$\{v_1,..,v_m\}$ basis? 
Let $V$ be a finite dimensional space over $\mathbb{R}$, with a positive definite scalar product. Let $\{v_1,..,v_m\}$ be the set of elements of $V$, of norm 1 , and mutually perpendicular (i.e$\langle v_i,v_j \rangle = 0, i \neq j$). Assume that for every $v\in V$ we have $||v||^2=\sum_\limits{i=1}^{m}{\langle v,v_i\rangle}^2$.
Show that $\{v_1,..,v_m\}$ is a basis of $V$.

What am I supposed to prove since the basis is an orthonormal basis?
 A: First we notice that if $\{ v_i \}_{i=1}^m$ is an orthonormal basis, then
\begin{align*}
v = \sum_{i=1}^m c_i v_i = \sum_{i=1}^m \langle v, v_i \rangle v_i.
\end{align*}
Then 
\begin{align*}
\| v \|^2 = \sum_{i=1}^m \langle v, v_i \rangle^2.
\end{align*}
Therefore it is reasonable to say that $\{ v_i \}_{i=1}^m$ should be a basis for $V$. Now we try to prove it. Since $V$ is of finite dimension, for now assume that it is $n \geq m$. We want to show that $n=m$. If $n > m$, $\{ v_i \}_{i=1}^m$ is already an orthonormal set, we can extend it to a basis for $V$ by adding $n-m$ mutually perpendicular vectors to $\{ v_i \}_{i=1}^m$. Let the basis be
\begin{align*}
\mathscr{B} = \{ v_1, \cdots, v_m, v_{m+1}, \cdots, v_n \}.
\end{align*}
Then given any $v \in V$, we can write it as
\begin{align*}
v = \sum_{i=1}^n \langle v, v_i \rangle v_i
\end{align*}
and it gives us
\begin{align*}
\| v \|^2 = \sum_{i=1}^n \langle v, v_i \rangle^2 = \sum_{i=1}^m \langle v, v_i \rangle^2.
\end{align*}
Hence, for $i > m$,
\begin{align*}
\langle v, v_i \rangle  = 0
\end{align*}
which implies that
\begin{align*}
v = \sum_{i=1}^n \langle v, v_i \rangle v_i = \sum_{i=1}^m \langle v, v_i \rangle v_i + \sum_{i=m+1}^n 0 \cdot v_i = \sum_{i=1}^m \langle v, v_i \rangle v_i.
\end{align*}
$\{ v_i \}_{i=1}^m$ is indeed an orthnormal basis for $V$.
A: You need to show that $V = span \{v_1, v_2, \ldots, v_m \}$.
If $\{v_1, v_2, \ldots, v_m \}$ do not form a basis for $V$ then there exists $w \in V$ which is not a linear combination of the $v_i$.  Set
$w_0 = \sum_1^n \langle w, v_j \rangle v_j; \tag{1}$
then 
$z = w - w_0 \ne 0, \tag{2}$
since $w \notin span \{v_1, v_2, \ldots, v_m \}$.  But
$\langle z, v_i \rangle = \langle w, v_i \rangle - \langle w_0, v_i \rangle = \langle w, v_i \rangle - \langle \sum_1^n \langle w, v_j \rangle v_j, v_i \rangle = \langle w, v_i \rangle - \langle w, v_i \rangle =0\tag{3}$
for all $v_i$, since the $v_i$ are orthonormal.  Then by (2), (3) and our hypothesis,
$0 \ne \Vert z \Vert^2 = \sum_1^m \langle z, v_i \rangle^2 = 0; \tag{4}$
this contradition shows that $\{v_1, v_2, \ldots, v_m \}$ spans $V$, so $\{v_1, v_2, \ldots, v_m \}$ is a basis.
