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Suppose that

$$ M = \begin{bmatrix}A & B \\ C & D\end{bmatrix}. $$

I know that if $D$ and $M\setminus D$ (where $M\setminus D$ is the Schur Complement of $D$ in $M$) are invertible, then $M$ is invertible.

However, suppose that we know that $M$ and $D$ are invertible, but we know nothing about the invertibility of $A$. Can we say that $M\setminus D$ is invertible?

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    $\begingroup$ Yes. If $D$ is invertible, then the matrix $M$ is congruent to the matrix $\left(\begin{array}{cc} A - BD^{-1}C & 0 \\ 0 & D \end{array}\right)$. Thus, $M$ is invertible if and only if $\left(\begin{array}{cc} A - BD^{-1}C & 0 \\ 0 & D \end{array}\right)$ is invertible, i.e., if and only if $A - BD^{-1}C$ is invertible (because $D$ is invertible by assumption). Here, we have used the fact that a block-diagonal matrix is invertible if and only if all of its diagonal blocks are invertible. $\endgroup$
    – darij grinberg
    Commented Jun 24, 2017 at 17:05

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Yes, if you take the determinants, you obtain with $$\operatorname{det}(M)=\operatorname{det}(M/ D)\cdot\operatorname{det}(D) $$ therefore if $\operatorname{det}(M)$ is non-zero then $\operatorname{det}(M/ D)$ non-zero.

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