# Uniqueness of a complex inner product, given a positive definite quadratic real-valued function

Let $$V$$ be a finite-dimensional complex vector space. Then we can regard $$V$$ also as a real vector space; let $$V_\Bbb R$$ denote the real vector space. Suppose $$\mu:V_\Bbb R\to \Bbb R$$ is a positive definite ($$\mu(v)>0$$ for all nonzero $$v\in V_\Bbb R$$) quadratic function, and suppose $$\mu(iv)=\mu(v)$$ for all $$v\in V$$. I want to show that there is a unique complex inner product $$V\times V\to \Bbb C$$ satisfying $$\langle v,v\rangle =\mu(v)$$. I have shown existence: define $$\langle v,w\rangle=\frac{1}{2}(\mu(v+w)-\mu(v)-\mu(w))+\frac{i}{2}(\mu(v+iw)-\mu(v)-\mu(iw)).$$ Then a straightforward (but little bit long) computation shows that this is indeed a complex inner product. But how can we show uniqueness?

Assuming your formula is correct: suppose that $$\langle \cdot,\cdot\rangle_1$$ and $$\langle \cdot, \cdot \rangle_2$$ induce the same quadratic form $$\mu$$. It follows that for all $$u,v$$, we have $$\langle v,w\rangle_1 = \langle v,w \rangle_2 = \frac{1}{2}(\mu(v+w)-\mu(v)-\mu(w))+\frac{i}{2}(\mu(v+iw)-\mu(v)-\mu(iw)).$$