Finding a matrix that satisfies a trace constraint If $U$ is a $p\times q$ matrix such that $U^TU = I_{q},$ $S$ is a $p \times p$ symmetric matrix (either positive definite or nonnegative definite), and $D$ is a $q \times q$ diagonal matrix (can assume all diagonal elements are greater than zero), can we find a symmetric $p \times p$ matrix $A$ such that 
$$\operatorname{tr}(U^T S U D) = \operatorname{tr}(U^TAU) \text{?}$$ 
If this were true, then it would allow me to write a conditional distribution arising in a statistical model in a simpler form, corresponding to a known distribution. I think the answer is no, but I'm having a hard time seeing why. 
If not, I wonder if we have an additional symmetric matrix $\Omega,$ can we find a symmetric $B$ and $A$ as before such that 
$$\operatorname{tr}(U^T S U D + U^T\Omega U) = \operatorname{tr}(BU^TAU) \text{?}$$ 
 A: Given


*

*a tall $p \times q$ matrix $\mathrm U$ with orthonormal columns  ($\mathrm U^{\top} \mathrm U = \mathrm I_q$)

*a $p \times p$ symmetric matrix $\mathrm S$

*a $q \times q$ diagonal matrix $\mathrm D$ whose entries on the main diagonal are positive


we would like to find a symmetric $p \times p$ matrix $\mathrm X$ such that 
$$\mbox{tr} (\mathrm U^{\top} \mathrm X \mathrm U) = \mbox{tr} (\mathrm U^{\top} \mathrm S \mathrm U \mathrm D)$$
Let us exploit the orthonormality of the columns of $\mathrm U$ and look for a symmetric solution of the form
$$\mathrm X = \gamma \, \mathrm U \mathrm U^{\top}$$
Hence, the scaling factor is
$$\gamma = \frac{1}{q} \, \mbox{tr} (\mathrm U^{\top} \mathrm S \mathrm U \mathrm D)$$
and, thus, a symmetric solution is
$$\boxed{\bar{\mathrm X} := \frac{1}{q} \, \mbox{tr} (\mathrm U^{\top} \mathrm S \mathrm U \mathrm D)\, \mathrm U \mathrm U^{\top}}$$
Lastly, do note that $\mathrm U \mathrm U^{\top}$ is the projection matrix that projects onto the $q$-dimensional column space of $\mathrm U$. Hence, $\mbox{tr} (\mathrm U \mathrm U^{\top}) = \mbox{rank} (\mathrm U \mathrm U^{\top}) = \mbox{rank} (\mathrm U) = q$ and, thus, $\mbox{tr} (\bar{\mathrm X}) = \mbox{tr} (\mathrm U^{\top} \mathrm S \mathrm U \mathrm D)$.
