2
$\begingroup$

Let's consider inner product space with vectors $x, y, z$ which satisfies:

$$\|x+y+z\|^2 = 14$$

$$\|x+y-z\|^2 = 2$$

$$\|x-y+z\|^2 = 6$$

$$\|x-y-z\|^2 = 10$$

I want to prove that $x$ is perpendicular to $y$.

My work so far

In other words we want to prove that $\langle x, y \rangle = 0$.

My first idea was to use Cauchy Schwarz inequality $ | \langle x , y \rangle |\le \|x\| \|y\|$ and to show that our conditions force that $\|x\|\cdot\|y\| = 0$. However I didn't manage to do anything sensible. Also I tried to somehow prove that under this conditions our norm has bo inducted by inner product - in other words we have that: $$2(\|x\|^2 + \|y\|^2) = \|x+y\|^2 + \|x-y\|^2$$

but also I didn't end up with something rational. Could you please give me a hand, what's the correct approach to this problem?

$\endgroup$
1
  • 3
    $\begingroup$ Using Cauchy-Schwarz is too strong, the condition $\|x\|\|y\|=0$ means either $x$ or $y$ must be $0$ and this is stronger than just $x$ being orthogonal to $y$. $\endgroup$ – TSF Jun 23 at 10:26
5
$\begingroup$

I assume you work with a real and not a complex vector space. Is that true?

Add the first and second one: by the parallelogram identity you get $$2(\|x+y\|^2+\|z\|^2)=16$$ so $\|x+y\|^2+\|z\|^2=8$. Now add the third and the fourth one and in the same way you get $\|x-y\|^2+\|z\|^2=8$. So $\|x+y\|^2+\|z\|^2=\|x-y\|^2+\|z\|^2$, so $$\|x-y\|^2=\|x+y\|^2$$ so $\|x\|^2+\|y\|^2-2\langle x,y\rangle=\|x\|^2+\|y\|^2+2\langle x,y\rangle$, so $\langle x,y\rangle=0$.

$\endgroup$
9
  • 1
    $\begingroup$ who downvoted and why? $\endgroup$ – JustDroppedIn Jun 23 at 10:33
  • $\begingroup$ How do you know that $x+y$ and $z$ are perpendicular to each other (assumption of pythagorean theorem)? I didn't downvote - just to add ;)) $\endgroup$ – Lucian Jun 23 at 10:33
  • $\begingroup$ @Lucian I mean the pythagorean theorem in its generalized form, as mentioned by OP: $\|x+y\|^2+\|x-y\|^2=2\|x\|^2+2\|y\|^2$. No orthogonality is assumed for this. $\endgroup$ – JustDroppedIn Jun 23 at 10:33
  • 2
    $\begingroup$ I've usually heard that called the parallelogram identity $\endgroup$ – Alan Jun 23 at 10:34
  • 1
    $\begingroup$ @Lucian An inner product space, by definition, is a vector space with an inner product that induces a norm $\endgroup$ – Alan Jun 23 at 10:36
3
$\begingroup$

Assuming that you are working over $\Bbb R$, let:

  • $s=\|x\|^2+\|y\|^2+\|z\|^2$;
  • $a=2\langle x,y\rangle$;
  • $b=2\langle x,z\rangle$;
  • $c=2\langle y,z\rangle$.

Then those four equalities tell you that$$\left\{\begin{array}{l}s+a+b+c=14\\s+a-b-c=2\\s-a+b-c=6\\s-a-b+c=10.\end{array}\right.$$In particular,$$\begin{split}(s+a+b+c)+(s+a-b-c)-(s-a+b-c)-(s-a-b+c) &=14+2-6-10 \\ &=0 \end{split}$$In other words, $4a=0$. But $a=2\langle x,y\rangle$.

$\endgroup$
2
  • $\begingroup$ this answer also assumes that the vector space is over the field of real numbers $\endgroup$ – JustDroppedIn Jun 23 at 10:44
  • $\begingroup$ @JustDroppedIn Indeed. I've added that assumption to my answer. $\endgroup$ – José Carlos Santos Jun 23 at 10:47

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.