I'm a new user in texmaker.
How can I produce this equation? Thank you for your help
[Tex/LaTex] Matrix equation
equationsmatrices
Related Solutions
With the code of this answer, and a bit of effort (manual tweaking). I got this (far from optimal, but if you have to fight with this kind of matrix only one time, it might work).
I used the \coolunder, \coolover, \coolrightbrace and \coolleftbrace from the answer I linked, but a bit tweaked to adapt the mtpro2 package. The reason to use the mtpro2 package is that it provides those curly long braces. If you don't have this font/package, just change the definitions of the commands to
\newcommand\coolover[2]{\mathrlap{\smash{\overbrace{\phantom{%
\begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2}
\newcommand\coolunder[2]{\mathrlap{\smash{\underbrace{\phantom{%
\begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2}
\newcommand\coolleftbrace[2]{%
#1\left\{\vphantom{\begin{matrix} #2 \end{matrix}}\right.}
\newcommand\coolrightbrace[2]{%
\left.\vphantom{\begin{matrix} #1 \end{matrix}}\right\}#2}
Code
Here it is the code. As I said, it is far from optimal (it has a lot of phantoms):
\documentclass{scrartcl}
\usepackage{mathtools}
\usepackage{newtxtext}
\usepackage[lite]{mtpro2}
\usepackage{multirow}
\usepackage[hmargin=1.5cm]{geometry}% You have to find the way to deal with the margins.
% You can comment this (only used to get the appearence of the image).
\setkomafont{captionlabel}{\scshape}
\setcounter{equation}{4}
\setcounter{figure}{1}
% The commands used to get the desired braces.
\newcommand\coolover[2]{\mathrlap{\smash{\overcbrace{\phantom{%
\begin{matrix} #2 \end{matrix}}}^{\mbox{$#1$}}}}#2}
\newcommand\coolunder[2]{\mathrlap{\smash{\undercbrace{\phantom{%
\begin{matrix} #2 \end{matrix}}}_{\mbox{$#1$}}}}#2}
\newcommand\coolleftbrace[2]{%
#1\LEFTRIGHT\{.{\vphantom{\begin{matrix} #2 \end{matrix}}}}
\newcommand\coolrightbrace[2]{%
\LEFTRIGHT.\}{\vphantom{\begin{matrix} #1 \end{matrix}}}#2}
\newcommand\Vdots{\vdots}% You can change the size/appearence of the dots in
\newcommand\Cdots{\cdots}% the matrixes easily changing this definitions.
\begin{document}
\begin{center}
\bfseries PHILIP E. PFEIFER AND STUART JAY DEUTCH
\end{center}
\begin{figure}[h!]
\small
\centering
\begin{equation}
\begin{matrix}
\coolleftbrace{s = 1}{\\ \\ \vphantom{\Vdots} \\ \\} \\
\coolleftbrace{s = 2}{\\ \\ \vphantom{\Vdots} \\ \\} \\
\vphantom{\Vdots} \\
\coolleftbrace{s = k}{\\ \\ \vphantom{\Vdots} \\ \\}
\end{matrix}%
\begin{bmatrix}
\gamma_{00}(1) \\
\gamma_{01}(1) \\
\Vdots \\
\gamma_{\lambda0}(1) \\ \hline
\gamma_{00}(2) \\
\gamma_{01}(2) \\
\Vdots \\
\gamma_{\lambda0}(2) \\ \hline
\Vdots \\ \hline
\gamma_{00}(1) \\
\gamma_{01}(1) \\
\Vdots \\
\gamma_{\lambda0}(1)
\end{bmatrix}
=
\left[
\begin{array}{@{} cccc|cccc|c|c @{}}
\gamma_{00}(0) & \gamma_{01}(0) & \Cdots & \gamma_{0\lambda}(0) & \gamma_{00}(-1) & \gamma_{01}(-1) & \Cdots & \gamma_{0\lambda}(-1) & \multirow{4}{*}{$\Cdots$} & \multirow{4}{*}{$(1 - k)$} \\
\gamma_{10}(0) & \gamma_{11}(0) & \Cdots & \gamma_{1\lambda}(0) & \gamma_{10}(-1) & \gamma_{11}(-1) & \Cdots & \gamma_{1\lambda}(-1) & & \\
\multicolumn{4}{c|}{\Vdots} & \multicolumn{4}{c|}{\Vdots} & & \\
\gamma_{\lambda0}(0) & \gamma_{\lambda1}(0) & \Cdots & \gamma_{\lambda\lambda}(0) & \gamma_{\lambda0}(-1) & \gamma_{\lambda1}(-1) & \Cdots & \gamma_{\lambda\lambda}(-1) & & \\ \hline
\multicolumn{4}{c|}{\multirow{4}{*}{$(1)$}} & \multicolumn{4}{c|}{\multirow{4}{*}{$(0)$}} & & \multirow{4}{*}{$(2 - k)$} \\
& & & & & & & & & \\
& & & & & & & \vphantom{\Vdots} & & \\
& & & & & & & & & \\ \hline
\multicolumn{4}{c|}{\Vdots} & \multicolumn{4}{c|}{\Vdots} & & \Vdots \\ \hline
\multicolumn{4}{c|}{\multirow{4}{*}{$(k - 1)$}} & \multicolumn{4}{c|}{\multirow{4}{*}{$(k - 2)$}} & \multirow{4}{*}{$\Cdots$} & \multirow{4}{*}{$(0)$} \\
& & & & & & & & & \\
& & & & & & & \vphantom{\Vdots} & & \\
\coolunder{j = 1}{\hphantom{\gamma_{00}(0)} & \hphantom{\gamma_{01}(0)} & \hphantom{\Cdots} & \hphantom{\gamma_{0\lambda}(0)}} & \coolunder{j = 2}{\hphantom{\gamma_{00}(-1)} & \hphantom{\gamma_{01}(-1)} & \hphantom{\Cdots} & \hphantom{\gamma_{0\lambda}(1)}} & & \coolunder{j = k}{\hphantom{(1 - k)}}
\end{array}
\right]
\begin{bmatrix}
\phi_{10} \\
\phi_{11} \\
\Vdots \\
\phi_{1\lambda} \\ \hline
\phi_{20} \\
\phi_{21} \\
\Vdots \\
\phi_{2\lambda} \\ \hline
\Vdots \\ \hline
\phi_{k0} \\
\phi_{k1} \\
\Vdots \\
\phi_{k\lambda}
\end{bmatrix}
\end{equation}\bigskip
\caption{The space-time analogue of the Yule-Walker equations}
\end{figure}
\end{document}
And this is how it looks:
very simple ... just add a \vphantom of the height of the tallest matrix to the other two.
\documentclass{article}
\usepackage[utf8]{inputenc}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{bm}
\usepackage{mathtools}
\begin{document}
\begin{equation}
\underset{\pmb{y}\in \mathbb{R}^m}{%
\vphantom{\begin{bmatrix}0\\0\\0\\0\\0\\0\end{bmatrix}}
\begin{bmatrix}
18\\5\\15\\-9\\-3\\
\end{bmatrix}}=
\underset{\pmb{A}\in \mathbb{R}^{m\times n}}{%
\vphantom{\begin{bmatrix}0\\0\\0\\0\\0\\0\end{bmatrix}}
\begin{bmatrix}
4 & 0 & 0 & -2 & 0 & 0 \\
0 & 0 & -1 & 0 & 3 & 0\\
0 & 5 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & -1 & 0 & -4 \\
1 & 0 & 0 & 0 & -5 & 0
\end{bmatrix}}
\underset{\pmb{x}\in \mathbb{R}^n}{\begin{bmatrix}
2\\ 3\\ -2\\ -5\\ 1\\ 4
\end{bmatrix}}
\end{equation}
\end{document}
a couple of comments:
- don't leave blank lines inside of any math; this will always result in an error.
- i removed an extra \begin{equation} from the provided example.
Best Answer