[Tex/LaTex] Rotate matrix in LaTeX

matricesrotating

I have a matrix size 25 and I need to rotate it in LaTeX so that it fits the page. How do I do??

PS.: I tried the landscape command, but it did not work because rotated the page but not its contents, that is, the array does not rotated.

Somebody help me??

My LaTex document:

\documentclass[a4paper, 12pt]{report}

\usepackage{graphicx,url}
\usepackage[active]{srcltx} %habilita busca reversa
\usepackage[brazil]{babel}
\usepackage[normalem]{ulem}
\usepackage{float}
\usepackage[latin1]{inputenc}
\usepackage{enumerate} %permite definir o tipo dos itens no ambiente enumerate
\usepackage{indentfirst, amsfonts, amsmath, amsthm, amssymb, amscd}
\usepackage{amsmath,amsfonts,amscd,bezier}
\usepackage[a4paper,left=1.5cm,right=1.5cm,top=2.5cm]{geometry}
\usepackage{booktabs}
\usepackage{multirow}
\usepackage{multicol}
\usepackage{lscape}
\usepackage{pdflscape}

\begin{document}


$$\textbf{D}_{euc} = \left[
\begin{array}{cccccccccccccccccccccccc}

23.18 & & & & & & & & & & & & & & & & & & & & & & & \\

15.47 & 31.95 & 32.68 & 4.87 & 23.98 & 39.87 & 33.28 & 39.33 & 33.72
& 18.75 & 17.61 & 37.95 & 20.72 & 37.97 \\

\end{array}\right].$$

\end{document}

Best Answer

The matrix in your MWE has size only 2x14, not 2x25, so it can be fit in normal page orientation:

\documentclass[a4paper, 12pt]{report}

\usepackage{graphicx}
%\usepackage[active]{srcltx} %habilita busca reversa
%\usepackage[brazil]{babel}
\usepackage[normalem]{ulem}
\usepackage{float}
\usepackage[latin1]{inputenc}
\usepackage{enumerate} %permite definir o tipo dos itens no ambiente enumerate
\usepackage{indentfirst, amsfonts, amsmath, amsthm, amssymb, amscd}
\usepackage{bezier}
\usepackage[a4paper,
            left=1.5cm,right=1.5cm,top=2.5cm,
            showframe]{geometry}
\usepackage{booktabs,multirow}
\usepackage{multicol}
\usepackage{lscape}
\usepackage{pdflscape}
\usepackage{url}

\setcounter{MaxMatrixCols}{20}% <---
\begin{document}

\[\setlength{\arraycolsep}{3pt}% <---
\mathbf{D}_{euc} = 
\begin{bmatrix}% <---
23.18 & & &   & & & &  & & & &  & & \\
15.47 & 31.95 & 32.68 & 4.87
    & 23.98 & 39.87 & 33.28 & 39.33
        & 33.72 & 18.75 & 17.61 & 37.95
            & 20.72 & 37.97         \\
\end{bmatrix}.
\]%end{equation}

\end{document}

In above MWE I commented out two of your packages which I haven't installed. As you can see, I use \bmatrix instead array. For which I increased the number of columns with \setcounter{MaxMatrixCols}{20} and also reduced inter-column space with \setlength{\arraycolsep}{3pt}. Result is:

If real matrices is really so big as you mentioned in your question, then you have two possibilities of how to write it:

in landscape page mode by enclosing equation by \begin{landscape} ... \end{landscape}
break the matrix into two parts, as it is shown for example in answers on question Shrinking or splitting some equations in a group of equations

Related Solutions

[Tex/LaTex] TikZ: How to rotate a matrix (of nodes)

Ah well - I guess this is the answer, pg 487 of the pgfmanual.pdf:

Rotations and scaling. The matrix node is never rotated or shifted, because the current coordinate transformation matrix is reset (except for the translational part) at the beginning of \pgfmatrix. This is intentional and will not change in the future. If you need to rotate the matrix, you must install an appropriate canvas transformation yourself.

However, nodes and stuff inside the cell pictures can be rotated and scaled normally.

... Also, from matrix nodes with sloped option? - pgf-users:

It does say in the manual that it isn't possible (in section 16.2 "Matrices are Nodes"), and that the transformation matrix is reset at the beginning of a matrix. The (internal) use of \halign precludes any kind of fancy transformations. It would be pretty hard to do matrices without using \halign.

... HOWEVER ...

... matrix nodes with sloped option? - pgf-users also says:

Indeed.

However, you can do the following: Put the whole node inside a
tikzpicture, which in turn you put in a node that has the sloped
option. Something like
... (A) -- (B) node[midway,sloped] {\tikz \matrix ...;};

which means that the code above can be modified like this:

\begin{tikzpicture}[font=\tt]
    \matrix (xA) [anchor=west,text ragged]
    {% 
        \node(xA1) [draw,right,minimum width=5em] {horizon1} ;  \\
        \node(xA2) [draw,right] {horiz2} ;  \\
%       \node(xA3) [draw,anchor=west,minimum width=5em] {\begin{minipage}{5em}hor3\end{minipage}} ; \\
        \node(xA3) [draw,anchor=west,minimum width=5em] {\parbox{5em}{hor3}} ;  \\
    };
    \matrix (yA) [below left=of xA]
    {% 
        \node(yA1) [draw,right,rotate=270] {vert1} ;    &
        \node(yA2) [draw,right,rotate=270] {vert2} ;    &
        \node(yA3) [draw,right,rotate=270] {vert3} ;    \\
    };
    \node (zzA) [rotate=270,below left=of yA] {
        \tikz \matrix (zA) 
        {% 
            \node(zA1) [draw,right] {horvert1} ;    \\
            \node(zA2) [draw,right] {horvert2} ;    \\
            \node(zA3) [draw,right] {horvert3} ;    \\
        };  
    };
\end{tikzpicture}

... which will finally result with the originally desired image:

tikz hor/vertical matrix rotation OK

[Tex/LaTex] Handling a too big matrix equation

Actually there is a strong structure in your matrix though I don't know how to express it without typing it out completely. So here is an attempt to describe it. By making it 8x8 and removing all the dots because currently they are wrong. 1 ... 0 doesn't make any sense and also ambiguous. The superscripted c terms and others can be grouped and that makes the matrix a block partitioned one.

\documentclass{article}
\usepackage{amsmath}
\usepackage{rotating}

\begin{document}

\begin{sideways}
\thispagestyle{empty}%
\makebox[\textheight][c]{\begin{minipage}{\paperheight}
\setlength\arraycolsep{4pt}
\begin{align*}
         &\left[ {\begin{array}{ccccccccccc}
         1                       & 0                       & 0                       &0              & -P(N_k^a|L_k^c) & 0                       & 0                      &0\\
         -P(N_{k+1}^d|L_{k+1}^c) & 1                       & 0                       &0              & 0               & -P(N_{k+1}^a|L_{k+1}^c) & 0                      &0\\
         0                       & -P(N_{k+2}^d|L_{k+2}^c) & 1                       &0              & 0               & 0                       &-P(N_{k+2}^a|L_{k+2}^c) &0\\
         0                       & 0                       & -P(N_{c-1}^d|L_{c-1}^c) & 1             & 0               & 0                       & 0                      & -P(N_{c-1}^a|L_{c-1}^c)\\
         -P(N_{k+1}^d|L_{k+1})   & 0                       & 0                       &0              & 1               & -P(N_{k+1}^a|L_{k+1})   & 0                      &0\\
         0                       & -P(N_{k+2}^d|L_{k+2})   & 0                       &0              & 0               & 1                       &-P(N_{k+2}^a|L_{k+2})   &0\\
         0                       & 0                       & -P(N_{c-1}^d|L_{c-1})   &0              & 0               & 0                       & 1                      &-P(N_{c-1}^a|L_{c-1})\\
         0                       & 0                       & 0                       & -P(N_c^d|L_c) & 0               & 0                       & 0                      & 1
         \end{array} } \right]
         \cdot\left[ {\begin{array}{c}
         E(\widetilde{A}_{kk}^{cn}|L_k^c)\\
         E(\widetilde{A}_{k k+1}^{cn}|L_{k+1}^c)\\
         E(\widetilde{A}_{k k+2}^{cn}|L_{k+2}^c)\\
         E(\widetilde{A}_{k c-1}^{cn}|L_{c-1}^c)\\
         E(\widetilde{A}_{k k+1}^{cn}|L_{k+1})\\
         E(\widetilde{A}_{k k+2}^{cn}|L_{k+2})\\
         E(\widetilde{A}_{k c-1}^{cn}|L_{c-1})\\
         E(\widetilde{A}_{kc}^{cn}|L_{c})\\
         \end{array} } \right]\\
         &= \begin{bmatrix}K &L\\M&N\end{bmatrix}\begin{bmatrix}E^c_1\\E_2\end{bmatrix}
\end{align*}
\end{minipage}}
\end{sideways}
\newpage
where 
\begin{align*}
K_{mm}    &= 1, \\
K_{m,m-1} &= -P(N_{k+m}^d|L_{k+m}^c),\\
N_{mm}    &= 1, \\
N_{m-1,m} &= -P(N_{k+m}^a|L_{k+m}),\\
\end{align*}
and zero otherwise for $m=1,\ldots,c-(k+1)$. Similarly $L,M$ are diagonal matrices such that
\begin{align*}
L &=\operatorname{diag}\{-P(N_{k+m}^a|L_{k+m}^c)\}\\
M &=\operatorname{diag}\{-P(N_{k+m}^d|L_{k+m}) \}
\end{align*}
for $m=0,\ldots,c-(k+1)$. 
\end{document}

enter image description here

So I would use K,L,M,N and put a description along the lines of

enter image description here

instead of fitting a beast in a journal paper. I need to warn you that this has the structure of a linear system Ax=b and I didn't include the operation on b in the document. But it is essentially a row permutation with 1,3,5,7,...2,4,6,....

Best Answer

Related Solutions

[Tex/LaTex] TikZ: How to rotate a matrix (of nodes)

[Tex/LaTex] Handling a too big matrix equation

Related Question