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Thread: Proving intersection is a subspace

  1. #1
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    Proving intersection is a subspace

    Let $\displaystyle W_1$ and $\displaystyle W_2$ be two subspaces of a vector space $\displaystyle V$ over the field $\displaystyle \mathbb{F}$. Prove that the intersection of $\displaystyle W_1$ and $\displaystyle W_2$ is a subspace of $\displaystyle V$.


    Can someone check my working for this question

    Let $\displaystyle x\in W_1, y\in W_1$

    Since $\displaystyle W_1$ is a subspace, it is closed under addition and so $\displaystyle x+y\in W_1$

    Similarly, let $\displaystyle x\in W_2, y\in W_2$

    Since $\displaystyle W_2$ is a subspace, it is closed under addition and so $\displaystyle x+y\in W_2$

    $\displaystyle \Rightarrow x+y\in W_1\cap W_2\Rightarrow$ Closure under addition


    Let $\displaystyle \lambda\in\mathbb{R}, x\in W_1, W_2$

    $\displaystyle \lambda x\in W_1$ since $\displaystyle W_1$ is a subspace

    $\displaystyle \lambda x\in W_2$ since $\displaystyle W_2$ is a subspace

    Clearly, $\displaystyle \lambda x\in W_1\cap W_2\Rightarrow$ Closure under scalar multiplication

    $\displaystyle \Therefore W_1 \cap W_2$ is a subspace of $\displaystyle V$




    What about the condition for non-empty? This condition always confuses me. Why is it that the zero vector has to be included for it to be non-empty?
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by acevipa View Post
    Let $\displaystyle W_1$ and $\displaystyle W_2$ be two subspaces of a vector space $\displaystyle V$ over the field $\displaystyle \mathbb{F}$. Prove that the intersection of $\displaystyle W_1$ and $\displaystyle W_2$ is a subspace of $\displaystyle V$.


    Can someone check my working for this question

    Let $\displaystyle x\in W_1, y\in W_1$

    Since $\displaystyle W_1$ is a subspace, it is closed under addition and so $\displaystyle x+y\in W_1$

    Similarly, let $\displaystyle x\in W_2, y\in W_2$

    Since $\displaystyle W_2$ is a subspace, it is closed under addition and so $\displaystyle x+y\in W_2$

    $\displaystyle \Rightarrow x+y\in W_1\cap W_2\Rightarrow$ Closure under addition


    Let $\displaystyle \lambda\in\mathbb{R}, x\in W_1, W_2$

    $\displaystyle \lambda x\in W_1$ since $\displaystyle W_1$ is a subspace

    $\displaystyle \lambda x\in W_2$ since $\displaystyle W_2$ is a subspace

    Clearly, $\displaystyle \lambda x\in W_1\cap W_2\Rightarrow$ Closure under scalar multiplication

    $\displaystyle \Therefore W_1 \cap W_2$ is a subspace of $\displaystyle V$




    What about the condition for non-empty? This condition always confuses me. Why is it that the zero vector has to be included for it to be non-empty?
    well, since $\displaystyle \displaystyle W_1$ and $\displaystyle \displaystyle W_2$ are subspaces, they each contain 0. so 0 will be in their intersection, and hence, the intersection is non-empty.

    containing zero is important for vector spaces and are hence important for subspaces. so the subspaces must have 0 in them. it is not to say that they have to have 0 to be non-empty per sae, but it is usually easy to verify that the zero vector is in there. so you always check that. and if 0 is not in there, then you're in trouble. you wouldn't have a vector space at all. just some other kind of set.
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  3. #3
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    Thanks a lot. I understand that well now.
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