We know that the conditional variance of a multivariate normal vector $(X,Y)$ is the Schur complement: $$V(X|Y=(y_1,...,y_n))=\Sigma_{XX}-\Sigma_{XY}\Sigma_{YY}^{-1}\Sigma_{YX}$$
I have the intuition that the conditional covariance has the same form but I don't know how to prove it: $$Cov(X_1,X_2|Y=(y_1,...,y_n))=\Sigma_{X_1X_2}-\Sigma_{X_1Y}\Sigma_{YY}^{-1}\Sigma_{YX_2}$$ Where $Cov(X_1,X_2|Y)=E(X_1X_2|Y)-E(X_1|Y)E(X_2|Y)$.
Does someone know how to prove it?
Thank you so much.
Since the first equation is true for any $X$ vector, we can define $X=\begin{bmatrix}X_1 \\ X_2\end{bmatrix}$, and $V(X|Y)$ will be $\begin{bmatrix}V(X_1,X_1|Y) & V(X_1,X_2|Y) \\ V(X_2,X_1|Y) & V(X_2,X_2|Y)\end{bmatrix}$. We want the upper right corner of this matrix. Also, we can define $\Sigma_{XX}=\begin{bmatrix}\Sigma_{X_1,X_1} & \Sigma_{X_1,X_2} \\ \Sigma_{X_2,X_1} & \Sigma_{X_2,X_2}\end{bmatrix}$ and $\Sigma_{XY}=\begin{bmatrix}\Sigma_{X_1,Y} \\ \Sigma_{X_2,Y}\end{bmatrix}$, $\Sigma_{YX}=\begin{bmatrix}\Sigma_{X_1,Y} & \Sigma_{X_2,Y}\end{bmatrix}$. Substituting these into the first equation yields: $$\begin{align}V(X|Y) & =\begin{bmatrix}\Sigma_{X_1,X_1} & \Sigma_{X_1,X_2} \\ \Sigma_{X_2,X_1} & \Sigma_{X_2,X_2}\end{bmatrix}-\begin{bmatrix}\Sigma_{X_1,Y} \\ \Sigma_{X_2,Y}\end{bmatrix}\Sigma_{YY}^{-1}\begin{bmatrix}\Sigma_{X_1,Y} & \Sigma_{X_2,Y}\end{bmatrix} \\ & = \begin{bmatrix}\Sigma_{X_1,X_1} & \Sigma_{X_1,X_2} \\ \Sigma_{X_2,X_1} & \Sigma_{X_2,X_2}\end{bmatrix}-\begin{bmatrix}\Sigma_{X_1Y}\Sigma_{YY}^{-1}\Sigma_{YX_1} & \Sigma_{X_1Y}\Sigma_{YY}^{-1}\Sigma_{YX_2} \\ \Sigma_{X_2Y}\Sigma_{YY}^{-1}\Sigma_{YX_1} & \Sigma_{X_2Y}\Sigma_{YY}^{-1}\Sigma_{YX_2}\end{bmatrix}\end{align}$$. The upper right corner is $\Sigma_{X_1,X_2}-\Sigma_{X_1Y}\Sigma_{YY}^{-1}\Sigma_{YX_2}$ just as your wrote.
