Let $(E,\langle \cdot,\cdot\rangle)$ an Euclidean space. We suppose there exists a set of vector $\big\{e_i\big\}_{i=1}^n\subset E$ such that:
$$\forall x\in E,\quad ||x||^2=\sum_{i=1}^n\langle e_i,x\rangle^2$$Show that $\big\{e_i\big\}_{i=1}^n$ is an orthonormal basis of $E$
Attempt :
At first I have to show $\forall i,j\in[\![1,n]\!]$ if $i\ne j$ we have $\langle e_i,e_j\rangle =0$, and if $i= j$, $\langle e_i,e_j\rangle =1$
And the second step is to show $\big\{e_i\big\}_{i=1}^n$ is linear independant.
With $k\in [\![1,n]\!]$, we have $$||e_k||^2=\langle e_k,e_k\rangle=\sum_{i=1}^n\langle e_i,e_k\rangle^2$$
Then $$\sum\limits_{i=1}^n\langle e_i,e_k\rangle^2-\langle e_k,e_k\rangle=0 \iff \sum\limits_{\substack{i=1\\i\ne k}}^n\langle e_i,e_k\rangle^2+\langle e_k,e_k\rangle^2-\langle e_k,e_k\rangle=0$$
And I'm already blocked …
I tried to solve this equation :
$X^2-X+\alpha=0$ with $X=\langle e_k,e_k\rangle$ and $\alpha =\sum\limits_{\substack{i=1\\i\ne k}}^n\langle e_i,e_k\rangle^2$
I tried to use the polar form :$\langle x,y\rangle=\dfrac{1}{4}\bigg(||x+y||^2-||x-y||^2\bigg)$, and I have found the result below:
$$\langle x,y\rangle =\sum_{i=1}^n\langle e_i,x\rangle\langle e_i,y\rangle$$
Best Answer
Let $f_1, \ldots, f_n$ be the standard orthonormal basis for $\mathbb{R}^n$, and consider the map $ T: \mathbb{R}^n \to E$ defined by $f_k \mapsto e_k$ for all $k$. We have, for all $x \in E$, $$\|x\|^2 = \sum_{i=1}^n\langle x, e_i \rangle = \sum_{i=1}^n\langle x, Tf_i)^2 = \sum_{i=1}^n ((T^*x) \cdot f_i) ^2 = \|T^*x\|^2.$$ Therefore, $T^* : E \to \mathbb{R}$ is an isometry. Also, the set $\{e_j\}$ is total: if $\langle e_j,x\rangle=0$ for all $j$, the identity gives $x=0$. Thus $T$ is surjective.
The inverse of a surjective isometry is its adjoint, which is also an isometry, hence $T$ is an isometry too. Invertible isometries preserve orthonormal bases, by the polarisation identity, so as $T$ maps the standard basis to $e_1, \ldots, e_k$, they necessarily form an orthonormal basis.