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Thread: [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 = {$\displaystyle v_1$, ..., $\displaystyle v_n$} be a basis for V. If $\displaystyle c_1$, ..., $\displaystyle c_n$ in K are scalars, then show that there is a unique v in V such that < v, $\displaystyle v_i$ > = $\displaystyle c_i$.

    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 = $\displaystyle c_1$$\displaystyle v_1$ + $\displaystyle c_2$$\displaystyle v_2$ + ... + $\displaystyle c_i$$\displaystyle v_i$ + ... + $\displaystyle c_n$$\displaystyle v_n$

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

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

    So this is where I am stuck. Somehow I want everything to drop off but $\displaystyle c_i$. If this was an orthogonal basis, all the $\displaystyle v_j$, 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:
    $\displaystyle <v,b_i>=c_i$, and
    $\displaystyle <w,b_i>=c_i $.

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

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

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