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?
Best Answer
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.