Show that a matrix $A \in M(n \times n, \mathbb{C})$ is hermitian iff $v^tA\overline{v} \in \mathbb{R}$ for all $v \in \mathbb{C}^n$.
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Hint: I think you need nothing more than knowing that $x=v^\intercal A\bar{v}$ is a scalar, whence $x^\intercal=x$. A number $x$ is real if $\bar{x}=x$.
EDIT: More thoroughly, $A$ is Hermitian iff $\overline{A^\intercal}=A$. Then $\overline{x}=\overline{x^\intercal}=\overline{\bar{v}^\intercal A^\intercal v}=v^\intercal\overline{A^\intercal}\bar{v}=v^\intercal A\bar{v}=x$. Hence $x$ is real.
Conversely, if $\bar{x}=x$, then $v^\intercal A\bar{v}=\bar{v}^\intercal A^\intercal v=v^\intercal \overline{A^\intercal}\bar{v}$, whence $\overline{A^\intercal}=A$ since the relation holds for all $v$.
mathse
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I think it will more relevant if you explain why $\bar{v}^{t}Bv=0 \Longrightarrow B=0$. – user10676 May 14 '14 at 14:20
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@user10676 But this implication, that $B=0$, does not hold. Or you mean $B=A-\overline{A^\intercal}$? – mathse May 14 '14 at 14:30
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1Links to external resources are encouraged, but please add context around the link so your fellow users will have some idea what it is and why it’s there. Always quote the most relevant part of an important link, in case the target site is unreachable or goes permanently offline. – Najib Idrissi May 14 '14 at 13:53