NUS-MATH530 Some 6.A Problems
Suppose \mathbb{F} = \mathbb{R}, and V \neq \{0\}. Replace the positivity condition with the condition that \langle v,v \rangle > 0 for some v \in V. Show that this change in definition does not change the set of functions from V \times V to \mathbb{R} that are inner products on V.
We hope to show that \langle v,v \rangle >0 for some v \in V implies that \langle v,v \rangle \geq 0 for all v \in V in real vector spaces.
Take some v_0 \in V such that \langle v_0,v_0 \rangle >0. Now, WLOG let v \in V and v = v_0+w. So:
\begin{align} 0 &< \langle v_0,v_0 \rangle \\ &= \langle v-w, v-w \rangle \\ &= \langle v,v \rangle + \langle w,w \rangle - 2\langle v,w \rangle \end{align}
Now, the last step is possible because symmetry becomes conjugate symmetry in reals.
We now have that:
\begin{equation} 2 \langle v,w \rangle - \langle w,w \rangle < \langle v,v \rangle \end{equation}