By definition we have for all $x_1,\ldots,x_n\in\C$: $$ \sum_{j,k}a_{jk}x_j\bar x_k\geq0 \quad\mbox{and by conjugation:}\quad \sum_{j,k}\bar a_{jk}\bar x_jx_k\geq0~. $$ Since $x_1,\ldots,x_n\in\C$ are arbitarary, $\bar A$ is positive definite. 2. Let $b_1,\ldots,b_n$ be an ONB for $\C^n$ such that for some $\l_j\in\R_0^+$: $Bb_j=\l_jb_j$. Then $$ \tr(AB) =\sum_j\la ABb_j,b_j\ra =\sum_j\l_j\la Ab_j,b_j\ra \in\R_0^+ $$ 3. If $\tr(AB)=0$ then for all $j$: $\l_j\la Ab_j,b_j\ra=0$. If $\la Ab_j,b_j\ra=0$ for all $j$, then $\tr(A)=0$, which can only occure if all eigen-values of $A$ vanish, i.e. $A=0$.