# Thread: [SOLVED] Proof with positive definite inner products question

1. ## [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.

2. 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...

3. I have no idea how to even start that.

4. 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:
$=c_i$, and
$=c_i$.

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:
$=-=c_i-c_i=0$

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