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

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 \$

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

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

I need only part (c)

(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

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

(c)

Proceeding in the same way as in part (b)

$e_i=c_1e_1'+....+c_ie_i'+........c_ne_n' \$

Operating $L_G \$ on both side , we get

$L_G(e_i)=c_1 L_G(e_1')+........+c_i L_G(e_i')+........+c_n L_G(e_n') \\ \Rightarrow L_G(e_i)(v)=c_1 L_G(e_1')(v)+.......+c_iL_G(e_i')(v)+.....+c_n L_G(e_n')(v)$

But Now I can not finish the proof

Help me finishing the proof

• Let $v =\sum v_j' e_j'$. Then $G(e_i,v) = G(e_i, \sum v_j' e_j') = \sum _{j} v_j' G(e_i,e_j') = \sum_jv_j' \delta_{ij} = v_i'$. So $v = \sum_i G(e_i,v) e_i'$ – Sou Jun 26 '18 at 14:07