# show that there exists a basis $\ \mathcal{B}'=\{e_1',e_2',…, e_n' \} \ \ of \ \ V$ such that $\ G(e_i,e_j')=\delta^i_j \$

Let $V \$ be a n-dimensional vector space over the field $\ \mathbb{F} \$ and let $\ G \$ be a non-degenerate billinear form on $\ V \$. Thus the map $\ L_G : V \to V^{*} \$ is a linear Isomorphism.

(a) Let $\ \mathcal{B}=\{e_1,e_2,......, e_n \} \$ be a basis of $\ V \$ . Using $\ L_G \$ , show that there exists a basis $\ \mathcal{B}'=\{e_1',e_2',......, e_n' \} \ \ of \ \ V$ such that $\ G(e_i,e_j')=\delta^i_j \$

(b) For any $\ v \in V \$ , show that $\ v=\sum_{i=1}^{n} G(v,e_i')e_i \$

where $\ V^{*} \$ is the dual space of $\ V \$

(a)

Since $\ L_G \ : \ V \to V^{*}$ is a isomorphism , then $\ \{L_G(e_1), L_G(e_2),......., L_G(e_n) \} \$ is a basis of $\ V^{*} \$ such that $\ L_G (e_i)(e_j')=G(e_i,e_j')=\delta^i_j \ \ , \ 1 \leq i,j \leq n$

(b)

Let $v \$ be any vector in $\ V \$ , then $\ V \$ can be written as

$v=c_1e_1+c_2e_2+........+c_ne_n \ , ...........(1)$ , where $\ c_i \in \mathbb{F} \$ for each $\ 1 \leq i \leq n \$

Then,

Operating $\ L_G \$ on both side , we get

$L_G(v)=c_1L_G(e_1)+c_2L_G(e_2)+....+c_jL_G(e_j)+.....+c_n L_G(e_n) \\ \Rightarrow L_G(v)(e_j')=c_j L_G(e_j)(e_j') \\ \Rightarrow G(v,e_j')=c_j \delta^j_j=c_j \ \ , \forall j \\ \Rightarrow c_j=G(v,e_j'), ...............(2)$

From (1) and (2) , we get

$v=G(v,e_1')e_1+G(v,e_2')e_2+.........+G(v,e_n')e_n \\ \Rightarrow v=\sum_{i=1}^{n} G(v,e_i')e_i$

But i am not sure of the part (a) answer.

I think we need more to conclude part (a) answer.

help me with the part (a) answer

Once we have that $L_G:V\to V^*$ is a linear isomorphism, we just need to use the definition of $L_G$ to find the requisite basis $\mathcal B'$ for part (a).
Recall that $L_G(v)$ is the linear map $V\to F$ which takes $u\in V$ to $G(v, u)$.
Since $L_G$ is an isomorphism, we have that for any $i$ we have a vector $e_i'\in V$ such that $L_G(v_i')(v_j)=\delta_{i}^j$.
The $e_i'$'s are linearly independent.
For if $x:=a_1e_1'+\cdots + a_ne_n'=0$ for some $a_n\in F$, we would then have $G(x, e_i)=0$, which gives $a_i=0$. Thus $\mathcal B'=\{e_1', \ldots, e_n'\}$ forms a basis of $V$ and satisfies the requirement of part (a).