Vtyjkyuk

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 $v_iin v_1 ,v_2 ,...,v_k$ such that $<v_i,v_i>neq0$. By the properties $||x||=sqrt<x,x>$ and $overlinesum_ell=1^n x_ell=sum_ell=1^n overlinex_ell$ ('the conjugate of a sum is the sum of the conjugates'), we have $$||sum_i=1^k a_i v_i||^2 =<sum_i=1^k a_i v_i,sum_i=1^k a_i v_i>$$
$$=sum_i=1^k a_i sum_i=1^k overlinea_i <v_i,v_i>.$$ Note that as each $sum_i=1^k a_i ,sum_i=1^k overlinea_i$ 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, $|sum_i=1^k a_i|^2neq sum_i=1^k |a_i|^2$ (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?

You should treat expression $leftlangle sum^k_i=1 a_iv_i , sum^k_i=1 a_iv_irightrangle$ more carefully:

$$
leftlangle sum^k_i=1 a_iv_i , sum^k_i=1 a_iv_irightrangle =
sum^k_i=1 leftlangle a_iv_i, sum^k_j=1 a_j v_j rightrangle =
sum^k_i=1 a_i leftlangle v_i, sum^k_j=1 a_jv_j rightrangle.
$$

and then each

$$
leftlangle v_i, sum^k_j=1 a_jv_j rightrangle = sum^k_j=1 bar a_j leftlangle v_i,v_j rightrangle
$$

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

$$
leftlangle sum^k_i=1 a_iv_i , sum^k_i=1 a_iv_irightrangle = sum^k_i=1 |a_i|^2| v_i|^2
$$

follows.

Note that

$left<sum^k_i=1a_i v_i,sum^k_j=1a_j v_j right>= sum_i=1^ksum_j=1^ka_ioverlinea_jleft< v_i,v_j right>$

Also note that, for each $i$,

$sum_j=1^ka_ioverlinea_jleft<v_i,v_j right>=a_ioverlinea_ileft< v_i,v_iright> $ since $left<v_i,v_j right>=0$ if $ineq j$.

Thus,

$left<sum^k_i=1a_i v_i,sum^k_j=1a_j v_j right>= sum_i=1^ksum_j=1^ka_ioverlinea_jleft< v_i,v_j right>= sum_i=1^ka_ioverlinea_ileft<v_i,v_i right>=sum_i=1^k|a_i|^2left| v_i right|^2$

I hope it helps :)