Blockwise inversion case when $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ is singular

What means in blockwise matrix inversion when $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ is singular but $\textbf{A}$ is not? is that necessary and sufficient for the whole composed matrix be singular as well? are there any cases where the whole matrix is not singular but blockwise matrix inversion fails because $\textbf{D} - \textbf{C}\textbf{A}^{-1}\textbf{B}$ is singular?

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1 Answer

It cannot happen. You are starting with a block matrix of the form $$\mathbf{M}=\left(\begin{array}{cc} \mathbf{A} & \mathbf{B}\\ \mathbf{C} & \mathbf{D} \end{array}\right),$$ which we are assuming is invertible; that is, $\det(\mathbf{M})\neq 0$.

But if $\mathbf{A}$ is invertible, then we also have $$\det(\mathbf{M}) = \det(\mathbf{A})\det(\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}),$$ so $\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}$ cannot be singular.

To see this, note first that $$\det\left(\begin{array}{cc} \mathbf{X}& \mathbf{Y}\\ \mathbf{0} & \mathbf{W}\end{array}\right) = \det(\mathbf{X})\det(\mathbf{W})$$ and $$\det\left(\begin{array}{cc} \mathbf{X}&\mathbf{0}\\ \mathbf{Z}&\mathbf{W}\end{array}\right) = \det(\mathbf{X})\det(\mathbf{W}).$$ If $\mathbf{A}$ is invertible, then we have $$\mathbf{M}=\left(\begin{array}{cc} \mathbf{A} & \mathbf{B}\\ \mathbf{C} & \mathbf{D}\end{array}\right) = \left(\begin{array}{cc} \mathbf{A} & \mathbf{0}\\ \mathbf{C} & \mathbf{I} \end{array}\right)\left(\begin{array}{cc} \mathbf{I} & \mathbf{A}^{-1}\mathbf{B}\\ \mathbf{0} & \mathbf{D}-\mathbf{CA}^{-1}\mathbf{B} \end{array}\right),$$ where $\mathbf{I}$ is a suitably sized identity matrix. Since the determinant of a product is the product of the determinants, we get the desired formula.

So if $\mathbf{M}$ and $\mathbf{A}$ are both nonsingular, then $\mathbf{D}-\mathbf{CA}^{-1}\mathbf{B}$ is nonsingular as well.

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thanks! its pretty clear now – lurscher Jun 21 '11 at 20:19