# Inverse of a partitioned matrix

Suppose I have a partitioned matrix $$\begin{pmatrix} 0 & F^T \\ F & R \\ \end{pmatrix}$$ where $0$ is $k \times k$, $F$ is $n \times k$ and $R$ is $n \times n$. I would much appreciate if someone could help me find the inverse of the matrix. I have seen some formula online but they require the first entry to be non-singular.

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You need assumptions on $F,R$. What if $F=0$, for instance? –  1015 Apr 6 '13 at 16:36
is $R$ invertible and $F\neq 0$ ? –  Dominic Michaelis Apr 6 '13 at 16:36
R is invertible and F is non zero but not invertible –  Reven Apr 6 '13 at 16:40

Wikipedia has the alternative formula for the inverse of a partitioned matrix if the leading submatrix is singular. Applied to your system, we then have

$$\begin{pmatrix}\mathbf 0&\mathbf F^\top\\\mathbf F&\mathbf R\end{pmatrix}^{-1}= \begin{pmatrix}-(\mathbf F^\top\mathbf R^{-1}\mathbf F)^{-1}&(\mathbf F^\top\mathbf R^{-1}\mathbf F)^{-1}\mathbf F^\top\mathbf R^{-1}\\\mathbf R^{-1}\mathbf F(\mathbf F^\top\mathbf R^{-1}\mathbf F)^{-1}&\mathbf R^{-1}-\mathbf R^{-1}\mathbf F(\mathbf F^\top\mathbf R^{-1}\mathbf F)^{-1}\mathbf F^\top\mathbf R^{-1}\end{pmatrix}.$$

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Thanks a lot. This solves the problem. –  Reven Apr 6 '13 at 16:56
Just a remark, you need $F^TR^{-1}F$ to be invertible for this formula to make sense. –  1015 Apr 6 '13 at 17:12
@julen, the Wikipedia entry says as much. Reven, also look up the "Schur complement" (before copper.hat beats me to it ;)). –  Ｊ. Ｍ. Apr 6 '13 at 17:14