2
$\begingroup$

Let $v$ and $w$ be elements of an inner product space. Prove that $\|v + w\|^2 = \|v\|^2 + \|w\|^2$ iff $v,w$ are orthogonal.

I know that if $v,w$ are orthogonal they are linearly independent. Now if I suppose $v_1 + ...+ v_k = 0$ and $w_1 + ... +w_k = 0$

I have that:

$$0 = \langle v_1 + ... + v_k, v_i\rangle = \langle v_1,v_i\rangle + ... + \langle v_k,v_i\rangle = \langle v_i,v_i\rangle = \|v_i\|^2$$

and this is where I am stuck because it doesn't complete the proof.

$\endgroup$

2 Answers 2

3
$\begingroup$

You are using it wrong. You don't care about linear independence, because there are linearly independent vectors which are not orthogonal. So it's not enough to use that.

Instead we can use the orthogonality directly, recall that $\|u\|^2=\langle u,u\rangle$, and now we have: $$\langle v+w,v+w\rangle = \langle v,v\rangle+\langle v,w\rangle+\langle w,v\rangle+\langle w,w\rangle$$

The assumption of orthogonality means that $\langle v,w\rangle =0$, and the conclusions follow.

$\endgroup$
1
  • $\begingroup$ Ah, of course! I made it more complicated then it really was. Thank you very much, Asaf!Toda Raba! $\endgroup$
    – diimension
    Oct 21, 2012 at 2:25
3
$\begingroup$

Use the fact that $ <u,v>=0 $, and notice that

$$||u+v||^2 =<u+v,u+v>=<u,u>+<u,v>+<v,u>+<v,v>=\dots$$

$\endgroup$
2
  • $\begingroup$ Of course, thank you very much, Mhenni! $\endgroup$
    – diimension
    Oct 21, 2012 at 2:25
  • 1
    $\begingroup$ @diimension: You are welcome. $\endgroup$ Oct 21, 2012 at 2:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .