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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

$$\operatorname{tr}(U^T S U D + U^T\Omega U) = \operatorname{tr}(BU^TAU) \text{?}$$

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    Yes, that's right. – 2017-01-18

1 Answers 1

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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, $\mbox{tr} (\bar{\mathrm X}) = \mbox{tr} (\mathrm U^{\top} \mathrm S \mathrm U \mathrm D)$.