TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I was using the following code for preparation of a poster in Latex using TeXstudio 2.6.6

\newcommand{\R}{{\mathbb R}}
\newcolumntype{Z}{>{\centering\arraybackslash}X} % centered tabularx columns
\newcommand{\pphantom}{\textcolor{ta3aluminium}} % phantom introduces a vertical space in p formatted table columns??!!
 \title{\huge Comparisons of   $B_{row}$-splittings  of Matrices}
\author{ \vspace{.2cm} {\bf \large{ author}}\\ \vspace{.2cm} Research Supervisor- prof john\\ \vspace{.2cm} School of statistics}
\institute{institute name}
    \begin{minipage}[T]{.95\textwidth}  % tweaks the width, makes a new \textwidth
In many practical problems we come across the problem of computing a 
solution to a system of linear equations in $n$ unknowns \textcolor{blue}
where $A$ is a real rectangular $m\times n$ matrix and $b$ is a real $m$-vector.
 In a wide variety of such problems, including the Neumann problem  and those 
for elastic bodies with free surfaces, the finite difference formulations lead to 
a singular, consistent linear system (\ref{eq0}) where $A$ is large and sparse.
 Here the general method of solution is iterative in nature. Iterative methods
 where $A$ is rectangular or inconsistent, have been studied
in \cite{bpcones}. The authors  used generalized matrix
inverses for computing least square solutions in the inconsistent
where $U^{\dag}$ is the Moore-Penrose inverse of $U$.
The above scheme is said to be convergent if the spectral radius of
$U^{\dag}V$ is less than 1. For a proper splitting, the authors of 
\cite{bpcones} have shown that  $x=A^{\dag}b$ for any initial
vector $x^{0}$ if and only if (\ref{eq01}) is convergent.
\textcolor{darkgreen}{Moore-Penrose inverse}\\
The  Moore-Penrose inverse of $A\in {\R}^{m\times n}$
is the unique matrix $A^{\dag}\in {\R}^{n\times m}$ that satisfies the following four    
If $A^{\dag}\geq 0$, then it is semimonotone. Berman and Plemmons, \cite{bpmonotono}  
showed that
 $A^{\dag}\geq 0$  if and only if $ Ax\in {\R}^m_+ +N(A^T) ~~\mbox{and}~~x \in
R(A^{T})$ imply  $x\geq 0.$ 

\textcolor{darkgreen}{Row monotone matrix}\\
 $A\in {\R}^{m\times n}$ is said to be row monotone \cite{bpmonotono} if  $ Ax\geq 0   
~~\mbox{and}~~x \in
R(A^{T})$ imply  $x\geq 0.$ \\  

$A$ is row monotone if and only if $A$ is 
$\{1,4\}$-monotone. ($\{1,4\}$-monotone means there is a nonnegative $G$ satisfying  
$AGA=A$ and $(GA)^T=GA$.)
If $A^{\dag} \geq 0$,  then $A$ and $A^T$ are  row monotone. 
However, the converse is not true. \\

 A  decomposition $A=U-V$ of $A\in {\R}^{m\times n}$ is called 
 \textcolor{blue}{{\it positive}}  if $U\geq 0$ and $V\geq 0$. \\

\textcolor{darkgreen}{ $B_{row}$-splitting (Definition 2.6, \cite{mis})}\\
A positive   proper splitting $A=U-V$ of $A\in {\R}^{m\times n}$ is called a $B_{row}$-  
splitting if it satisfies the following conditions: \\
(i) $VU^{\dag}\geq 0$, and \\
(ii) $ Ax,~ Ux\geq 0 ~~\mbox{and}~~x \in R(A^{T})$ imply  $x\geq 0.$\\

\textcolor{darkgreen}{Theorem 1} (Theorem 2.7, \cite{mis})\\
Let $A\in {\R}^{m\times n}$. Suppose that $R(A) \cap int(\mathbb{R}^m_+) \neq   
\emptyset$. Consider the following statements:\\
(a) $A$ is row monotone.\\
(b) $ {\R}_{+}^{m}\cap R(A) \subseteq A{\R}_{+}^{n}$.\\
(c) There exists $x^{0}\in {\R}_{+}^{n}$ such that $Ax^{0} \in int({\R}_{+}^{m})$.\\
Then, we have (a) $\Rightarrow$ (b) $\Rightarrow$ (c).\\
Suppose that $A$ has a $B_{row}$-splitting. Then each of the above is equivalent to the   
(d) $\rho(VU^{\dag})<1$.


\textcolor{darkgreen}{Theorem 2} (Theorem 2.12, \cite{mis})\\
Suppose that $A$ is row monotone and $R(A) \cap int({\R}^{m}_{+}) \neq \emptyset$ for   
$A\in {\R}^{m\times n}$.
 Further, let $A^{\dag}A\geq 0$. Then $A$ possesses a  $B_{row}$-splitting $A=U-V$  with     
To  present a more general
  convergence theorem for $B_{row}$-splitting and to  compare two $B_{row}$-splittings.
% ---------------------------------------------------------%
% end the column

% ---------------------------------------------------------%
% Set up a column 
    \begin{minipage}[T]{.95\textwidth} % tweaks the width, makes a new \textwidth
      \parbox[t][\columnheight]{\textwidth}{ % must be some better way to set the the height, width and textwidth simultaneously
        % Since all columns are the same length, it is all nice and tidy.  You have to get the height empirically
        % ---------------------------------------------------------%
        % fill each column with content

       \begin{block}{MAIN RESULTS}
We begin with the following lemma which is useful to prove our main results of this section.
\textcolor{yellow}{LEMMA 3}\\
\textcolor{blue}{(a) If $A$ is row monotone, $V\geq 0$ and $R(V)\subseteq R(A)$, then  
$A^{\dag}V\geq 0$.\\
(b) If $A^T$ is row monotone, $V\geq 0$ and $N(A)\subseteq N(V)$, then $VA^{\dag}\geq 

Now, we obtain a new  convergence theorem for  $B_{row}$-splittings which holds
 even without the assumption $R(A) \cap int({\R}^{m}_{+}) \neq \emptyset$. Thus, the  
present one is more general  than the earlier one (Theorem 2).

\textcolor{yellow}{THEOREM  4}\\
\textcolor{blue}{ Let  $A=U-V$ be a  $B_{row}$-splitting of $A\in {\R}^{m\times n}$. If  
$A$ is row monotone,then\\
(a) $A^{\dag}\geq U^{\dag}$;\\
(b) $\rho(VA^{\dag})\geq \rho(VU^{\dag})$;\\
(c) $\rho(VU^{\dag})=\rho(U^{\dag}V)=\frac{\rho(A^{\dag}V)}{1+\rho(A^{\dag}V)}<1$.}

Theorem 2 enables us that there exist several $B_{row}$-splittings of a given matrix.
In this direction, we  present  two comparison theorems for  $B_{row}$-splittings.

\textcolor{yellow}{THEOREM 5}\\
\textcolor{blue}{ Let  $A=U_{1}-V_{1}=U_{2}-V_{2}$ be two  $B_{row}$-splittings of
$A$. If  $A$ is  row monotone and $V_{2}\geq V_{1}$, then
$$1> \rho(U_{2}^{\dag}V_{2}) \geq \rho(U_{1}^{\dag}V_{1}).$$}

\textcolor{yellow}{EXAMPLE 6 }
Let $A=\left(
       1 & 1 \\
       1 & 1 \\
$. Clearly $A$ is row monotone. Setting
$U_{1}=3A$ and $U_{2}=4A$.  We  then have $0\leq V_{1}=2A\leq 3A=V_{2}$.  Hence
$\rho(V_{1}U_{1}^{\dag})=\frac{2}{3}\leq \frac{3}{4}=\rho(V_{2}U_{2}^{\dag})<1$.

The condition $V_2\geq V_1$ can not be dropped. For example, set $V_{1}=3A$ and   
$V_{2}=2A$.Then the implication $$\rho(U_{1}^{\dag}V_{1}) \leq \rho(U_{2}^{\dag}V_{2}) <  
1$$ does not hold. Similarly,the assumption  $U_{1}^{\dag}\geq U_{2}^{\dag}$ in the  
Theorem given below can not be dropped.

\textcolor{yellow}{THEOREM 7}\\
\textcolor{blue}{ Let $A\in {\R}^{m\times n}$ be such that $A$ and $A^T$ are row   
monotone. Let $A=U_{1}-V_{1}=U_{2}-V_{2}$ also be   two  $B_{row}$-splittings of
$A$.  If $U_{1}^{\dag}\geq U_{2}^{\dag}$,then $$1>\rho(U_{2}^{\dag}V_{2}) \geq  
\rho(U_{1}^{\dag}V_{1}).$$ }
A convergence theorem and comparison theorems for  $B_{row}$-splittings are presented. 
{Jena, L. and Mishra, D.},
 \emph{Comparisons of   $B_{row}$-splittings and $B_{ran}$-splittings of Matrices},
Linear and Multilinear Algebra, DOI 10.1080/03081087.2012.661426
I thank my research supervisor  prof. john    for his  encouragement. 

\bibitem{bpcones} {Berman, A.; Plemmons, R. J.},
 \emph{Cones and iterative methods for best square least squares solutions of linear  
systems},SIAM J. Numer. Anal., 11 (1974) 145-154.

\bibitem{bpmonotono} {Berman, A.; Plemmons, R. J.},
\emph{Monotonicity and the generalized inverse},SIAM J. Appl. Math.  22 (1972) 155-161.

\bibitem{mis} {  Mishra, D.;  Sivakumar, K. C.},
\emph{Generalizations of matrix monotonicity and their relationships with 
certain subclasses of proper splittings},
Linear Algebra Appl. DOI:10.1016/j.laa.2011.11.016 .

\bibitem{per} {Peris, J. E.},
 \emph{A new characterization of inverse-positive matrices}, 
Linear Algebra Appl.  154/156  (1991) 45-58.

\bibitem{var} {Varga, R. S.},
\emph{Matrix Iterative Analysis},  Springer-Verlag, Berlin, 2000.

% ---------------------------------------------------------%
% end the column

I am getting the following errors

File `deselaers/logos/rwthaachenuniversity-whitegray' not found. 
Package xcolor Error: Undefined color `darkgreen' \end{frame}
Package xcolor Error: Undefined color `darkgreen' \end{frame}
Package xcolor Error: Undefined color `darkgreen' \end{frame}
Package xcolor Error: Undefined color `darkgreen' \end{frame}
Package xcolor Error: Undefined color `darkgreen' \end{frame}
File `deselaers/logos/rwthaachenuniversity-whitegray' not found. \end{frame}

I don't know why these errors are there. I need the help of experts.

share|improve this question
up vote 3 down vote accepted

Be sure to have the latest version of beamerthemeI6pd2.sty downloadable at https://github.com/jtanderson/Fish-Poster/blob/master/beamerthemeI6pd2.sty.

Then pass the option svgnames to xcolor through beamer, that means load beamer as


After that, replace all instances of darkgreen with DarkGreen in your document and it should compile fine.

Then you'll probably have to adjust something in your document.

This is the result:

enter image description here

share|improve this answer
Thank you sir It works fine now. Sir I want to change the background color what command i should give in the code? Also I want it in ao size and landscape form. – Litun John Feb 7 '14 at 15:21
@Litun put the line \setbeamercolor*{normal text}{fg=tachameleon, bg=ta3gray} in your document and change ta3gray to whatever you like. The document is already in a0 landscape size. BTW. remember that you can accept (and upvote) the answer if it helped you. – karlkoeller Feb 7 '14 at 15:41
I've accepted and upvoted your answer. Thank you for your quick reply. what are the alternatives for ta3gray that i can use to change color. – Litun John Feb 7 '14 at 15:47
@Litun for example the ones you find in the xcolor documentation, section 4.3, p.38. – karlkoeller Feb 7 '14 at 15:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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