Let $(\cdot,\cdot):\mathbb{R}^n\times \mathbb{R}^n\to\mathbb{R}^n$ be any map that fulfills the properties

$u,v,w\in\mathbb{R}^n;\; \lambda\in\mathbb{R};\; \langle u;v\rangle:=(u_1v_1)+\cdots+(u_nv_n)$; $\| u \|:=\sqrt{u_1^2+\cdots +u_n^2}$ \begin{align} \langle u,v+w\rangle&=\langle u,v\rangle+\langle u,w\rangle\\ \langle u,\lambda v\rangle&=\lambda\langle u,v\rangle=\langle \lambda u,v\rangle\\ \langle u,v\rangle&=\langle v,u\rangle\\ \langle u,u\rangle&>0 \qquad \forall u\neq 0 \end{align}

Also be $\| w\|:=\sqrt{\langle w,w\rangle};\; \forall w\in\mathbb{R}^n$. Show that the following statements apply to all $u,v\in\mathbb{R}^n$

1) $\| u+v\|^2 + \| u-v\|^2=2\| u\|^2+2\| v\|^2$ This one is clear, I was already able to prove it.

2) $\| u \|=\| v \|$ iff $\langle u+v,u-v\rangle=0$

\begin{align} \langle u+v,u-v\rangle &=\langle u_1+v_1,\cdots ,u_n+v_n;u_1-v_1,\cdots u_n-v_n\rangle&=0\\ &=(u_1+v_1)(u_1-v_1)+\cdots +(u_n+v_n)(u_n-v_n)&=0\\ &=(u_1^2-v_1^2)+\cdots +(u_n^2-v_n^2)&=0 \end{align} How do I go on? I know that this equation only holds for ${\| u \|} = {\| v \|}$ or $u=0=v$ which is in fact $\| u \|^2 =\| v\|^2$, which is what I need to show. The other way around is easier:

"$\Rightarrow$" \begin{align} \| u \|^2 = \| v\|^2&=\left(\sqrt{u_1^2+\cdots +u_n^2}\right)^2=\left(\sqrt{v_1^2+\cdots +v_n^2}\right)^2\\ &=u_1^2+\cdots +u_n^2=v_1^2+\cdots +v_n^2\\ &\implies \mid u_i \mid=\mid v_i\mid\qquad \forall i\in\{1,\cdots ,n\}\\ &\implies u-v=0; \text{ Let } u+v =w \text{ with } w\in\mathbb{R}^n\\ &\implies\langle u+v,u-v\rangle=\langle w,0\rangle\\ &=w_10+\cdots +w_20\\ &=0 \end{align}

  • $\begingroup$ I don't think 2) is true... Take for example $v=0$. $\endgroup$ – Pink Panther Apr 19 at 11:56
  • $\begingroup$ @PinkPanther If $v=0$ it becomes $\|u\|=0\iff \left<u,u\right>=0$ which is true. $\endgroup$ – Yanko Apr 19 at 11:58
  • $\begingroup$ oh right, i don't know what i was thinking there... $\endgroup$ – Pink Panther Apr 19 at 12:03


You can check, using bilinearity and symmetry, that $$\langle u+v,u-v\rangle = \langle u,u\rangle-\langle v,v\rangle=\|u\|^2-\|v\|^2. $$ Note this works even for non-finite dimensional vector spaces.

  • $\begingroup$ Could I write: $\begin{align} \langle u+v,u-v\rangle &=\langle u_1+v_1,\cdots ,u_n+v_n;u_1-v_1,\cdots u_n-v_n\rangle&=0\\ &=(u_1+v_1)(u_1-v_1)+\cdots +(u_n+v_n)(u_n-v_n)&=0\\ &=(u_1^2-v_1^2)+\cdots +(u_n^2-v_n^2)&=0\\&=\langle u,u\rangle -\langle v,v\rangle&=0\\ \end{align}$ $\endgroup$ – Doesbaddel Apr 19 at 13:14
  • $\begingroup$ or do this needs to be more detailed? $\endgroup$ – Doesbaddel Apr 19 at 13:21
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    $\begingroup$ You don't need to use coordinates (it is true even in the infinite dimensional case). Just expand the l.h.s. by bilinearity. $\endgroup$ – Bernard Apr 19 at 13:23
  • $\begingroup$ Oh I understood that now. Thank you so much! You mean, that $\langle u+v, u-v\rangle =(u+v)(u-v)=u^2-uv+uv-v^2=u^2-v^2\iff u^2=v^2 \implies \|\| u\|\|^2=\|\| v\|\|^2\implies \|\| u\|\|=\|\| v\|\|$ right? $\endgroup$ – Doesbaddel Apr 19 at 13:36
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    $\begingroup$ Almost. You forgot that what you denote $u^2$ is just $\langle u,u\rangle$, in other words it is $\|u\|^2$, so it's even shorter. $\endgroup$ – Bernard Apr 19 at 13:38

It would be simpler to prove or use the fact that inner products are sesquilinear. That is to say, $\langle a+b,c\rangle = \langle a,c\rangle + \langle b,c\rangle$, that $\langle a,b+c\rangle = \langle a,b\rangle + \langle a, c\rangle$ and that $\langle a,b\rangle = \overline{\langle b,a\rangle}$

We have as a result:

$\langle u+v, u-v\rangle = \langle u+v,u\rangle - \langle u+v,v\rangle = \langle u,u\rangle + \langle v,u\rangle - \langle u,v\rangle - \langle v,v\rangle$

$ = \|u\|^2-\|v\|^2 + \left(\langle v,u\rangle - \overline{\langle v,u\rangle}\right) = \|u\|^2 - \|v\|^2$

where here we notice that since we are working in $\Bbb R^n$ you have $\langle v,u\rangle$ is real and so taking the complex conjugate of it doesn't change the value.

  • $\begingroup$ There is a typo at the end. $\endgroup$ – Kavi Rama Murthy Apr 19 at 12:00
  • $\begingroup$ You have typed $\|u\|^{2}+\|v\|^{2}$ instead of $\|u\|^{2}-\|v\|^{2}$ $\endgroup$ – Kavi Rama Murthy Apr 19 at 12:02
  • $\begingroup$ @KaviRamaMurthy Ah, yes. Of course. Thanks for the correction. $\endgroup$ – JMoravitz Apr 19 at 12:02

$\sum u_i^{2}=\sum v_i^{2}$ does not imply that $u_i=v_i$ for all $i$ so your argument doesn't work even for the 'easy' part.

From your calculation we have $ \langle (u+v),(u-v) \rangle =\sum u_i^{2}-\sum v_i^{2}$. This is $0$ iff $\sum u_i^{2}=\sum v_i^{2}$. But this is same as $\|u\|^{2}=\|v\|^{2}$ or $\|u\|=\|v\|$.

  • $\begingroup$ Sry, I wanted to write $\mid u \mid = \mid v\mid$ Well, the argument won't work anyways, you're right. $\endgroup$ – Doesbaddel Apr 19 at 11:53

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