Expectation of the product of two independent random vectors and a positive-definite matrix I am trying to compute the following: $\mathbb{E}[X^T\Omega^{-1}\epsilon]$, where $X$ is a  random matrix, $\epsilon$ is a random vector, $\Omega$ is a real positive-definite matrix, and $\mathbb{E}[X^T\epsilon]=0$ (since each row of $X$ is independent from each row of $\epsilon$, and $\mathbb{E}[\epsilon]=0)$.
I conjecture that $\mathbb{E}[X^T\Omega^{-1}\epsilon]=0$ as well, as a linear transformation of independent random vectors is also independent. Is this correct?

The expectation is zero, using that expectation and trace commutes (what some calles the trace trick), and the cyclic property of the trace operator.
$$   \DeclareMathOperator{\E}{\mathbb{E}}  \DeclareMathOperator{\Tr}{Tr}  
\E \left[ X^T\Omega^{-1}\epsilon \right]= \\
\Tr \E \left[ X^T\Omega^{-1}\epsilon \right]= \\
\E \Tr \left[ X^T\Omega^{-1}\epsilon \right]= \\
\E \Tr \left[ \Omega^{-1}\epsilon X^T \right]= \\ 
\Omega^{-1} \E \Tr \left[ \epsilon X^T \right]= \\ 
\Omega^{-1} \E \Tr \left[ X^T \epsilon \right]= \\ 
\Omega^{-1} \E 0= 0.
$$
