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We have $A \in \mathbb{R}^{n \times n}$ which is symmetric and positive-definite. Also, $A$ is a block matrix:

$$A = \begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \\ \end{pmatrix}$$

I have managed to show that both $A_{11}$ and $A_{22}$ are symmetric and positive-definite. Also, it is easy to show $S = A_{22} -A_{12}^T A_{11}^{-1} A_{12}$ (Schur complement) is symmetric. What I cannot do is to show that $S$ is positive definite as well.

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2 Answers 2

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Let $S$ be the Schur complement. For any (column) vector $v$ define $$ \tilde{v}=\begin{bmatrix}-A_{11}^{-1}A_{12}v\\v\end{bmatrix}. $$ Then $\tilde{v}^TA\tilde{v} = v^TSv$.

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Show that $A$ $$A = \begin{bmatrix}I \\ A_{12}^\top A_{11}^{-1} & I\end{bmatrix} \begin{bmatrix}A_{11} \\ & A_{22} - A_{12}^\top A_{11}^{-1} A_{12}\end{bmatrix} \begin{bmatrix}I & A_{11}^{-1} A_{12} \\ & I\end{bmatrix} =: NDN^\top.$$

Show that $N$ is invertible.

$$N^{-1} = \begin{bmatrix}I \\ -A_{12}^\top A_{11}^{-1} & I\end{bmatrix}.$$

Show that consequently, $D$ is positive-definite.

$D = N^{-1} A (N^{-1})^\top$ so for any $v \ne 0$ we have $v^\top D v = v^\top N^{-1} A (N^{-1})^\top v = w^\top A w > 0$ where $w = (N^{-1})^\top v$.

Show that $A_{22} - A_{12}^\top A_{11}^{-1} A_{12}$ is positive-definite.

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  • $\begingroup$ D being positive-definite is a consequence of N being invertible? How is that? $\endgroup$
    – Uppermost
    Commented Nov 5, 2019 at 19:47
  • $\begingroup$ @Uppermost See my edit $\endgroup$
    – angryavian
    Commented Nov 5, 2019 at 20:25
  • $\begingroup$ Got it, thanks a lot! $\endgroup$
    – Uppermost
    Commented Nov 6, 2019 at 8:36

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