Prove that the distribution of $Q$ is chi-squared with $p_2$ degrees of freedom 
Suppose $X$ is a $p$-dimensional vector following $N_p(\mu,\Sigma)$ distribution, where $\mu$ is $p$-dimensional and $\Sigma$ is $p\times p$. Let $X=\left(\begin{array}{ccc}X_1\\X_2\end{array}
\right)$ and $\Sigma=\left( \begin{array}{ccc}
\Sigma_{11} & \Sigma_{12} \\
\Sigma_{21}&\Sigma_{22} \end{array} \right)$ where $X_1$ is $p_1\times1$ and $X_2$ is $p_2\times1$ with $p_1+p_2=p$. Also we have by usual notations $\Sigma_{12}=\text{Cov}(X_1,X_2)=\Sigma_{21}^T$. Show that $$Q:=X^T\Sigma^{-1}X-X_1^T\Sigma_{11}^{-1}X_1$$follows $\chi^2(p_2)$.

I tried to proceed making the changes $Y=\Sigma^{-1/2}X$ and $Z=\Sigma_{11}^{-1/2}X_1$ but am not really sure if it is in the right direction. Also, I tried to proceed by considering the transformation $X_2'=X_2-\Sigma_{21}\Sigma_{11}^{-1}X_1$ but things are getting too messy and I don't know how to proceed.
Please prove some hint(s) only and not a complete solution.
 A: Hint (really, really good hint):
The Schur Complement is your friend.
Section A.5.5, section C.4 and numerous other occurrences in "Convex Optimization", Stephen Boyd and Lieven Vandenberghe
A: There is a simple solution that was perhaps not considered.
We have
$$Q=X^T\Sigma^{-1}X-X_1^T\Sigma_{11}^{-1}X_1=X^TAX\,,$$
where $A=\Sigma^{-1}-\begin{pmatrix}\Sigma_{11}^{-1} & 0 \\ 0 & 0\end{pmatrix}$ is a symmetric matrix.
$Q$ has a $\chi^2$ distribution if and only if $A\Sigma$ is idempotent (the 'if part' is discussed here).
Now, $$A\Sigma=I_p-\begin{pmatrix}\Sigma_{11}^{-1} & 0 \\ 0 & 0\end{pmatrix}\Sigma$$
But $$\begin{pmatrix}\Sigma_{11}^{-1} & 0 \\ 0 & 0\end{pmatrix}\Sigma=\begin{pmatrix}\Sigma_{11}^{-1} & 0 \\ 0 & 0\end{pmatrix}\begin{pmatrix}\Sigma_{11} & \Sigma_{12} \\  \Sigma_{21} &  \Sigma_{22}\end{pmatrix}=\begin{pmatrix}I_{p_1} & \Sigma_{11}^{-1}\Sigma_{12} \\  0 &  0\end{pmatrix}\,,$$ which is symmetric idempotent. Hence $A\Sigma$ is also symmetric idempotent.
Degrees of freedom of $Q$ would be $\operatorname{rank}(A\Sigma)=\operatorname{tr}(A\Sigma)$. There would also be a noncentrality parameter if $\mu\ne 0$.
A: Too long for a comment, I just wanted to point out the conclusion stated in the OP's question is inaccurate, if not incorrect.  The correct conclusion should be stated as follows (source:  Aspects of Multivariate Statistical Theory by Muirhead, Robb J.)

Specifically, the conclusion cannot ignore the role of the mean vector $\mu$.  OP's conclusion is correct only if $\mu = 0$. But it looks to me OP did not want to restrict himself to this special case.
