# Nearest positive semi definite matrix to a complex valued Hermitian matrix

How would you find the nearest (via Hilbert distance), PSD matrix (with trace = 1) to a Hermitian matrix? I found an answer to a similar question here. However, as I understand, Hingham's work only applies to real matrices.

(I am looking at reconstructing a density matrix from experimental data which needs to be a PSD with trace = 1 )

Isn't (as was originally asked) the closest psd matrix to $$H=U^* \Delta U$$ where $$U$$ is unitary and $$\Delta$$ real diagonal with diagonal entries $$d_i$$ given by $$G=U^* \Delta_+ U$$ where $$\Delta_+$$ is the diagonal matrix with diagonal entries $$d_i'= \max(0,d_i)$$? Because $$X\mapsto U^* X U$$ is an isometry with respect to the Hilbert distance?
Added after the question was changed. If the closest psd matrix with trace 1 to $$H$$ is sought, chose $$\Delta_+$$ to be closest to $$\Delta$$ subject to the trace 1 and non-negativity condition. The $$U$$ stuff above reduces your problem to a $$\mathbb R^n$$ problem from an $$n\times n$$ matrix problem. I think this is attained when $$d_i'$$ of the form $$d_i'=0$$ if $$d_i<0$$ and $$d_i' = d_i+\lambda$$ for $$d_i\ge0$$, where $$\lambda$$ is chosen to make $$\sum d_i' = 1.$$