A Hermitian form $\langle,\rangle$ on a vector space over the complex field $\mathbb{C}$ is a function $$\langle\cdot,\cdot\rangle: V \times V \rightarrow \mathbb{C}$$

that satisfies conditions that need not be listed for the purpose of my question. A Hermitian form is positive-definite if for every $v \neq 0 \in V$, $$\langle v,v\rangle > 0$$

If a Hermitian form returns a complex number with an imaginary part, do we only look at the real part to see if the Hermitian is positive-definite? I'm under the impression that positive and negative are only defined only on the real line.

  • 3
    $\begingroup$ 1. Usually when, for a $z\in\mathbb{C}$, one writes $z>0$ it means "$z\in\mathbb{R}$ and $z>0$". 2. Anyway, in this case it can be show that $\langle v,v\rangle\in\mathbb{R}$, for all $v$. $\endgroup$ – Veridian Dynamics Nov 30 '17 at 4:24
  • $\begingroup$ The usual Hermitian product on $\Bbb C^n$, that is $\sum_i z_i\overline{w_i}$ manages quite easily to be positive definite. $\endgroup$ – Lord Shark the Unknown Nov 30 '17 at 4:43
  • $\begingroup$ @VeridianDynamics is right. The "conditions that need not be listed" are in fact extremely relevant to understanding why ;) $\endgroup$ – user7530 Nov 30 '17 at 6:47

$\langle v,v\rangle >0$ is just a statement about real numbers. One of the two axioms for a Hermitian form is $$\langle u,v\rangle=\overline{\langle v,u\rangle}.$$ For $u=v$ this implies $$\langle v,v\rangle=\overline{\langle v,v\rangle}\in\mathbb R$$ since a complex number equal to its own conjugate must be real.


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