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Math Help - Proving orthogonal vectors are a subspace

  1. #1
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    Proving orthogonal vectors are a subspace

    Problem:
    For a set of vectors { u_1,....,u_k} \subset R^n show that the set of vectors X \subset R^n that are orthogonal to each u_i is a subspace of R^n

    Solution:
    Alright, I'm thinking that since the set of vectors X is orthogonal to u_i then I can say something like
    { u_1x_1, u_2x_2,....,u_kx_k} = 0
    therefore,
    u_1x_1 = 0 , and u_2x_2 = 0
    so
    u_1x_1 + u_2x_2 = 0

    Am I thinking this through properly?
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  2. #2
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    Think about what it means to be a subspace.

    You're notation is confusing. You should be dealing with an inner product, and show that when pairing u_i with a linear combination of vectors in the orthogonal complement, that you still get 0.
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  3. #3
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    Quote Originally Posted by MattMan View Post
    Think about what it means to be a subspace.

    You're notation is confusing. You should be dealing with an inner product, and show that when pairing u_i with a linear combination of vectors in the orthogonal complement, that you still get 0.
    So you're saying it should look more like this?

    Xu_1 = 0 and Xu_2 = 0

    X(u_1 + u_2) = 0

    Xu_1 + Xu_2 = 0
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  4. #4
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    ***Reply got posted twice***
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  5. #5
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    If we let U=\{u_1,...,u_k\}\subset\mathbb{R}^n
    Clearly 0 \in U^{\perp} since (0,u_i)=0,\: 1<i<k
    So 0 \in U^{\perp}\neq \emptyset
    Now let x_1,x_2 \in U^{\perp}; \alpha,\beta \in \mathbb{R}
    Now consider  (\alpha x_1+\beta x_2,u_i)=\alpha (x_1,u_i)+\beta (x_2,u_i)=0+0=0, \text{ for } 1<i<k, hence \alpha x_1 + \beta x_2 \in U^{\perp}. So U^{\perp} is closed under linear combinations and hence a subspace of \mathbb{R}^n.
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  6. #6
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    Thank you very much for clearing that up!
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