I have a system of equations with an arbitrary number of equations (k). I'd like to use \vdots to compactly describe the system, like so:

  R(-1) &= \sum_{i=1}^m A(i)R(i-1) \\
  R(-2) &= \sum_{i=1}^m A(i)R(i-2) \\
        &\vdots                    \\
  R(-k) &= \sum_{i=1}^m A(i)R(i-k)

LaTeX output of above code with only minimal.cls and amsmath.sty

I'd like the dots, however, to be centered with = or the entire equation. Is there an elegant way of centering a column or row with AMSMath? I'm currently using an unholy combination of whitespace operators (\; \, etc.) to get the job done.


  R(-1) &= \sum_{i=1}^m A(i)R(i-1) \\
  R(-2) &= \sum_{i=1}^m A(i)R(i-2) \\
        &\mathrel{\makebox[\widthof{=}]{\vdots}} \\
  R(-k) &= \sum_{i=1}^m A(i)R(i-k)

| improve this answer | |
  • 3
    You beat me to it … anyhow, in case there is a wish to avoid using the calc package, here is my version: \setbox0\hbox{=}\mathrel{\makebox[\wd0]{\hfil\vdots\hfil}}. – Harald Hanche-Olsen Dec 23 '10 at 15:18
  • I like it! I had been looking at this answer, but I didn't like the need for all the temporary width definitions. \widthof is a great new tool for me. I'll leave the question open for a bit longer, but I think this is the answer. – Matt B. Dec 23 '10 at 15:18
  • @Harald the \hfil are not needed – user2478 Dec 23 '10 at 15:49
  • You're right; instead of \makebox[\wd0] I had \hbox to \wd0 and then I forgot to remove the \hfil after I changed it. – Harald Hanche-Olsen Dec 23 '10 at 20:39

Consider the package mathtools, which provides several corrections for and additions to amsmath.


It also provides a comfortable solution for your problem. You can even choose between a normal (\vdotswithin) and a short (\shortvdotswithin) distance.

  a &= b \\
  & \vdotswithin{=} \\
  & = c \\
  & = d

The result convinces.

Example showing vdotswithin and shortvdotswithin

More details can be found in the documentation of the package, section "Centered \vdots", where also the example above is taken from.

| improve this answer | |
  • This solution worked for me while Herbert's led to a misalignement, I'm not sure why. – anderstood Oct 20 '15 at 19:04

Here is another option (not very popular among TeXans):

  R(-1) &=& \sum_{i=1}^m A(i)R(i-1) \\
  R(-2) &=& \sum_{i=1}^m A(i)R(i-2) \\
        &\vdots& \\
  R(-k) &=& \sum_{i=1}^m A(i)R(i-k)
| improve this answer | |
  • 2
    the spacing is lousy in an eqnarray environment. You can it improve wit \arraycolsep=1.4pt – user2478 Dec 25 '10 at 9:30
  • Yes, I know... this is why I said it is not a popular option... – user1999 Dec 25 '10 at 13:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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