Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

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

I'm working on a homework problem, and I'm stuck. I guess my linear algebra still needs some work...

I've arrived at

$\mathbf{D}= \left[ \begin{matrix} \mathbf{C} & \mathbf{1}^T \\ \mathbf{1} & 0 \end{matrix} \right] $

where $\mathbf{C}$ is a $n$ by $n$ matrix and $\mathbf{1}$ is a $n$ by $1$ vector of all ones. I need to find $\mathbf{D}^{-1}$. Can I express it in terms of $\mathbf{C}^{-1}$? Can I proceed at all? Does it help if $\mathbf{C}$ is symmetric?


share|cite|improve this question
Have you tried any examples? Make up some $2\times2$ matrix $C$, and see how/whether $D^{-1}$ relates to $C^{-1}$? – Gerry Myerson Sep 18 '12 at 6:11
I have tried that. It's difficult to see any relation... but I guess I don't know what to look for. – Paul Accisano Sep 18 '12 at 6:13
Look up the "Schur complement", or see my comment on… – copper.hat Sep 18 '12 at 6:36
Computing the Schur complement requires that $D$ (ie, the component in the lower right) be invertable, which it isn't... EDIT: Or at least that's what my cursory reading of the wiki article says. Your linked comment is much more helpful. Thanks! – Paul Accisano Sep 18 '12 at 6:40
How is it that you're asking how to find $\mathbf D^{-1}$ if you believe that $\mathbf D$ isn't invertible? – joriki Sep 18 '12 at 6:42

Your Answer


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

Browse other questions tagged or ask your own question.