Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

Thanks in advance.

share|cite|improve this question
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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.