# Positive definiteness in an inner product over complex numbers

From Wikipedia

In this article, the field of scalars denoted F is either the field of real numbers $$\mathbb{R}$$ or the field of complex numbers $$\mathbb{C}$$.

Formally, an inner product space is a vector space V over the field F together with an inner product, i.e., with a map

$${\displaystyle \langle \cdot ,\cdot \rangle :V\times V\to F}$$ that satisfies the following three axioms for all vectors $$x$$, $$y$$, $$z ∈ V$$ and all scalars $$a ∈ F$$

$$\cdots$$

Positive-definiteness: {\displaystyle {\begin{aligned}\langle x,x\rangle &\geq 0\\\langle x,x\rangle &=0\Leftrightarrow x=\mathbf {0} \,.\end{aligned}}}

If $$F = \mathbb{C}$$, then $$\langle \cdot ,\cdot \rangle :V\times V\to \mathbb{C}$$, so $$\langle x,x \rangle$$ could be non-real. But then $$\langle x,x \rangle \geq 0$$ is meaningless because there is no order on $$\mathbb{C}$$.

How is the positive-definiteness condition meaningful in the case $$F = \mathbb{C}$$?

Conjugate symmetry: $${\displaystyle \langle x,y\rangle ={\overline {\langle y,x\rangle }}}$$
it follows that $$\langle x,x \rangle$$ is real for all $$x$$. Hence $$\langle x,x \rangle \geq 0$$ is meaningful.
• Because ${\displaystyle \langle x,x\rangle ={\overline {\langle x,x\rangle }} \implies \operatorname{Im} \, \langle x,x\rangle } = 0$. Oct 31, 2020 at 6:41