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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 $F = \mathbb{C}$?

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Because of one of the conditions on an inner product I didn't include in the original post

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.

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  • $\begingroup$ Because ${\displaystyle \langle x,x\rangle ={\overline {\langle x,x\rangle }} \implies \operatorname{Im} \, \langle x,x\rangle } = 0$. $\endgroup$ – Mateen Ulhaq Oct 31 '20 at 6:41

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