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In my studies of the Schur complement within matrix theory I have come across this problem which seems tough:

Let us consider the block matrix $ A = \left( \begin{array}{ccc} B & C \\ D & E \end{array} \right) $ where $ E $ is an invertible principal square submatrix of $ A $ and we are asked to prove the following equality of ranks $\operatorname{rank}\left( \begin{array}{ccc} CE^{-1}D & C \\ D & E \end{array} \right) = \operatorname{rank}(E) $ using the Schur complement

I have thought about using the identity $\operatorname{rank}(A)=\operatorname{rank}(E)+\operatorname{rank}(A/E) $ and then using the definition of the Schur complement in this case $ A/E = B-CE^{-1}D $ but I have no idea how to show equality of ranks in this case. I appreciate all help.

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This follows easily from the following identity $$ \begin{pmatrix} I & -CE^{-1} \\ 0 & I \end{pmatrix} \begin{pmatrix} CE^{-1}D & C \\ D & E \end{pmatrix}\begin{pmatrix} I & 0 \\ -E^{-1}D & I\end{pmatrix} = \begin{pmatrix} 0 & 0 \\ 0 & E \end{pmatrix}. $$

Or in terms of Schur's complement if $E$ is invertible then $\texttt{rank}\begin{pmatrix} B & C \\ D & E \end{pmatrix} = \texttt{rank}(E) + \texttt{rank}(B - CE^{-1}D),$ putting $B=CE^{-1}D$ leads to the result.

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