Let $\{v_1,v_2,...,v_k\}$ be an orthogonal set in $V$. Prove that $\|\sum_{i=1}^k a_i v_i\|^2 = \sum_{i=1}^k |a_i|^2 \|v_i\|^2$. Let ${v_1,v_2,...,v_k}$ be an orthogonal set in V, and let $a_1, a_2,..,a_k$ be scalars. Prove that $||\sum_{i=1}^k a_i v_i||^2= \sum_{i=1}^k |a_i|^2 ||v_i||^2$.
The original source is Proof of this problem 
$$
 \left\langle \sum^k_{i=1} a_iv_i , \sum^k_{i=1} a_iv_i\right\rangle = 
 \sum^k_{i=1} \left\langle a_iv_i, \sum^k_{j=1} a_j v_j \right\rangle = 
 \sum^k_{i=1} a_i \left\langle v_i, \sum^k_{j=1} a_jv_j \right\rangle.
$$
and then each
$$
   \left\langle v_i, \sum^k_{j=1} a_jv_j \right\rangle = \sum^k_{j=1} \bar a_j \left\langle v_i,v_j  \right\rangle 
$$
As vectors are orthogonal last expression simplifies to $\bar a_i \| v_i \|^2$ and so the result
$$
 \left\langle \sum^k_{i=1} a_iv_i , \sum^k_{i=1} a_iv_i\right\rangle = \sum^k_{i=1} |a_i|^2\| v_i\|^2
$$

My question for the proof is:  I know from orthogonal set that $<v_i, v_j>=0$ for $i \neq j$. The reason why the equation $\left\langle \sum^k_{i=1} a_iv_i , \sum^k_{i=1} a_iv_i\right\rangle = 
 \sum^k_{i=1} \left\langle a_iv_i, \sum^k_{j=1} a_j v_j \right\rangle$ is switching from $v_i$ to $v_j$ is to take out the scalars $a_i$ and $a_j$, why only that way, can we take out the scalars? I mean if we stick to $<\sum_{i=1}^k a_i v_i, \sum_{i=1}^k a_i v_i>$, why can't we arrive at $\sum_{i=1}^k |a_i|^2 ||v_i||^2$?

 A: Here it is a way to extend you reasoning. Given an  orthogonal basis $\mathcal{B} = \{v_{1},v_{2},\ldots,v_{n}\}$ for the finite dimensional vector space $V$ where $\dim V = n$, and $v = a_{1}v_{1} + a_{2}v_{2} + \ldots + a_{n}v_{n}$, one has that
\begin{align*}
\|v\|^{2} = \langle v,v\rangle & = \langle a_{1}v_{1} + a_{2}v_{2} + \ldots +a_{n}v_{n},a_{1}v_{1} + a_{2}v_{2} + \ldots + a_{n}v_{n} \rangle\\\\
& = a_{1}\overline{a_{1}}\langle v_{1},v_{1}\rangle + \sum_{j\neq 1}\langle a_{1}v_{1},a_{j}v_{j}\rangle + \ldots + a_{n}\overline{a_{n}}\langle v_{n},v_{n}\rangle + \sum_{j\neq n}\langle a_{n}v_{n},a_{j}v_{j}\rangle\\\\
& = |a_{1}|^{2}\langle v_{1},v_{1}\rangle + \sum_{j\neq 1}a_{1}\overline{a_{j}}\langle v_{1},v_{j}\rangle + \ldots + |a_{n}|^{2}\langle v_{n},v_{n}\rangle + \sum_{j\neq n}a_{n}\overline{a_{j}}\langle v_{n},v_{j}\rangle\\\\
& = |a_{1}|^{2}\langle v_{1},v_{1}\rangle + |a_{2}|^{2}\langle v_{2},v_{2}\rangle + \ldots + |a_{n}|^{2}\langle v_{n},v_{n}\rangle\\\\
& = |a_{1}|^{2}\|v_{1}\|^{2} + |a_{2}|^{2}\|v_{2}\|^{2} + \ldots + |a_{n}|^{2}\|v_{n}\|^{2}
\end{align*}
as desired. In order to prove it, it has been used the fact that $\langle v_{i},v_{j}\rangle = 0$ for $i\neq j$ as well as the sesquilinearity of the inner product. Hopefully this helps.
