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Math Help - Proof related to subspaces

  1. #1
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    Proof related to subspaces

    Hi guys,

    I have a problem from my homework and am not sure where to start. (I hate proof questions). Could some1 just point me in the direction, I'm sure I can do it myself once I get some pointers, thanks!

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  2. #2
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    You better get over hating proofs, because that's pretty much what math is.

    Clearly each of the v_i\in U_1. If it were also in U_2, we would get v_i\in U_1\cap U_2=\mathrm{Span}(u_1,\ldots,u_m), which would imply that the set \{u_1,\ldots,u_m,v_i\}\subseteq\{u_1,\ldots,u_m,v_  1,\ldots,v_k\} is not linearly independent, contradicting the fact that it is a basis for something.

    Thus, for each 1\leq i\leq k, v_i \notin U_2. Since subspaces are closed under taking linear combinations of their elements, this implies \{0\}=\mathrm{Span}(v_1,\ldots,v_k)\cap U_2=W\cap U_2.

    For the second part,
    U_1+U_2=V\Rightarrow [(U_1\cap U_2)+W]+U_2=V
    \Rightarrow W+[(U_1\cap U_2)+U_2]=V
    \Rightarrow W+U_2=V.

    But since we already know that the intersection of these two subspaces is trivial, this means that the sum is actually a direct sum: W\oplus U_2=V.
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  3. #3
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    thanks alot! yeah I really should, any tips on how to improve on the thought process when tactling a proof problem?
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  4. #4
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    Why does [U1 intersection U2)+ U2] = U2? This implies U1 intersection U2 = 0
    Last edited by holaboo; February 13th 2011 at 10:16 AM.
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