Let $$A=\begin{bmatrix}A_{11}&A_{12}\\A_{21}&A_{22}\end{bmatrix} \text{be hermitian}$$
Given: $A_{11} $ and $A_{22}-A_{21}A_{11}^{-1}A_{12}$ are both positive definite, then show that $A$ is positive definite.
Since $A$ is hermitian $A_{11}^*=A_{11}, \; A_{22}^*=A_{22}, \;A_{12}^*=A_{21}\;\; ->>(1)$
Let $$B=\begin{bmatrix}A_{11}&0\\0&A_{22}-A_{21}A_{11}^{-1}A_{12}\end{bmatrix}$$
Let $U^*=\begin{bmatrix}I&0\\-A_{21}A_{11}^{-1}&I\end{bmatrix}$
Then, using $(1)$, please verify that
$U^*AU=B$
How can I make use of this to prove that $A$ is also positive definite?
Best Answer
Let us first check that $S$ has hermitian symmetry :
$$S^*=A_{22}^*-(A_{21}A_{11}^{-1}A_{12})^*=A_{22}-(A_{12}^*(A_{11}^{-1})^*A_{21}^*)=$$ $$=A_{22}-(A_{21}(\underbrace{A_{11}^{*}}_{A_{11}})^{-1}A_{12})=S.\tag{1}$$
(please note that we have used the fact that the inverse of a transpose-conjugate is the transpose-conj. of the inverse).
$B$ is positive definite because its eigenvalues are the union of
those of $A_{11}$ which are real positive, because $A_{11}$ is symmetric definite positive with hermitian symmetry, and
those of $S:=A_{22}-A_{21}A_{11}^{-1}A_{12}$, all of them being as well $>0$ for the same reason ($S$ is assumed positive definite and we just verified in (1) that it has hermitian symmetry).
Besides, a consequence of conjugation relationship $U^*AU=B$ is that $A$ and $B$ have the same (positive !) eigenvalues.
Therefore, $A$, having all its eigenvalues $>0$ is (symmetric) positive definite.