Let $V$ be a vector space over $\mathbb{C}$ equipped with a mapping $\langle , \rangle \: V \times V \to \mathbb{C}$ such that
-
$\langle , \rangle$ is linear in the first argument and conjugate linear in the second argument.
-
$\langle , \rangle$ 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?
Best Answer
$ \newcommand\form[1]{\langle#1\rangle} \newcommand\conj\overline $
Yes, conjugate-symmetry does in fact follow.
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.