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?

  • 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

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 $\langle x,y\rangle=0$.

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    $\begingroup$ 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
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    $\begingroup$ I've usually heard that called the parallelogram identity $\endgroup$ – Alan Jun 23 at 10:34
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    $\begingroup$ @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

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 $a=2\langle x,y\rangle$.

  • $\begingroup$ 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

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