Let $M_n(\mathbb{C})$ the matrix space, $\tau:M_n(\mathbb{C})\to\mathbb{C}$ defined by $$ \tau(A)=trace(AB) $$ We consider $M_n(\mathbb{C})$ with norm $\|A\|^2=\sup\{|\lambda|\in\mathbb{C}:\lambda$ is a eigenvalue of $A^*A\}$ (in other words, a norm of $C^*$-algebra).
We need resolve two problems:
$\|\tau\|=trace(|B|)$; (where $|B|=\sqrt{B^*B}$)
$\tau$ is positive iff $B$ is positive. (where $\tau$ is positive if $\tau(A^*A)\geqslant0$ for all $A\in M_n(\mathbb{C})$ and $B$ is positive if $B=C^*C$ for any $B\in M_n(\mathbb{C}$).
I was studying by Murphy's book of Operator Algebras, but I didn't found a way to solve this problem. I also did a counts, but it was long and useless. If there is any idea to start...
--- EDIT ---
I first wanna to thank Martin Argerami for awnser, the part 2. to me it's ok, but part 1. I know a awnser that convinced me:
Define $\widetilde{\tau}$ by $\widetilde{\tau}(A)=\operatorname{tr}(A|B|)$. Consider the polar form of $B$ by $B=U|B|$, where $U$ is a isometry (not only partial, since is a matrix $n\times n$). Hence, \begin{align*} |\tau(A)| =|\operatorname{tr}(AB)| =|\operatorname{tr}(AU|B|)| =|\widetilde{\tau}(AU)| \leqslant\|\widetilde{\tau}\|\|AU\| \leqslant\|\widetilde{\tau}\|\|A\|\|U\| \leqslant\|\widetilde{\tau}\|\|A\|. \end{align*} Let the sup of matrices of norm $\leqslant1$, we get \begin{align*} \|\tau\|\leqslant\|\widetilde{\tau}\|. \end{align*} But, by (ii), we know that $\widetilde{\tau}$ é positive, since $|B|$ is positive. Also, we have that $\|\widetilde{\tau}\|=\widetilde{\tau}(I)$, since $\widetilde{\tau}$ is positive and $M_n(\mathbb{C})$ is unital. Hence, \begin{align*} \|\tau\|\leqslant\|\widetilde{\tau}\|=\widetilde{\tau}(I)=\operatorname{tr}(I|B|)=\operatorname{tr}(|B|). \end{align*} By other hand, $U^*$ is a matrix of norm $\leqslant1$ such that \begin{align*} \tau(U^*)=\operatorname{tr}(U^*B)=\operatorname{tr}(|B|). \end{align*} Therefore, $\|\tau\|=\tau(U^*)=\operatorname{tr}(|B|)$.
You have, using Hölder's Inequality for the trace $$ |\tau(A)|=|\operatorname{tr}(AB)|\leq\|A\|\,\operatorname{tr}(|B|). $$ So $\|\tau\|\leq\operatorname{tr}(|B|)$. If write $B=V|B|$ the polar decomposition, then you have $|B|=V^*B$. As $V$ is a partial isometry $\|V\|=1$, and so $$ \tau(V^*)=\operatorname{tr}(V^*B)=\operatorname{tr}(|B|). $$ Hence $\|\tau\|=\tau(|B|)$.
If $B=C^*C$, and $A=D^*D$, then $$ \tau(A)=\operatorname{tr}(D^*DC^*C)=\operatorname{tr}(CD^*DC^*)=\operatorname{tr}((DC^*)^*DC^*)\geq0. $$ Conversely, if $\tau(A)\geq0$ for all $A$ positive, fix any unit vector $x\in\mathbb C^n$. Let $P$ be the rank-one projection onto the span of $x$. Then $$ \langle Bx,x\rangle=x^*Bx=\operatorname{tr}(PBP)=\operatorname{tr}(BP)=\tau(P)\geq0. $$ Then $B$ is selfadjoint and $B$ positive.