# When is conjugate symmetry of the inner product implied by other conditions?

Let $$V$$ be a vector space over $$\mathbb{C}$$ equipped with a mapping $$\langle , \rangle \: V \times V \to \mathbb{C}$$ such that

1. $$\langle , \rangle$$ is linear in the first argument and conjugate linear in the second argument.

2. $$\langle , \rangle$$ is positive definite in that $$\langle v , v\rangle \geq 0$$ for all $$v \in V$$ with $$\langle v , v\rangle=0$$ iff $$v=0$$.

Does the above two assumptions make $$\langle , \rangle$$ a genuine inner product? That is, conjugate symmetry $$\langle v , w\rangle=\overline{\langle w , v\rangle}$$ is implied by the above two assumptions?

It is always confusing...Could anyone please clarify?

$$\newcommand\form[1]{\langle#1\rangle} \newcommand\conj\overline$$
Every sesquilinear form $$\form{{-},{-}}$$ (i.e., linear in the first argument and conjugate-linear in the second argument) can be decomposed as $$\form{v, w} = \form{v, w}_+ + i\form{v, w}_-,$$$$\form{v, w}_+ = \frac12(\form{v, w} + \conj{\form{w, v}}),\quad \form{v, w}_- = \frac1{2i}(\form{v, w} - \conj{\form{w, v}}),$$ where $$\form{{-},{-}}_\pm$$ are both conjugate-symmetric sesquilinear forms. $$\form{{-},{-}}$$ is then positive-definite iff $$\form{{-},{-}}_+$$ is positive-definite and $$\form{v,v}_- = 0$$ for all $$v$$. This latter condition is equivalent to $$0 = \form{v+w,v+w}_- = \form{v, w}_- + \form{w, v}_-$$$$\implies \form{v, w}_- = -\conj{\form{v, w}_-},$$ meaning $$\form{{-},{-}}_-$$ is always imaginary-valued. But then $$i\form{v,w} = \form{iv,w} = 0$$ since it must be both real and imaginary, so in fact $$\form{v,w} = \form{v,w}_+$$ and every positive-definite sesquilinear form is conjugate-symmetric. More generally we have shown that $$\form{v,v} \in \mathbb R$$ for $$v$$ implies conjugate-symmetry.