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.