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Math Help - URGENT: Subspaces spanned by independent vectors in an inner product space.

  1. #1
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    Exclamation URGENT: Subspaces spanned by independent vectors in an inner product space.

    I have no idea where to go with this.

    Show that if x1,...,xn are independent vectors in an inner product space, and y≠ 0 is such that y is orthogonal to xi for each i, then y is not a linear combination of x1,...,xn (i.e., is not in the subspace spanned by x1,...,xn )

    I know that the x's being independent means that a1x1 + a2x2+...+anxn = 0 iff a1=a2=...=an=0

    and that <y,xi> = 0 for each i

    And i know that we have to show by the contrapositive that y ≠ a linear combination of the x's so we assume

    y = b1x1 + ... + bnxn

    I just dont know where to go from here.
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  2. #2
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    Quote Originally Posted by Mandi_Moo View Post
    I have no idea where to go with this.

    Show that if x1,...,xn are independent vectors in an inner product space, and y≠ 0 is such that y is orthogonal to xi for each i, then y is not a linear combination of x1,...,xn (i.e., is not in the subspace spanned by x1,...,xn )

    I know that the x's being independent means that a1x1 + a2x2+...+anxn = 0 iff a1=a2=...=an=0

    and that <y,xi> = 0 for each i

    And i know that we have to show by the contrapositive that y ≠ a linear combination of the x's so we assume

    y = b1x1 + ... + bnxn

    I just dont know where to go from here.
    Hint: If y = b_1x_1 + \ldots + b_nx_n then \langle y,y\rangle = \langle b_1x_1 + \ldots + b_nx_n,y\rangle.

    (The information about x_1,\,\ldots\,,x_n being independent is irrelevant. The result is still true if these vectors are not independent.)
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  3. #3
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    So after i do that i get <y,y> = 0 which i have said implys that y is orthogonal to y, which dosent make sense so our assumption that y = b1*x1..... is false and b is not a linear combination of x1,...,xn?
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  4. #4
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    Quote Originally Posted by Mandi_Moo View Post
    So after i do that i get <y,y> = 0 which i have said implys that y is orthogonal to y, which dosent make sense so our assumption that y = b1*x1..... is false and b is not a linear combination of x1,...,xn?
    That's right: <y,y> = 0 implies that y = 0, which contradicts the initial information that y ≠ 0.
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