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Math Help - Subspace proof, just need a little push

  1. #1
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    Subspace proof, just need a little push

    Let W_1 , W_2 be non-empty subspaces of V. Prove that if W_1 \cup W_2 is a subspace of V, than either (I don't know how to do this in Latex) W_1 \le W_2 or W_2 \le W_1. ( \le symbol should mean proper subset here).

    ------------------
    This should be done by contradiction I think. So assume that there exists x_1 \in W_1 and x_2 \in W_2, wuch that x_1 , x_2 \notin W_1 \cap W_2.

    Still assume now that W_1 \cup W_2 is a subspace of V.

    Now I'm having a brain fade. Hint perhaps?
    Last edited by mr fantastic; December 8th 2009 at 11:59 AM.
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  2. #2
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    Quote Originally Posted by hatsoff View Post
    I don't think that is true (and thus can't be proven). I'm rusty on subspaces but...

    Let V=R^3...

    Let W_1 be the set spanned by \left[\begin{array}{c}0\\0\\1\end{array}\right]. Let W_2 be the set spanned by \left[\begin{array}{c}0\\1\\0\end{array}\right]. Then W_1\cup W_2 is the set spanned by \left[\begin{array}{c}0\\1\\1\end{array}\right]. So W_1, W_2 and W_1\cup W_2 are subspaces of V, but W_1 and W_2 are non-overlapping.

    The vector O must be a member of both subspaces so they will overlap. Also you have defined W_1\cup W_2 incorrectly, what you have given is the sum of W_1 and W_2. the union is not a subspace.

    Bobak
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  3. #3
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    So bobak, what's your take on this problem?

    Now is overlapping what we don't want here? For instance, if the zero vector is in both subspaces or any vector in both, does the union and overlap take away the uniqueness of terms?
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  4. #4
    Super Member PaulRS's Avatar
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    Suppose, <br />
W_1  \not\subset W_2 <br />
and <br />
W_2  \not\subset W_1 <br />
then there's a vector, call it <br />
v_1  \in W_1 /v_1  \notin W_2 <br />

    There's also <br />
v_2  \in W_2 /v_2  \notin W_1 <br />

    We have <br />
v_1 ,v_2  \in \left( {W_1  \cup W_2 } \right)<br />
so if the union is a subspace we should get <br />
v_1  + v_2  \in \left( {W_1  \cup W_2 } \right)<br />

    However <br />
v_1  + v_2  \notin W_1 <br />
and <br />
v_1  + v_2  \notin W_2 <br />
since <br />
v_1  + v_2  \in W_1 <br />
would imply ( by definition of subspace, summing -v_1 ) that <br />
v_2  \in W_1 <br />
analog. with <br />
 v_1  + v_2  \notin W_2 <br />

    So <br />
v_1+v_2  \notin \left( {W_1  \cup W_2 } \right)<br />
absurd

    So our assumption <br />
W_1  \not\subset W_2 <br />
and <br />
W_2  \not\subset W_1 <br />
, must have been wrong

    If one is contained in the other, the union is obviously a subspace, and so we are done
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