1

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?

Thanks

Paul Accisano
  • 235
  • 1
    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 06:11
  • 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 06:13
  • 3
    Look up the "Schur complement", or see my comment on http://math.stackexchange.com/questions/182309/block-inverse-of-symmetric-matrices – copper.hat Sep 18 '12 at 06:36
  • 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 06:40
  • 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 06:42
  • Have you tried block wise inversion? I think $D$ will be invertible provided $\sum_{i,j}C_{i,j}^{-1}\neq0$. – Daryl Sep 18 '12 at 06:44
  • Sorry, I meant $D$ in the wiki article, ie, the the component in the lower right, ie 0. The linked question is more than sufficient to answer my question. Thanks for the helpful replies. – Paul Accisano Sep 18 '12 at 06:48
  • @PaulAccisano I think you'll find that you need to permute blocks (at least mentally) so that $\mathbf C$ rather than $0$ comes into the lower-right position of $D$ in the WP article (with all the mention of Schur complement, I thought at least one link to it would be useful). – Marc van Leeuwen Sep 18 '12 at 08:06

0 Answers0