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I have a invertible matrix P which is constructed by:

$\mathbf{P} = \begin{bmatrix} \mathbf{R_z} & \mathbf{0} & \mathbf{A} \\ \mathbf{0} & \mathbf{0} & \mathbf{C} \\ \mathbf{A}^T & \mathbf{C}^T & \mathbf{0} \\ \end{bmatrix}$

Where, $\mathbf{R_z}$ is a square and invertible matrix and $\mathbf{A}$ and $\mathbf{B}$ are rectangular matrix with full rank. I want to find a expression for $\mathbf{P}^{-1}$, so I tried finding a matrix $\mathbf{B}$ such as:

$ \begin{bmatrix} \mathbf{B_1}' & \mathbf{B_2}' & \mathbf{B_3}' \\ \mathbf{B_4}' & \mathbf{B_5}' & \mathbf{B_6}' \\ \mathbf{B_7}' & \mathbf{B_8}' & \mathbf{B_9}' \\ \end{bmatrix} \begin{bmatrix} \mathbf{R_z} & \mathbf{0} & \mathbf{A} \\ \mathbf{0} & \mathbf{0} & \mathbf{C} \\ \mathbf{A}^T & \mathbf{C}^T & \mathbf{0} \\ \end{bmatrix} = \begin{bmatrix} \mathbf{I} & \mathbf{0} & \mathbf{0}\\ \mathbf{0} & \mathbf{I} & \mathbf{0} \\ \mathbf{0} & \mathbf{0} & \mathbf{I} \\ \end{bmatrix} $

Writting the equations, we have:

$ \begin{cases} \mathbf{B_1}'\mathbf{R_z} + \mathbf{B_3}'\mathbf{A}^T = \mathbf{I} \\ \mathbf{B_3}'\mathbf{C}^T = \mathbf{0}\\ \mathbf{B_1}'\mathbf{A} + \mathbf{B_2}'\mathbf{C} = \mathbf{0} \\ \mathbf{B_4}'\mathbf{R_z} + \mathbf{B_6}'\mathbf{A}^T = \mathbf{0} \\ \mathbf{B_6}'\mathbf{C}^T = \mathbf{I} \\ \mathbf{B_4}'\mathbf{A} + \mathbf{B_5}'\mathbf{C} = \mathbf{0} \\ \mathbf{B_7}'\mathbf{R_z} + \mathbf{B_9}'\mathbf{A}^T = \mathbf{0} \\ \mathbf{B_9}'\mathbf{C}^T = \mathbf{0}\\ \mathbf{B_7}'\mathbf{A} + \mathbf{B_8}'\mathbf{C} = \mathbf{I} \\ \end{cases} $

Solving this, we have:

$ \mathbf{B}= \begin{bmatrix} \mathbf{R_z}^{-1} & - \mathbf{R_z}^{-1} \mathbf{A} \left( \mathbf{C}^T\mathbf{C}\right)^{-1} \mathbf{C}^T & \mathbf{0} \\ - \left( \mathbf{C}\mathbf{C}^T\right)^{-1} \mathbf{C} \mathbf{A}^T \mathbf{R_z}^{-1} & \left( \mathbf{C}\mathbf{C}^T\right)^{-1} \mathbf{C} \mathbf{A}^T \mathbf{R_z}^{-1} \mathbf{A} \left( \mathbf{C}^T\mathbf{C}\right)^{-1} \mathbf{C}^T & \left( \mathbf{C}\mathbf{C}^T\right)^{-1} \mathbf{C} \\ \mathbf{0} & \left( \mathbf{C}^T\mathbf{C} \right)^{-1} \mathbf{C}^T & \mathbf{0} \\ \end{bmatrix} $

The problem is that $\left( \mathbf{C}^T\mathbf{C} \right)$ and $\left( \mathbf{C}\mathbf{C}^T\right)$ are not $\underline{both}$ invertible.

What am I doing wrong? The matrix is invertible, but I can't find a expression for it.

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    It is not contradictory. It just means you cannot operate by using this $3 \times 3$ block splitting. Turn to a $2 \times 2$ block splitting, the upper left block being $R_Z$. – 2017-02-01
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    I tried, but it falls on the same thing. Thanks! – 2017-02-02
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    I'm trying to find a matrix $\alpha$ that: $ \begin{bmatrix} \mathbf{R_z} & \mathbf{0} & \mathbf{A} \\ \mathbf{0} & \mathbf{0} & \mathbf{C} \\ \mathbf{A}^T & \mathbf{C}^T & \mathbf{0} \\ \end{bmatrix} ^{-1} = \begin{bmatrix} \mathbf{I} & \mathbf{0} & \mathbf{A} \\ \mathbf{0} & \mathbf{0} & \mathbf{C} \\ \mathbf{A}^T & \mathbf{C}^T & \mathbf{0} \\ \end{bmatrix} ^{-1} \alpha $ – 2017-02-02

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