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Math Help - [SOLVED] Proof with positive definite inner products question

  1. #1
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    [SOLVED] Proof with positive definite inner products question

    Let V be a vector space and < , > be a positive definite inner product and B = { v_1, ..., v_n} be a basis for V. If c_1, ..., c_n in K are scalars, then show that there is a unique v in V such that < v, v_i > = c_i.

    This is what I have tried so far.

    Since we have a basis for V, we can write v as a linear combination of the basis elements.

    v = c_1 v_1 + c_2 v_2 + ... + c_i v_i + ... + c_n v_n

    < v, v_i > = ( c_1 v_1 + c_2 v_2 + ... + c_i v_i + ... + c_n v_n) * v_i

    < v, v_i > = c_1 v_1 v_i + c_2 v_2 v_i + ... + c_i v_i v_i + ... + c_n v_n v_i we want all of this to equal c_i

    So this is where I am stuck. Somehow I want everything to drop off but c_i. If this was an orthogonal basis, all the v_j, where j is not equal to i, would drop out but it is a regular basis.

    Any help would be appreciated.
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  2. #2
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    Life would be easier with an orthonormal basis. Nonetheless...

    How about trying the opposite approach? Suppose there were a vector u not identical to v s.t. your inner products were all the same. Then you can create the vector w = v-u. The inner product of this vector with the basis vectors are all 0. Assuming the kernel of the transformation defined by this basis is only the null vector...
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  3. #3
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    I have no idea how to even start that.
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  4. #4
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    This uses the idea of linear independence property of a basis:
    [ http://en.wikipedia.org/wiki/Basis_(linear_algebra) ]

    In other words, the only vector whose inner product with each of the basis vectors = 0 is the 0 vector.

    So let's assume the contrary in your problem. There are 2 vectors v and w such that:
    <v,b_i>=c_i, and
    <w,b_i>=c_i .

    Here b_i is the i'th basis vector, and c_i is a scalar.

    Look at the vector v-w. For every basis vector:
    <v-w,b_i>=<v,b_i>-<w,b_i>=c_i-c_i=0

    This means v-w is the 0 vector. That means v=w.
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