TITLE: How to Express the Summation of an Inner Product Space
QUESTION [5 upvotes]: I'm having an issue with the following problem from Ch. 6.1 (Inner Product Spaces and Norms) of Friedberg's Linear Algebra, 4th ed.

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

My attempt at a solution:
$"$Recall that since $\{v_1 ,v_2 ,...,v_k\}$ is orthogonal, any two distinct vectors $v_1 ,v_j \in \{v_1 ,v_2 ,...,v_k\}$ would form the equality $<v_i ,v_j>=0$.  Let us pick a vector $v_i\in \{v_1 ,v_2 ,...,v_k\}$ such that $<v_i,v_i>\neq0$.  By the properties $||x||=\sqrt{<x,x>}$ and $\overline{\sum_{\ell=1}^n x_\ell}=\sum_{\ell=1}^n \overline{x_\ell}$ ('the conjugate of a sum is the sum of the conjugates'), we have $$||\sum_{i=1}^k a_i v_i||^{2} =<\sum_{i=1}^k a_i v_i,\sum_{i=1}^k a_i v_i>$$
$$=\sum_{i=1}^k a_i \sum_{i=1}^k \overline{a_i} <v_i,v_i>.$$ Note that as each $\sum_{i=1}^k a_i ,\sum_{i=1}^k \overline{a_i}$ simply denotes a scalar, we can use the properties $<cx,y>=c<x,y>$ and $<x,cy>=\bar c <x,y>."$

At this point I am basically stuck.  If I make the argument... $$\sum_{i=1}^k a_i \sum_{i=1}^k \bar a_i=(a_1 + a_2 +...+ a_k)(\bar a_1 + \bar a_2 +...+\bar a_k)$$
$$=(a_1 + a_2 +...+ a_k) \overline{(a_1 + a_2 +...+ a_k)}$$
$$=|a_1 + a_2 +...+ a_k|^{2}$$
$$=|\sum_{i=1}^k a_i|^{2}$$
... I get no further, because to the best of my knowledge, $|\sum_{i=1}^k a_i|^{2}\neq \sum_{i=1}^k |a_i|^{2}$ (because of the triangle inequality).  I feel I am probably making a few fatal errors throughout this proof.  Could anybody knowledgeable in Linear Algebra please shed some light for me?

REPLY [3 votes]: You should treat expression $\left\langle \sum^k_{i=1} a_iv_i , \sum^k_{i=1} a_iv_i\right\rangle$ more carefully:
$$
 \left\langle \sum^k_{i=1} a_iv_i , \sum^k_{i=1} a_iv_i\right\rangle = 
 \sum^k_{i=1} \left\langle a_iv_i, \sum^k_{j=1} a_j v_j \right\rangle = 
 \sum^k_{i=1} a_i \left\langle v_i, \sum^k_{j=1} a_jv_j \right\rangle.
$$
and then each
$$
   \left\langle v_i, \sum^k_{j=1} a_jv_j \right\rangle = \sum^k_{j=1} \bar a_j \left\langle v_i,v_j  \right\rangle 
$$
By properties of inner product you know.
As vectors are orthogonal last expression simplifies to $\bar a_i \| v_i \|^2$ and so the result
$$
 \left\langle \sum^k_{i=1} a_iv_i , \sum^k_{i=1} a_iv_i\right\rangle = \sum^k_{i=1} |a_i|^2\| v_i\|^2
$$
follows.

REPLY [2 votes]: Note that
$\left<\sum^{k}_{i=1}a_i v_i,\sum^{k}_{j=1}a_j v_j  \right>= \sum_{i=1}^{k}\sum_{j=1}^{k}a_i\overline{a_j}\left< v_i,v_j \right>$
Also note that, for each $i$,
$\sum_{j=1}^{k}a_i\overline{a_j}\left<v_i,v_j  \right>=a_i\overline{a_i}\left< v_i,v_i\right> $  since $\left<v_i,v_j  \right>=0$  if  $i\neq j$.
Thus,
$\left<\sum^{k}_{i=1}a_i v_i,\sum^{k}_{j=1}a_j v_j  \right>= \sum_{i=1}^{k}\sum_{j=1}^{k}a_i\overline{a_j}\left< v_i,v_j \right>= \sum_{i=1}^{k}a_i\overline{a_i}\left<v_i,v_i \right>=\sum_{i=1}^{k}|a_i|^2\left\| v_i \right\|^2$
I hope it helps :)