Determine a unique inner product with respect to which the given basis is orthonormal. Let $V$ be a finite dimensional vector space over a field $F$ with basis $X$. Then show that there is a unique inner product on $V$ with respect to which $X$ is an orthonormal basis.
If $V$ be a inner product space then we know that $V$ must have an orthonormal basis, but in this case we have to find a unique inner product for which $X$ is an orthonormal basis of $V$.
Please help me for this question.
 A: Suppose your basis is $\{e_1,\dots,e_n\}$. An inner product $g$ is a bilinear form, thus is completely determined by the image of the pairs $(e_i,e_j), i,j=1,\dots,n$. You can set 
$$g(e_i,e_j) = \delta_{ij}$$
(Kronecker delta, gives 1 if $i = j$, 0 otherwise) which gives the inner product wanted. 
As an example: take $\mathbb{R}^2$ with basis $e_1 = (1,1), e_2 = (1,-1)$ (note that these are orthogonal but not unit length) and impose the condition above:
$$g(e_1,e_1) = 1$$
$$g(e_1,e_2) = 0$$
$$g(e_2,e_1) = 0$$
$$g(e_2,e_2) = 1.$$
$g$ will be a matrix like
$$
\left(
\begin{matrix}
a & b\\
c & d
\end{matrix}
\right)
$$
so you get
$$(1,1)\left(
\begin{matrix}
a & b\\
c & d
\end{matrix}
\right)
\left(
\begin{matrix}
1\\
1 
\end{matrix}
\right) = 1 \rightarrow a+c+b+d=1$$
$$(1,1)\left(
\begin{matrix}
a & b\\
c & d
\end{matrix}
\right)
\left(
\begin{matrix}
1\\
-1 
\end{matrix}
\right) = 0 \rightarrow a+c-(b+d)=0$$ 
$$(1,-1)\left(
\begin{matrix}
a & b\\
c & d
\end{matrix}
\right)
\left(
\begin{matrix}
1\\
1 
\end{matrix}
\right) = 0 \rightarrow a-c+b-d=0$$ 
$$(1,-1)\left(
\begin{matrix}
a & b\\
c & d
\end{matrix}
\right)
\left(
\begin{matrix}
1\\
-1 
\end{matrix}
\right) = 1 \rightarrow a-c-(b-d)=1$$
and this gives
$$g = \left( \begin{matrix}
\frac{1}{2} & 0 \\
0 & \frac{1}{2}
\end{matrix}
\right).$$
