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Math Help - Dimension problem

  1. #1
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    Dimension problem

    Let V be a vector space and U and W subspaces of V.
    dimV = n, dimU = dimW = n-1
    Show that:
    U+W=V is equivalent to: U≠W

    I think one direction of the equivalence is easy to prove:
    If U = W then U+W=2U=U≠V.
    I hope this is right. I don't know how to prove the other direction though.

    Any help would be greatly appreciated.
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by durrrrrrrr View Post
    I think one direction of the equivalence is easy to prove:
    If U = W then U+W=2U=U≠V I hope this is right.
    Right, but don't write 2U. Simply U+W=U+U=U\neq W

    I don't know how to prove the other direction though.
    Suppose U\neq W then,

    \exists x\in W\;(x\neq 0): x\not\in V or \exists x\in V\;(x\neq 0): x\not\in W

    Could you continue?.

    Regards.

    Fernando Revilla
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by durrrrrrrr View Post
    Let V be a vector space and U and W subspaces of V.
    dimV = n, dimU = dimW = n-1
    Show that:
    U+W=V is equivalent to: U≠W

    I think one direction of the equivalence is easy to prove:
    If U = W then U+W=2U=U≠V.
    I hope this is right. I don't know how to prove the other direction though.

    Any help would be greatly appreciated.
    There is another way which gives us more information:
    Spoiler:


    Note that from a common theorem we have that


    \begin{aligned}\dim\left(U+W\right) &=\dim\left(U\right)+\dim\left(W\right)-\dim\left(U\cap W\right)\\ &=(n-1)+(n-1)+\dim\left(U\cap W\right)\\ &=2n-2+\dim\left(U\cap V\right)\end{aligned}\quad\quad\mathbf{(1)}


    Note though that since U+W\leqslant V we must have that \dim\left(U+W\right)\leqslant n and so by \mathbf{(1)} we know that \dim\left(U\cap W\right)\geqslant n-2. But, since U\cap W\leqslant U we know that \dim\left(U\cap W\right)\leqslant n-1. Thus, putting these two together gives


    n-2\leqslant\dim\left(U\cap W\right)\leqslant n-1


    Thus, \dim\left(U\cap W\right) equals either n-2 or n-1. And so plugging this into \mathbf{(1)} gives


    \displaysytle \dim\left(U+W\right)\begin{cases}n & \mbox{if}\quad \dim\left(U\cap W\right)\ne n-1\\ n-1 & \mbox{if}\quad \dim\left(U\cap W\right)=n-1\end{cases}


    But note that \dim\left(U\cap V\right)\ne n-1\implies U\cap V\ne U\implies U\ne V and \dim\left(U\cap V\right)=n-1\implies U=V (the first is clear, and the second follows since U\cap V\leqslant U and a subspace which has the same dimension as the ambient space must be the full space). Thus, this may be rewritten as


    \dim\left(U+W\right)=\begin{cases}n & \mbox{if}\quad U\ne W\\ n-1 & \mbox{if}\quad U=W\end{cases}


    Thus, if U\ne W then U+W is an n-dimensional subspace of V and thus equal to V and if U=V then U+W is an n-1-dimensional subspace and not equal to V. The conclusion follows.
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