Math & Science ⇒ Array dimension problem

[adnaniut]

Hi,

For the following array,
\left[
\underbrace{\begin{array}{c c c c}
p_2^m & p_1 p_2^m & \cdots & p_1^{n-1} p_2^m
\end{array}
}_n
\right.\underbrace{\left. \begin{array}{c c c c}
p_1^n & p_2 p_1^n &\cdots & p_2^{m-1} p_1^n \end{array}\right]}_m \times \left[
\begin{array}{c}
\frac{\displaystyle \sum_{i=0}^{n-1} q_2^i - q_2^n }{q_2^n} \\
\frac{\displaystyle \sum_{i=0}^{n-2} q_2^i - q_2^n }{q_2^n} \\
\vdots \\
\frac{\displaystyle \sum_{i=0}^{n-(n+1)} q_2^i - q_2^n }{q_2^n} \\
\frac{1 - q_2^n }{q_2^n} \\
0 \\
\vdots \\
0\\
\frac{\displaystyle q_1 \sum_{i=0}^{n-1} q_2^i}{q_2^n}
\end{array}\right]

I want to impose a curly bracket outside the main bracket to show dimension size. I could not do it for column vector
For row vector also its not perfect.

Please help me on how to do it also some tips on how to do the same for a matrix.

Thanks
Adnan

[localghost]

Please, always post full examples so that everybody instantly can comprehend the problem. If I understood you right, the code below should do what you want.

Code: Select all

\documentclass[11pt,a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{geometry}
\usepackage{amsmath}

\begin{document}
\[
  \left\{
  \left[
  \underbrace{%
    \begin{array}{cccc}
      p_2^m & p_1 p_2^m & \cdots & p_1^{n-1} p_2^m
    \end{array}
  }_n
  \underbrace{%
    \begin{array}{cccc}
      p_1^n & p_2 p_1^n &\cdots & p_2^{m-1} p_1^n
    \end{array}
  }_m
  \right]
  \times
  \left[
  \begin{array}{c}
    \frac{\displaystyle \sum_{i=0}^{n-1} q_2^i - q_2^n }{q_2^n} \\
    \frac{\displaystyle \sum_{i=0}^{n-2} q_2^i - q_2^n }{q_2^n} \\
    \vdots \\
    \frac{\displaystyle \sum_{i=0}^{n-(n+1)} q_2^i - q_2^n }{q_2^n} \\
    \frac{1 - q_2^n }{q_2^n} \\
    0 \\
    \vdots \\
    0\\
    \frac{\displaystyle q_1 \sum_{i=0}^{n-1} q_2^i}{q_2^n}
  \end{array}
  \right]
  \right\}
\]
\end{document}

If that is not what you are after, please be more precise in your descriptions.


Best regards and welcome to the board
Thorsten

[adnaniut]

Hi Thorsten,

Thanks a lot for your reply. Actually I am new into these things and thats why was a little confused about the whole matter.

Your reply solved my problem partially; for the underbrace part. The other part is,
as you divided the whole row vector into two underbraces - n and m, I want to do the same for column vector. I want to use similarly two side braces for the column vector and indicate n and m [only in one side]. So that someone can connect first n number of components in row vector with first n number of components in column vector.

I hope you can understand my problem now.

Thanks again
Adnan

[gmedina]

Hi, Adnan

you could try something like the following:

Code: Select all

\documentclass[11pt,a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{geometry}
\usepackage{amsmath}

\begin{document}
\[
  \left[
  \underbrace{%
    \begin{array}{cccc}
      p_2^m & p_1 p_2^m & \cdots & p_1^{n-1} p_2^m
    \end{array}
  }_n
  \underbrace{%
    \begin{array}{cccc}
      p_1^n & p_2 p_1^n &\cdots & p_2^{m-1} p_1^n
    \end{array}
  }_m
  \right]
  \times
  \left[
  \begin{array}{c}
    \dfrac{\displaystyle\sum_{i=0}^{n-1} q_2^i - q_2^n }{q_2^n} \\[3mm]
    \dfrac{\displaystyle\sum_{i=0}^{n-2} q_2^i - q_2^n }{q_2^n} \\
    \vdots \\
    \dfrac{\displaystyle\sum_{i=0}^{n-(n+1)} q_2^i - q_2^n }{q_2^n} \\[6mm]
    \dfrac{1 - q_2^n }{q_2^n} \\[4mm]
    0 \\
    \vdots \\
    0\\
    \dfrac{\displaystyle q_1 \sum_{i=0}^{n-1} q_2^i}{q_2^n}
  \end{array}
  \right]\hspace*{-6mm}
  \begin{array}{l}
    \left.
    \begin{array}{l}
      \rule{0pt}{6.3cm} % change the value here
    \end{array}
    \right\rbrace n\\
    \left.
    \begin{array}{l}
      \rule{0pt}{4.3cm} % change the value here
    \end{array}
    \right\rbrace m
  \end{array}
\]
\end{document}

Feel free to change the lengths used to control the lengths of the brackets (the corresponding lines are commented in my code with "change the value here")

[CrazyHorse]

Feel free to change the lengths used to control the lengths of the brackets (the corresponding lines are commented in my code with "change the value here")

put it into a \vphantom:

Code: Select all

\documentclass[11pt,a4paper]{article}
\usepackage[T1]{fontenc}
\usepackage[utf8]{inputenc}
\usepackage{geometry}
\usepackage{amsmath}
\newcommand\dsum{\displaystyle\sum}

\begin{document}
\[
  \left[
  \underbrace{%
    \begin{array}{cccc}
      p_2^m & p_1 p_2^m & \cdots & p_1^{n-1} p_2^m
    \end{array} }_n
  \underbrace{%
    \begin{array}{cccc}
      p_1^n & p_2 p_1^n &\cdots & p_2^{m-1} p_1^n
    \end{array} }_m
  \right]
  \times
  \left[
  \begin{array}{c}
    \dfrac{\dsum_{i=0}^{n-1} q_2^i - q_2^n }{q_2^n} \\[3mm]
    \dfrac{\dsum_{i=0}^{n-2} q_2^i - q_2^n }{q_2^n} \\
    \vdots \\
    \dfrac{\dsum_{i=0}^{n-(n+1)} q_2^i - q_2^n }{q_2^n} \\[6mm]
    \dfrac{1 - q_2^n }{q_2^n} \\[4mm]
    0 \\
    \vdots \\
    0\\
    \dfrac{\dsum_{i=0}^{n-1} q_2^i}{q_2^n}
  \end{array}
  \right]\mkern-15mu
  \begin{array}{l}
    \left.
     \vphantom{
      \begin{array}{c}
      \dfrac{\dsum_{i=0}^{n-1} q_2^i - q_2^n }{q_2^n} \\[3mm]
      \dfrac{\dsum_{i=0}^{n-2} q_2^i - q_2^n }{q_2^n} \\
      \vdots \\
      \dfrac{\dsum_{i=0}^{n-(n+1)} q_2^i - q_2^n }{q_2^n} \\[6mm]
      \end{array} }
    \right\} n\\
    \left.
     \vphantom{
      \begin{array}{l}
      \dfrac{1 - q_2^n }{q_2^n} \\[4mm]
      0 \\
      \vdots \\
      0\\
      \dfrac{\dsum_{i=0}^{n-1} q_2^i}{q_2^n}
      \end{array} }
    \right\rbrace m
  \end{array}
\]
\end{document}

Herbert

[adnaniut]

Gmedina and Herbert,

Thanks very much for helping me out.

Adnan

[adnaniut]

Just for consistency in coding, can you guys please advise me on if there is any way of doing the first part of problem similar to that in 2nd part specifying underbrace length.

Thanks
Adnan