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Thread: dimension

  1. #1
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    dimension

    Let V be a finite dimensional vector space. Show that if W1,....,Wn are subspaces of V such that none of these subspaces are qeual to V, then Union of all these subspaces does not equal V.

    i am thinking induction on dim V might help but do not really have idea how to go bout it. so please help.
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  2. #2
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    Quote Originally Posted by Kat-M View Post
    Let V be a finite dimensional vector space. Show that if W1,....,Wn are subspaces of V such that none of these subspaces are qeual to V, then Union of all these subspaces does not equal V.
    The union of two subpaces is a subspace if and only one is contained in another.
    Thus if $\displaystyle W_1\cup W_2$ is a subspace thus $\displaystyle W_1\subseteq W_2$ and $\displaystyle W_2\subseteq W_1$.
    And so $\displaystyle \dim (W_1\cap W_2) = \max \{ \dim (W_1),\dim (W_2) \} < \dim (V)$.
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    Quote Originally Posted by ThePerfectHacker View Post
    The union of two subpaces is a subspace if and only one is contained in another.
    The reason that is so: Suppose u is in subspace $\displaystyle U_1$ but NOT subspace $\displaystyle U_2$ and v is in subspace $\displaystyle U_2$ but not in $\displaystyle U_1$. The u+ v cannot be in $\displaystyle U_1\cup U_2$. If it were, then it would have to be in either $\displaystyle U_1$ or $\displaystyle U_2$ (or both). If u+ v were in $\displaystyle U_1$, then, because $\displaystyle U_1$ is closed under addition and scalar multiplication u+ v+ (-u)= v would be in $\displaystyle U_1$, a cotradiction. If u+ v were in $\displaystyle U_2$, simlarly u+ v+ (-v)= u would be in $\displaystyle U_2$, again a contradiction.
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