1
$\begingroup$

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

  • 1
    Have you tried any examples? Make up some $2\times2$ matrix $C$, and see how/whether $D^{-1}$ relates to $C^{-1}$?2012-09-18
  • 0
    I have tried that. It's difficult to see any relation... but I guess I don't know what to look for.2012-09-18
  • 2
    Look up the "Schur complement", or see my comment on http://math.stackexchange.com/questions/182309/block-inverse-of-symmetric-matrices2012-09-18
  • 0
    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!2012-09-18
  • 0
    How is it that you're asking how to find $\mathbf D^{-1}$ if you believe that $\mathbf D$ isn't invertible?2012-09-18
  • 0
    Have you tried block wise inversion? I think $D$ will be invertible provided $\sum_{i,j}C_{i,j}^{-1}\neq0$.2012-09-18
  • 0
    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.2012-09-18
  • 0
    @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](http://en.wikipedia.org/wiki/Schur_complement) (with all the mention of Schur complement, I thought at least one link to it would be useful).2012-09-18

0 Answers 0