 Solution

For a Hermitian matrix $A$, the spectral theorem gives $A=U\mathrm{diag}({\lambda }_1,\dots ,{\lambda }_d)U^{†}$, with real eigenvalues and a unitary matrix $U$. HenceThe square root here is the unique positive square root of an operator. Taking the trace, which is unchanged by unitary similarity, provesFor a density matrix, all eigenvalues are nonnegative and sum to one. Its trace norm is thereforeThe absolute values cannot be omitted for a general Hermitian matrix, since positive and negative eigenvalues can cancel in its trace but never in its trace norm.