5
$\begingroup$

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?

$\endgroup$
  • 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
2
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 :)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.