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.
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
$$g = \left( \begin{matrix} \frac{1}{2} & 0 \\ 0 & \frac{1}{2} \end{matrix} \right).$$