Independence of two multivariate normals. Suppose we have two multivariate normals $X_1 \sim N(u_1, \Sigma_{11}\Sigma_{22}$)  and $X_2 \sim N(u_2, \Sigma_{21} \Sigma_{22})$ .
Why are $X_2 $ and  $X_1-\Sigma_{12} \Sigma_{22}^{-1}X_2$ independent from each other? How would I go for proving that?
I am trying to prove the last bit of this page on wikipedia. Is it implicitly assumed that $X_1$ and $X_2$ are jointly normally distributed?
 A: $\newcommand{\var}{\operatorname{var}}\newcommand{\cov}{\operatorname{cov}}$You have not specified that the pair $\begin{bmatrix} X_1 \\ X_2 \end{bmatrix}$ is normally distributed, and that does not follow from the fact that the two components separately are normally distributed, nor have you said what the covariance between $X_1$ and $X_2$ is.
You say $X_1 \sim N(u_1, \Sigma_{11}\Sigma_{22}$ and $X_2 \sim N(u_2, \Sigma_{21} \Sigma_{22})$ and I was puzzled.  But I'm guessing you mean
$$
\begin{bmatrix} X_1 \\ X_2 \end{bmatrix} \sim N\left(\begin{bmatrix} \mu_1 \\  \mu_2 \end{bmatrix}, \begin{bmatrix} \Sigma_{11}, & \Sigma_{12} \\  \Sigma_{21}, & \Sigma_{22} \end{bmatrix}\right),
$$
which would imply that $X_1\sim N(\mu_1,\Sigma_{11})$ and $X_2\sim N(\mu_2,\Sigma_{22})$ and would further specify their covariance.
I think the article has in mind that $\var(X_1)=\Sigma_{11}$, $\var(X_2)=\Sigma_{22}$, $\cov(X_1,X_2)=\Sigma_{12}$ (so that $\cov(X_2,X_1)=\Sigma_{12}^T=\Sigma_{21}$.
Then we have
\begin{align}
\cov(X_2,X_1-\Sigma_{12} \Sigma_{22}^{-1}X_2) & = \cov(X_2,X_1)-\cov(X_2,\Sigma_{12} \Sigma_{22}^{-1}X_2) \\[8pt]
& = \Sigma_{21} - \cov(X_2,X_2)\Big(\Sigma_{12}\Sigma_{22}^{-1}\Big)^T \\[8pt]
& = \Sigma_{21} - \Sigma_{22}\Big(\Sigma_{22}^{-1}\Sigma_{21}\Big) = \cdots\cdots
\end{align}
Can you take it from there?
BTW, you seem to have used the vowel $u$ where the article uses the consonant $\mu$.
