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 $\sum_{i=1}^k |a_i|^2 ||v_i||^2$?
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.