In an inner product space over $\mathbb R$, prove $ (u,w)=0 \Leftrightarrow \left \| u+w \right \|=\left \| u-w \right \| $

Question

Let $V$ be an inner product space over field $F$ and $u,w\in V$.
Prove that if $F=\mathbb{R}$ then:
$$ (u,w)=0 \Leftrightarrow \left \| u+w \right \|=\left \| u-w \right \| $$
Is it also true for $F=\mathbb{C}$ ?

Answer 1

Hint: Try expanding $\|u + w\|^2 = \langle u + w, u + w \rangle$ and $\|u - w\|^2 = \langle u - w, u - w \rangle$. Now notice how some terms cancel out. Does this cancelation also happen in $\mathbb C$?

---

By following the hint, one gets: $$ \|u + w\| = \|u - w\| \iff \langle u, w \rangle + \overline{\langle u, w\rangle} = 0 $$

Since $\overline{\langle u, w\rangle} = \langle u, w\rangle$ in $F = \mathbb R$, this is equivalent to $\langle u, w\rangle = 0$.

On the other hand, if $F = \mathbb C$, then $u = (1, 2)$, $w = (2i, i)$ is a counterexample. $\|u + w\| = \sqrt{10} = \|u - w\|$. Yet, $\langle u, w \rangle = -4i \ne 0$.

Answer 2

Hint: $$ ||u+v||^2 = \langle u+v, u+v \rangle = \langle u, u \rangle+\langle u, v \rangle+\langle v, u \rangle+\langle v, v \rangle$$

$$=||u||^2+0+0+||v||^2, $$

since $\langle u, v \rangle=0$