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Let $M= \begin{bmatrix}A&B\\C&D\end{bmatrix}$ be a block matrix and $A$ is square invertible.

Since $A$ is nonsingular, $$M= \begin{bmatrix}A&B\\C&D\end{bmatrix}=\begin{bmatrix}I&0\\CA^{-1}&I\end{bmatrix}\begin{bmatrix}A&0\\0&M/A\end{bmatrix}\begin{bmatrix}I&A^{-1}B\\0&I\end{bmatrix}$$ My question is: why $\text{rank}(M)=\text{rank}(A)+\text{rank}(M/A)$, where $M/A$ is Schur - complement?

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    What is $A$? And where do you get $\left[\matrix{B & C\cr C & D}\right] = \left[\matrix{I & 0\cr CA^{-1} & I}\right]$ ? They're certainly not equal unless $B=I$, $C=0$ etc. – 2017-02-06
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    I edited it again. Thanks. – 2017-02-06

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The two triangular matrices are invertible, so the rank is that of $$\tag{*} \begin{bmatrix}A&0\\0&M/A\end{bmatrix}. $$ As this is a block-diagonal matrix, the rank is the sum of the ranks of the two diagonal entries: the rank is the dimension of the image, and the image of $(*)$ is the direct sum of the images of $A$ and $A/M$.