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Math Help - Inner product space with orthonormal subset

  1. #1
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    Inner product space with orthonormal subset

    Let V be an inner product space, let S = \{ v_{1},v_{2},...,v_{n} \} be an orthonormal subset of V. Prove that for any  x \in V , we have  ||x||^2 \geq \sum ^{n} _{i=1} |<x,v_{i}>|^2

    proof so far.

    by a theorem, there exist unique vectors u in span(S) and v in the orthogonal complement of span(S) such that x = u + v.

    Now, since u is orth. to v, meaning  ||x||^2 = ||u||^2 + ||v||^2

    Since u \in \ span{S} , write  u = \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n}

    then  x = u + v = v + \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n}

    Since they are all orthogonal to one another, we can write ||x||^2 = ||v||^2 + || \alpha _{1} v_{1} ||^2 + || \alpha _{2} v_{2} ||^2 + . . . + || \alpha _{n} v_{n} ||^2

    Now, I'm a bit stuck... Am I at least doing this right so far?
    Last edited by tttcomrader; April 4th 2008 at 10:10 AM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let V be an inner product space, let S = \{ v_{1},v_{2},...,v_{n} \} be an orthonormal subset of V. Prove that for any  x \in V , we have  ||x||^2 \geq \sum ^{n} _{i=1} |<x,v_{i}>|^2

    proof so far.

    by a theorem, there exist unique vectors u in span(S) and v in the orthogonal complement of span(S) such that x = u + v.

    Now, since u is orth. to v, meaning  ||x||^2 = ||u||^2 + ||v||^2

    Since u \in \ span{S} , write  u = \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n}

    then  x = u + v = v + \alpha _{1} v_{1} + \alpha _{2} v_{2} + . . . + \alpha _{n} v_{n}

    Since they are all orthogonal to one another, we can write ||x||^2 = ||v||^2 + || \alpha _{1} v_{1} ||^2 + || \alpha _{2} v_{2} ||^2 + . . . + || \alpha _{n} v_{n} ||^2

    Now, I'm a bit stuck... Am I at least doing this right so far?
    Yes, that's the right approach. Now you need to use the fact that since \{ v_{1},v_{2},...,v_{n} \} is an orthonormal basis for S, and u∈S, it follows that \alpha_j = \langle u,v_j\rangle = \langle x,v_j\rangle for each j. Together with the fact that each v_j is a unit vector, that will give you the result.
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  3. #3
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    I'm sorry, but I really don't follow your last post.

    Why <br />
\alpha_j = \langle u,v_j\rangle = \langle x,v_j\rangle<br />
?
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    I'm sorry, but I really don't follow your last post.

    Why <br />
\alpha_j = \langle u,v_j\rangle = \langle x,v_j\rangle<br />
?
    We know that S is orthonormal, so \langle v_i,v_j\rangle = 0 if i≠j, and \langle v_j,v_j\rangle = 1.

    We also know that  u = \alpha _{1} v_{1} + \alpha _{2} v_{2} + \ldots + \alpha _{n} v_{n} . Take the inner product with v_j, and we get \langle u,v_j\rangle = \langle  \alpha _{1} v_{1} + \alpha _{2} v_{2} + \ldots + \alpha _{n} v_{n},v_j\rangle =  \alpha _{1}\langle v_1,v_j\rangle + \alpha _{2}\langle v_2,v_j\rangle + \ldots + \alpha _{n}\langle v_n,v_j\rangle = \alpha_j because of the orthonormality.

    Finally, we know that x = u + v, where v\in S^{\perp}, and so \langle x,v_j\rangle = \langle u,v_j\rangle + \langle v,v_j\rangle. But \langle v,v_j\rangle = 0 because v_j∈S.
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