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Math Help - Prove that vj must be the zero vector

  1. #1
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    Prove that vj must be the zero vector

    Let S = { v_1, v_2,....v_k} be an orthogonal set of vectors in R^n. If S is linearly dependent, prove that one of the v_j must be the zero vector.



    Find an orthonormal basis for the column space of the matrix A =
    [2 5 7]
    [3 1 8]
    [6 6 10]
    [0 6 -9] and obtain the QR factorisation of A.
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  2. #2
    Junior Member nimon's Avatar
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    Let n=3, and let v_{1}=(1,0,0),\,\, v_{2}=(0,1,0). Then \{v_{1},v_{2}\}\subset \mathbb{R}^{n} are orthogonal and neither are zero. Did you mean to assume that k>n?
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  3. #3
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    That set is not linearly dependent.

    wopashui, if the set is linearly dependent, then there exist numbers, a_i, not all 0, such that a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_kv_k= 0. Now take the dot product of both sides of that with v_1, v_2, etc.

    As for the second problem, think of the three columns of A as three vectors and use "Gram-Schmidt" to find an orthonormal basis.
    Last edited by HallsofIvy; April 7th 2010 at 12:58 PM.
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  4. #4
    Junior Member nimon's Avatar
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    Woops! Note to self: read the question!
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  5. #5
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    Quote Originally Posted by nimon View Post
    Woops! Note to self: read the question!
    Always a good suggestion!
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    That set is not linearly dependent.

    wopashui, if the set is linearly dependent, then there exist numbers, a_i, not all 0, such that a_1v_1+ a_2v_2+ \cdot\cdot\cdot+ a_kv_k= 0. Now take the dot product of both sides of that with v_1, v_2, etc.

    As for the second problem, think of the three columns of A as three vectors and use "Gram-Schmidt" to find an orthonormal basis.

    but none of the vector of A is orthonormal, whhich vector do I start with? And what is the QR factorisation of A?
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  7. #7
    Junior Member nimon's Avatar
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    It doesn't matter that the columns aren't orthogonal. The Gram-Schmidt procedure takes any set of basis vectors and turns them into an orthogonal basis of the same set, which can then be normalised to give an orthonormal set.
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