I want to apply the Schur complement to one element of a block matrix. What I do not know is how to organize the resulting values inside the new matrix. E.g. Given the following block matrix with appropriate dimensions: \begin{equation} \begin{bmatrix} A-BD^{-1}B^T & E \\ E^T & C \end{bmatrix}\succ 0 \end{equation}

with $C=C^T$,$A=A^T$, $D=D^T$. Applying Schur complement with respect to $A-BD^{-1}B^T$ yields to: \begin{equation} \begin{bmatrix} A & B & E \\ B^T & D & 0\\ E^T & 0 & C \end{bmatrix}\succ 0 \end{equation} or to\begin{equation} \begin{bmatrix} A & E & B \\ E^T & C & 0\\ B^T & 0 & D \end{bmatrix}\succ 0 \end{equation} Which one of the two is correct? How to determine the order of the resulting matrix?

Thanks in advance!


The two $3 \times 3$ block matrices are similar via a block similarity, so one is positive definite if and only if the other is as well. In particular: $$ \pmatrix{I&0&0\\0&0&I\\0&I&0} \pmatrix{A & B & E \\ B^T & D & 0\\ E^T & 0 & C} \pmatrix{I&0&0\\0&0&I\\0&I&0}^T = \pmatrix{A & E & B \\ E^T & C & 0\\ B^T & 0 & D} $$ To find the size of the resulting matrices, it suffices to know the sizes of $A,C,D$.

  • $\begingroup$ Thank you very much! that is the answer I was looking for. $\endgroup$ – Robinson Medina Jan 27 '17 at 13:01
  • $\begingroup$ @RobinsonMedina great! If that answers everything, then please click that $\checkmark$ by the up/down arrows. $\endgroup$ – Omnomnomnom Jan 27 '17 at 14:20

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