I want to understand why the following equivalency holds.
$$\begin{bmatrix}A - BD^{-1}B^{\top} & E\\ E^{\top} & C\end{bmatrix}\succ 0 \iff \begin{bmatrix}A & B & E \\ B^{\top} & D & 0\\ E^{\top} & 0& C\end{bmatrix}\succ 0$$ What is the justification for opening up the matrix $A - BD^{-1}B^{\top}$ inside the original matrix? How does the mechanics of this work if I want to do this repeatedly? I tried to prove this myself but I think there is some "fact" that I don't know yet that is blocking me.
Presumably $D$ is positive definite. The matrix on the left is just the Schur complement of $D$ in the $3\times3$ block matrix on the right. If you fail to see this, you may aid yourself by permuting the blocks first: \begin{align} &\pmatrix{A&B&E\\ B^T&D&0\\ E^T&0&C}\simeq\left(\begin{array}{cc|c}A&E&B\\ E^T&C&0\\ \hline B^T&0&D\end{array}\right),\\ &\pmatrix{A&E\\ E^T&C}-\pmatrix{B\\ 0}D^{-1}\pmatrix{B^T&0} =\pmatrix{A-BD^{-1}B^T&E\\ E^T&C}. \end{align} Now, recall that the blockwise direct sum of $D$ and its Schur complement can be obtained from the original matrix by matrix congruence. Hence the $3\times3$ block matrix is positive definite iff both $D$ and its Schur complement are positive definite.