Proof that $\| u \|=\| v\|\iff\langle u+v,u-v\rangle=0$

Question

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

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$u,v,w\in\mathbb{R}^n;\; \lambda\in\mathbb{R};\; \langle u;v\rangle:=(u_1v_1)+\cdots+(u_nv_n)$; $\| u \|:=\sqrt{u_1^2+\cdots +u_n^2}$ \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}

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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 $\langle u+v,u-v\rangle=0$

"$\Leftarrow$"
\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}

Answer 1

Hint:

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.

Answer 2

$\sum u_i^{2}=\sum v_i^{2}$ 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 $\|u\|=\|v\|$.

Answer 3

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.