0
$\begingroup$

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 $S=A_{22}-A_{12}^TA_{11}^{-1}A_{12}$ (Shure complement) is symmetric. What I cannot do is to show that S is positive-definite aswell.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ D being positive-definite is a consequence of N being invertible? How is that? $\endgroup$ – Uppermost Nov 5 '19 at 19:47
  • $\begingroup$ @Uppermost See my edit $\endgroup$ – angryavian Nov 5 '19 at 20:25
  • $\begingroup$ Got it, thanks a lot! $\endgroup$ – Uppermost Nov 6 '19 at 8:36
0
$\begingroup$

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 $\tilde{v}^TA\tilde{v} = v^TSv$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.