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Math Help - prove or disprove direct sum question

  1. #1
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    prove or disprove direct sum question

    Prove or disprove

    W, U_1, U_2 are subspaces of V.

    If V=U_1 \oplus U_2

    then W = (U_1 \cap W) \oplus (U_2 \cap W)

    Attempt:

    False

    let's say dimV = 8, dimU_1 = dimU_2  = 4 (dimV = dimU_1 + dimU_2)
    and dimW = 5

    then dim(U_1 \cap W) may equal 4
    and dim(U_2 \cap W) may equal 4

    and then we get 5 \neq 4+4
    Last edited by jayshizwiz; November 7th 2010 at 12:31 PM.
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  2. #2
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    **Is the question understandable? Please let me know if there is something I need to explain better.

    Thanks!
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  3. #3
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    Quote Originally Posted by jayshizwiz View Post
    Prove or disprove

    W, U_1, U_2 are subspaces of V.

    If V=U_1 \oplus U_2

    then W = (U_1 \cap W) \oplus (U_2 \cap W)

    Attempt:

    False

    let's say dimV = 8, dimU_1 = dimU_2  = 4 (dimV = dimU_1 + dimU_2)
    and dimW = 5

    then dim(U_1 \cap W) may equal 4
    and dim(U_2 \cap W) may equal 4

    and then we get 5 \neq 4+4

    You haven't disproved anything: "may" it's not existence. You have to come up with a particular example that shows

    clearly that the claim is false.

    Hint: Take a look at \mathbb{R}^2\,,\,\,U_1=\left\{\,\binom{x}{x}\in\ma  thbb{R}^2\right\}\,,\,U_2=\left\{\,\binom{x}{-x}\in\mathbb{R}^2\right\}\,,\,\,W=\left\{\binom{x}  {0}\in\mathbb{R}^2\right\}

    Tonio
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  4. #4
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    Quote Originally Posted by tonio View Post
    You haven't disproved anything: "may" it's not existence. You have to come up with a particular example that shows

    clearly that the claim is false.

    Hint: Take a look at \mathbb{R}^2\,,\,\,U_1=\left\{\,\binom{x}{x}\in\ma  thbb{R}^2\right\}\,,\,U_2=\left\{\,\binom{x}{-x}\in\mathbb{R}^2\right\}\,,\,\,W=\left\{\binom{x}  {0}\in\mathbb{R}^2\right\}

    Tonio
    I truly hope you aren't grading my exam (;

    i sohld get at least partial credit
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  5. #5
    Senior Member Tinyboss's Avatar
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    The misconception that some students have, and that this problem is attempting to dispel, is that the direct sum is like a (set) union. But in fact, the direct sum of U1 and U2 contains a whole lot of elements that are not in U1 or U2. In case Tonio's excellent example wasn't clear, think about this and read it again.
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