问一个优化问题

min trace{(R+U)^(-1)}
s.t.  [U]_{ii}<=p_i
其中R已知，不是对角阵。 U是要求得。 [U]_{ii}是U的第i个对角元素。
请问有没有closed form解？

Further constraints are required (such as R and U are psd, and so on). Based on the current conditions, the infimum of that can be -inf. And for complex matrices, we cannot define the infimum.
Consider R=0, p_1=p_2=1, and U=[1 0.01; 101 1]. trace(inv(R+U))=-200.

you are right. R and U are PSD. All are complex.

If R is psd and U is Hermitian, I guess the answer is sum(1/(p_i+r_ii)) where r_ii are the diagonal entries of R. No strict proof till now.
If R and U are both psd, which means R+U cannnot be diagonalized in some cases, it will be very messy.

ic
if the constraint is trace(U)<=P. Then, we can always diagonalize R.
But now, we cannot.
I also tried many methods and referred to cioffi's paper regarding the diagonal constraint. But it seems that there is no way or I don't know the way to solve the problem.

Something that may help:
For a matrix A with eigenvalues e_1,...,e_n, and diagonal entries a_11,...,a_nn, we have sum(e_i)=sum(a_ii).
If A is psd, then by Hadamard inequality, product(e_i)<=product(a_ii). Equality holds only when some a_ii=0 or A is diagonal.