Consider first general matrices in $\mathbb{C}^{n \times m}$. Using norm duality, the nuclear norm (sum of singular values) can be expressed as
$$\|A\|_* = \max \{ | \langle A, B \rangle | : \|B\|_2 \leq 1 \}$$
where $\|\|_2$ denotes the spectral norm (largest singular value) and $\langle A,B \rangle = \mathrm{Tr} AB^*$.
How can I show that, if $A$ is self-adjoint, it is sufficient to consider self-adjoint $B$ in the above?
When $A$ is self-adjoint, it is diagonalizable, and the nuclear norm becomes the sum of the absolute values of the eigenvalues, say, $|\lambda_1| + \cdots + |\lambda_n|$ where $A = UDU^*$ and the diagonal entries of $D$ are $\lambda_1, \ldots, \lambda_n$.
Let $\tilde{D}$ be diagonal with diagonal entries $\lambda_1 / |\lambda_1|, \ldots, \lambda_n / |\lambda_n|$ and let $B = U\tilde{D} U^*$. You can check that $\|A\|_* = |\langle A, B \rangle|$ and that $\|B\|_2 \le 1$.
Thus, if $A$ is self-adjoint, there exists a self-adjoint $B$ with $\|B\|_2 \le 1$ such that $\|A\|_* = |\langle A, B \rangle|$, so the maximization can be restricted to self-adjoint $B$.