I encountered a problem that I could not figure out.
Suppose that ${U}_{k \times 1}$ is a k-variate (multivariate) normal distribution with positive definite variance $\Sigma_{}$. Let $U^{(1)}_{l \times 1}$ and $U^{(2)}_{(k-l)\times 1}$ be a partition of $U$ with variances $\Sigma_{11}$,$\Sigma_{22}$ and covariance $\Sigma_{12}=\text{Cov}(U^{(1)},U^{(2)})_{l\times(k-l)}$. Show that
$$ \text{Cov}(\Sigma_{12}\Sigma^{-1}_{22}U^{(2)},U^{(1)}-\Sigma_{12}\Sigma_{22}^{-1}U^{(2)}) = 0 $$
How could I go about this?
Without loss of generality we may assume the mean of $U$ is zero. For a dimensionally correct matrix $A$ you wish to evaluate $$\text{Cov}(AU_2,U_1 - AU_2) = \text{E}\left[AU_2(U_1-AU_2)^\top\right]=\text{E}\left[AU_2U_1^\top-AU_2U_2^\top A^\top\right] = A\left(\Sigma_{21}-\Sigma_{22}A^\top\right).$$
Your specific choice of $A=\Sigma_{12}\Sigma_{22}^{-1}$ yields $$A^\top =\Sigma_{22}^{-1} \Sigma_{21}$$ and the desired result follows.