq U_2 [/tex]
     
    Last edited: Aug 23, 2010
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