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Is there any way to show that a non-singular matrix A can be partitioned as follows: \begin{eqnarray*} A&=&\left[ \begin{array}{cc} \underset{\left( k\times k_{1}\right) }{{B}_{11}}\underset{\left( k_{1}\times {n_{1}}\right) }{C_{1}} & \underset{\left( k\times k_{2}\right) }{{B}_{12}}\underset{\left( k_{2}\times n_{2}\right) }{% D_{1}} \\ \underset{\left( \left( n-k\right) \times \left( n_{1}-k_{1}\right) \right) }{{B}_{21}}\underset{\left( \left( n_{1}-k_{1}\right) \times n_{1}\right) }{C_{2}} & \underset{\left( \left( n-k\right) \times \left( n_{2}-k_{2}\right) \right) }{{B}_{22}}\underset{% \left( \left( n_{2}-k_{2}\right) \times n_{2}\right) }{D_{2}}% \end{array}% \right] \end{eqnarray*}

where $n=n_{1}+n_{2}, k=k_{1}+k_{2}$ and the matrices \begin{equation*} \begin{array}{c} \left[ \begin{array}{cc} {{B}_{11}} & {{B}_{12}}% \end{array}% \right] ,\text{ } \left[ \begin{array}{cc} {{B}_{21}} & {{B}_{22}}% \end{array}% \right] \end{array}% \end{equation*}

\begin{equation*} \left[ \begin{array}{c} C_{1} \\ C_{2}% \end{array}% \right] ,\text{ }\left[ \begin{array}{c} D_{1}\\ D_{2}% \end{array}% \right] \end{equation*} are all non-singular square matrices?

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    $\begingroup$ Do you have a reason to think this is true? Have you tried some obvious seeming possible counterexamples such as the identity matrix? Also, are $n_1,n_2,k_1,k_2$ supposed to be given or free? $\endgroup$ Nov 1, 2014 at 6:41

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