Let $A$ be a given symmetric positive definite $N\times N$ matrix. I need to find a symmetric positive semi-definite matrix $S$ which is the solution to the following optimization problem
\begin{align}
\max\_{S}~&\det(A+S) \\s.t.~&\sum\_{i}^{N}\sigma\_i(S)\,=\,c \\&S\geq0
\end{align}where $\sigma\_i(S)$ are the singular values of $S$ and $c$ is a given positive constant. 


Note that I already know that this can be converted into a standard convex problem by replacing the constraint on sum of singular values by trace and also considering log-determinant. However, I am interested in a different approach.


My hunch is that following is the solution.


Let $A=U\_a\Sigma\_aU\_a^H$ be its Eigen-decomposition (EVD). Since both $A$ and $S$ are symmetric positive semi-definite, their EVD and SVD are same. Let $S= U\_s\Sigma\_sU\_s^H$ be its EVD which we are seeking. Then first part of the solution is $$U\_s = U\_a$$ It only remains find $\Sigma\_s$ which contains the singular values of $S$. Substituting this in the original optimization problem, it breaks down into the simple form of finding the variables $\sigma\_i(S)$ from the optimization problem
\begin{align}
\max\_{\sigma\_1(S),\dots,\sigma\_N(S)}~&\prod\_{i=1}^{N}(\sigma\_i(A)+\sigma\_i(S)) \\s.t.&\sum\_{i=1}^N\sigma\_i(S)=c
\end{align}
This is also equivalent to maximizing the $\log$ of the objective since the objective is a increasing function of each variable. Let us substitute $$x\_i=\frac{\sigma\_i(S)}{\sigma\_i(A)}$$ Also, we can use the fact that determinant is the product of singular values for a symmetric positive definite matrix. Combining all this, we have the optimization problem 
\begin{align}\max\_{x\_i}\sum\_{i=1}^{N}~&\log(1+x\_i)+\log\det(A)\\s.t.~&\sum\_{i=1}^{N}\sigma\_i(A)x\_i\,=\,c\end{align}
The constant term in the objective can be neglected. This is an instance of a very famous problem in wireless communications where one tries to maximize the capacity of set of $N$ wireless channels subject to power constraints in each. Its solution is very well known and is typically referred to as the water filling solution. So I know the process after reaching this point. Note that all this was facilitated by the assumption that $U\_s = U\_a$. Is that true? How do I prove this?. If not, what are other approaches I can try?



Optimization problem with determinant as objective
One can verify that $U\_{A}=U\_{S}$ as follows. Note that $$\det(U\_{A}\Sigma\_{A}V\_{A}^{T}+U\_{S}\Sigma\_{S}V\_{S}^{T})=\det(\Sigma\_{A}+U\_{A}^{-1}U\_{S}\Sigma\_{S}V\_{S}^{T}V\_{A}^{-T})$$ 
Let $Y=U\_{A}^{-1}U\_{S}\Sigma\_{S}V\_{S}^{T}V\_{A}^{-T}$ so that the problem can be rephrased as:


$$\max\limits\_{Y} \det(\Sigma\_{A}+Y) \text{ subject to } Tr(Y)=c,\ Y\succ 0$$


Now, let $Z=\Sigma\_{A}+Y$ so that the problem is reformulated: 


$$\max\limits\_{Z} \det(Z) \text{ subject to } Tr(Z)=c+\sum\limits\_{i=1}^{n} \sigma\_{A}(i),\ Z\succ 0,\ Z\_{i,i}\geq \sigma\_{A}(i)$$ 


Since $Z$ can be written $Z=UT$ where $U$ is orthogonal and $T$ upper triangular, $\det(Z)=\det(T)$ and the off-diagonal terms of $T$ do not contribute so that $T$ may be assumed diagonal. Thus, finally one arrives at:


$$\max\limits\_{T} \det(T) \text{ subject to } Tr(T)=c+\sum\limits\_{i=1}^{n} \sigma\_{A}(i),\ T\succ 0,\ T\_{i,i}\geq \sigma\_{A}(i),\ \text{T is diagonal}$$


This problem's solution follows from solving the water filling problem above where $t\_{k}=T\_{k,k}$: 


$$\max\limits\_{t\_{k}} \sum\limits\_{k=1}^{n} \log(t\_{k}+\sigma\_{k}) \text{ subject to } t\_{k}\geq 0 \text{ and }\sum\limits\_{k=1}^{n} t\_{k}=c$$


