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?

  • 1
    $\begingroup$ You have one algebraic mistake: it should be $\langle \sum_{i=1} a_iv, \sum_{i = 1} a_iv \rangle = \sum_{i=1} \sum_{j=1} a_i \bar a_j \langle v_i, v_j \rangle$. Think about how to apply properties of inner product iteratively. $\endgroup$ – Nik Pronko Aug 7 '18 at 17:38
  • $\begingroup$ @NikPronko: Thank you. Why though do you employ different subscripts, especially when orthogonality is an issue? If vi=/=vj, then wouldn't <vi,vj>=0, causing the entire expression to equal zero? Anyways, that is the assumption I proceeded with, possibly causing many errors. $\endgroup$ – greycatbird Aug 7 '18 at 17:46
  • $\begingroup$ This is true, but if you work out this sum just for general vectors it would be evedent that $\langle v_i, v_i\rangle$ are multiplied by $a_i \bar a_j$ and not by the product of sums. Your reasoning is essentially correct. $\endgroup$ – Nik Pronko Aug 7 '18 at 17:53

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

$$ \left\langle \sum^k_{i=1} a_iv_i , \sum^k_{i=1} a_iv_i\right\rangle = \sum^k_{i=1} |a_i|^2\| v_i\|^2 $$



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 $i\neq j$.


$\left<\sum^{k}_{i=1}a_i v_i,\sum^{k}_{j=1}a_j v_j \right>= \sum_{i=1}^{k}\sum_{j=1}^{k}a_i\overline{a_j}\left< v_i,v_j \right>= \sum_{i=1}^{k}a_i\overline{a_i}\left<v_i,v_i \right>=\sum_{i=1}^{k}|a_i|^2\left\| v_i \right\|^2$

I hope it helps :)


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