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?
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 $A/M$.