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Math Help - Direct sum of trivial subspaces

  1. #1
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    Direct sum of trivial subspaces

    Hi guys.
    i need help to prove the following:

    "Let W be subspace of V, such that there exist one and only one subspace W' for wich W\oplus W'=V.
    Prove that W must be a trivial subspace."

    this is what i've tried so far:

    let's suppose there exist only one subspace W'.
    and let's suppose it is not trivial.

    let B_w=\left \{ w_1, w_2... w_n \right \} be the base for W.
    let B_{w}'=\left \{ w'_1, w'_2... w'_m \right \} be the base for W'.

    let v\in V,
    since W\oplus W'=V, there exist one and only one linear combination such that:

    v=\alpha_1w'_1+\alpha_2w'_2+...+\alpha_mw'_m+\alph  a_{m+1}w_1+\alpha_{m+2}w_2+...+\alpha_{n}w_n

    according to the assumption, there exist only one W', but there are infinite number of bases for W.
    therefor, i can define new base for W and use other scalares...

    now, i'm not sure if i'm in the right way or where, if at all, lies the contradiction.
    i could really use some direction here.

    thanks in advanced!
    Last edited by Stormey; January 14th 2013 at 11:46 AM.
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  2. #2
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    Re: Direct sum of trivial subspaces

    Suppose W\ne V and W\ne <0>.
    let B_w=\left \{ w_1, w_2... w_n \right \} be a base for W and \left \{ y_1, y_2... y_m \right \} a basis for W^{\prime}. Set W_1^{\prime}=<w_1+y_1, y_2, ... y_m>.
    You can easily show, I hope, that V=W\oplus W_1^{\prime} and W^{\prime}\ne W_1^{\prime}
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  3. #3
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    Re: Direct sum of trivial subspaces

    Hi johng, thanks for your help.

    don't i need to show first that W_1^{\prime}=<w_1+y_1, y_2, ... y_m> is linear independent?
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  4. #4
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    Re: Direct sum of trivial subspaces

    Yes, that's part of the proof.
    Suppose a_1(w_1+y_1)+a_2y_2+...a_my_m=0, i.e., a_1w_1+a_1y_1+...a_my_m=0. Since w_1,y_1,...y_m are part of a basis of V, they are linearly independent. Hence all a's are 0 and so w_1+y_1, y_2,...y_m are linearly independent.
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