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.

3 Answers 3


  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)

  • 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}}. Dec 23, 2010 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.
    – mbauman
    Dec 23, 2010 at 15:18
  • @Harald the \hfil are not needed
    – user2478
    Dec 23, 2010 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. Dec 23, 2010 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.

  • This solution worked for me while Herbert's led to a misalignement, I'm not sure why.
    – anderstood
    Oct 20, 2015 at 19:04
  • this is a nice solution.
    – yanpengl
    Apr 29 at 23:09

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)
  • 2
    the spacing is lousy in an eqnarray environment. You can it improve wit \arraycolsep=1.4pt
    – user2478
    Dec 25, 2010 at 9:30
  • Yes, I know... this is why I said it is not a popular option...
    – user1999
    Dec 25, 2010 at 13:55

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