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Let $A$ and $D$ be two square matrices. Let $P$ be defined as

$P= \left( \begin{array}{ccc} A & B \\ C & D \end{array} \right)$. If $A$ is an invertible matrix of order $k$, show that rank $P$= $k+$rank$(D-CA^{-1}B)$. Deduce from this that if rank $P = $ rank $A$, then $D=CA^{-1}B$.

I would appreciate any hints or ideas to show this.

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Hint: What is $\begin{pmatrix}I_n&0\\-CA^{-1}&I_m\end{pmatrix} P\begin{pmatrix} A^{-1}&-A^{-1}B\\0&I_m\end{pmatrix}$, where $A$ is an $n\times n$-matrix and $D$ is an $m\times m$-matrix?

Edit: To find the above matrices we note successively $$P\begin{pmatrix}A^{-1}&0\\0&I_m\end{pmatrix}=\begin{pmatrix}I_n&B\\CA^{-1}&D\end{pmatrix}=:Q,$$ $$\begin{pmatrix}I_n&0\\-CA^{-1}&I_m\end{pmatrix}Q=\begin{pmatrix}I_n&B\\0&D-CA^{-1}B\end{pmatrix}=:R,$$ $$R\begin{pmatrix}I_n&-B\\0&I_m\end{pmatrix}=\begin{pmatrix}I_n&0\\0&D-CA^{-1}B\end{pmatrix}.$$These matrices correspond to elementary row/column operations written in block matrix form, for example multiplication from the right by the matrix $\begin{pmatrix}A^{-1}&0\\0&I_m\end{pmatrix}$ corresponds to "apply the elementary column operation $A^{-1}$ to the first $n$ columns".

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  • $\begingroup$ Actually, it's $\begin{pmatrix}I_n&0\\0&D-CA^{-1}B\end{pmatrix}$. Now what does that tell you about the rank of $P$? $\endgroup$ – walcher Sep 2 '13 at 18:28
  • $\begingroup$ How did you find the matrices $\begin{pmatrix}I_n&0\\-CA^{-1}&I_m\end{pmatrix}$ and $\begin{pmatrix} A^{-1}&-A^{-1}B\\0&I_m\end{pmatrix}$? $\endgroup$ – Twink Sep 2 '13 at 18:39
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Hint: Consider $$\pmatrix{I & 0\cr -CA^{-1} & I \cr} P$$

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