Prove that $\|v \|^2= |\langle v, e_1 \rangle |^2 + \cdots + | \langle v, e_m\rangle |^2$ 
Suppose $(e_1,\cdots, e_m)$ is an orthonormal basis in $V$. Let $v \in V$ . Prove that
  $\|v\|^2= |\langle v, e_1 \rangle |^2 + \cdots + | \langle v, e_m\rangle  |^2$


Let $v\in V$  and $(e_1,e_2,\cdots,e_m )$ be an orthonormal basis in $V$. Then $v\in \operatorname{span} (e_1,e_2,\cdots,e_m )$. So, there exist $a_i\in F$ where $i=1,2,\cdots,m$ such that $$v=a_1e_1+a_2e_2+\cdots+a_me_m$$ Since $(e_1,e_2,\cdots,e_m )$ is linear independent, this implies each $a_i=0$, thus $$\|v\|^2=\sum_{i=1}^m |\langle v,e_i\rangle |^2 $$

I don't understand why: $(e_1,e_2,\cdots,e_m )$ is linearly independent implies $$\|v\|^2=\sum_{i=1}^m |\langle v,e_i\rangle |^2 $$
can anyone why explain it to me? thanks
 A: I don't think this makes much sense. However, we can write: $$v = \sum_{i=1}^na_ie_i \implies \langle v, e_j \rangle  = \left\langle \sum_{i=1}^na_ie_i,e_j\right\rangle = \sum_{i=1}^na_i\delta_{ij}=a_j,$$ so that $$v = \sum_{i=1}^n\langle v,e_i\rangle e_i.$$Hence: $$\|v\|^2 = \langle v,v\rangle = \left\langle \sum_{i=1}^n \langle v, e_i\rangle e_i, \sum_{j=1}^n \langle v,e_j\rangle e_j\right\rangle = \sum_{i,j=1}^n \langle v,e_i\rangle\langle v,e_j\rangle \delta_{ij} = \sum_{i=1}^n |\langle v, e_i\rangle|^2.$$

In any case... $$\delta_{ij} = \begin{cases} 1, \text{ if }i=j \\ 0, \text{ otherwise}\end{cases}$$ Saying that $\{e_i\}_{i=1}^n$ is an orthonormal set means that $\langle e_i,e_j\rangle = \delta_{ij}$.
A: Othonormality implies that $\left \langle e_{i},v \right \rangle=a_{i}$. Doing the calculation, you get 
$\left \langle e_{i},v \right \rangle =\left \langle e_{i}, (a_{1}e_{1},\cdots ,a_{n}e_{n} )\right \rangle=0+0\cdots +a_{i}\left \langle e_{i},e_{i} \right \rangle+\cdots 0+0=a_{i}$.
Can you take it from here?
A: Since $(e_i)_{1\le i\le n}$ is an orthogonal basis,
$$v=\sum_{i=1}^n\langle v,e_i\rangle e_i\quad\text{and}\quad \lVert v\rVert^2=\langle v,v\rangle=\sum_{i=1}^n\langle v,e_i\rangle^2\langle  e_i, e_i\rangle$$
Now $ \lVert e_i\rVert^2=\langle  e_i, e_i\rangle=1$, so
$$\lVert v\rVert^2=\sum_{i=1}^n\langle v,e_i\rangle^2.$$
A: Hint : The scalar products $e_ie_j$ vanish for $i\ne j$, if $e_1,...e_n$ is a
       orthogonal basis.
The linear independence is not sufficient for the given equality.
