Taking $A=\begin{pmatrix} 2 & -1 & 0 \\-1 & 2 & 1 \\ 0 & 1 & 2 \end{pmatrix}$ and running cvx via MATLAB on the problem replacing the singular value condition with the equivalent trace condition with $c=1$ does roughly give $U_{A}=U_{S}$.
cvx_begin
variable S(3,3) symmetric;
S==semidefinite(3);
minimize(-log_det(S+A));
subject to
trace(S) ==1;
cvx_end
[U,S,V]=svd(A)
[U1,S1,V1]=svd(S)
The output is
Successive approximation method to be employed.
For improved efficiency, SDPT3 is solving the dual problem.
SDPT3 will be called several times to refine the solution.
Original size: 40 variables, 16 equality constraints
1 exponentials add 8 variables, 5 equality constraints
-----------------------------------------------------------------
Cones | Errors |
Mov/Act | Centering Exp cone Poly cone | Status
--------+---------------------------------+---------
1/ 1 | 2.305e-01 3.496e-03 0.000e+00 | Solved
1/ 1 | 2.716e-02 4.906e-05 0.000e+00 | Solved
1/ 1 | 2.918e-03 5.657e-07 0.000e+00 | Solved
0/ 1 | 3.172e-04 6.511e-09 0.000e+00 | Solved
-----------------------------------------------------------------
Status: Solved
Optimal value (cvx_optval): -2.38217
ans =
3.4142
2.0000
0.5858
U =
-0.5000 -0.7071 -0.5000
0.7071 -0.0000 -0.7071
0.5000 -0.7071 0.5000
S =
3.4142 0 0
0 2.0000 0
0 0 0.5858
V =
-0.5000 -0.7071 -0.5000
0.7071 0.0000 -0.7071
0.5000 -0.7071 0.5000
U1 =
-0.5000 0.7068 -0.5004
-0.7071 0.0004 0.7071
0.5000 0.7074 0.4996
S1 =
1.0000 0 0
0 0.0000 0
0 0 0.0000
V1 =
-0.5000 0.7068 -0.5004
-0.7071 0.0004 0.7071
0.5000 0.7074 0.4996
Edit: 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 is just the water filling problem above:
$\max\limits_{t_{k}} \sum\limits_{k=1}^{n} \log(t_{k}+\sigma_{k}) \text{ subject to } t_{k}\geq 0 \text{ and }\sum\limits_{k=1}^{n} t_{k}=c$