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Math Help - Subspace combination question

  1. #1
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    Subspace combination question

    We've done questions a bit like this before, but nothing with subspaces. Basically, for a vector space V there exist subspaces X, Y and Z in V (with dim V > 1), determine, either by proof or counter-example, whether the following statements are true or false:

    <br />
X + \left( {Y \cap Z} \right) = \left( {X + Y} \right) \cap \left( {X + Z} \right)<br />

    and

    <br />
X \cap \left( {Y + \left( {X \cap Z} \right)} \right) = \left( {X \cap Y} \right) + \left( {X \cap Z} \right)<br />

    Help?!
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  2. #2
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    Quote Originally Posted by StandardToaster View Post

    We've done questions a bit like this before, but nothing with subspaces. Basically, for a vector space V there exist subspaces X, Y and Z in V (with dim V > 1), determine, either by proof or counter-example, whether the following statements are true or false:

    <br />
X + \left( {Y \cap Z} \right) = \left( {X + Y} \right) \cap \left( {X + Z} \right)<br />
    false! counter-example: V=\mathbb{R}^2, \ X=\{(x,2x): \ x \in \mathbb{R} \}, \ Y=\{(x,0): \ x \in \mathbb{R} \} , \ Z=\{(0,x): \ x \in \mathbb{R}\}. obviously X + Y \cap Z = X, because Y \cap Z=(0).

    now see that (1,0) \in (X+Y) \cap (X+Z) but (1,0) \notin X=X+Y \cap Z. hence the equality does not hold.



    <br />
X \cap \left( {Y + \left( {X \cap Z} \right)} \right) = \left( {X \cap Y} \right) + \left( {X \cap Z} \right)<br />
    true by modular law.
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  3. #3
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    How does the second one work? I'm utterly confused. I thought I had a proof all written out, but I made a stupid mistake at the start which renders it essentially useless...
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