TikZ two-blocks matrix

I would like to get a plot like the one I sketched below, basically two blocks of the same matrix with arrows going up and anti-diagonally for the red block and arrows going up for the green one. The dots in the middle of the green one means to repeat the same pattern .... And I have as starting point the TikZ code below but I am not sure which would be the easiest way to do ... I struggle still to get conditions in the for loops of TikZ with the complicated markup. As starting point I have this including a commented attempt to create the upper triangular effect omitting the arrows for elements below the diagonal but can't get this code to work.

\begin{figure}[!h]
\centering
\begin{tikzpicture}[scale=1.5]
\clip (-0.3,-1.3) rectangle (8.3,4.3);
%   \foreach \x in {0,...,8} {
%       \foreach \y [evaluate=\y as \z using int(4-\y)] in {0,...,4} {
%           \pgfmathtruncatemacro{\myresult}{\x<\z ? 1 : 0}
%           \ifnum\myresult=0%
%           \node at (\x cm,\y cm) (g-\x-\y) {$a_{\x,\z}$};\fi
%       }
%   }
\foreach \x in {0,...,8} {
\foreach \y [evaluate=\y as \z using int(4-\y)] in {0,...,4} {
\node at (\x cm,\y cm) (g-\x-\y) {$a_{\x,\z}$};
}
}
% vertical dependency
\foreach \x in {0,...,8} {
\foreach \y in {0,...,3} {
\draw[->,shorten <=2pt,shorten >=7pt] (g-\x-\y) -- +(0,20pt);
}
}
% horizontal
\foreach \x in {0,...,7} {
\foreach \y in {0,...,4} {
\draw[<-,shorten >=2pt,shorten >=7pt] (g-\x-\y) -- +(20pt,0);
}
}
% diagonal
\foreach \x in {1,...,8} {
\foreach \y in {0,...,3} {
\draw[->] (\x-.3,\y+.3) -- (\x - .6, \y+.6);
}
}
\end{tikzpicture}
\caption{Triangularize}
\label{fig:triangularize1}
\end{figure}

and the generated image in this case is (but actually I dont need the horizontal arrows): Maybe I took it a little too literal.

\documentclass{article}
\usepackage{emerald,tikz}
\usepackage[T1]{fontenc}

\usetikzlibrary{calc,decorations.pathmorphing,fit}

\makeatletter

\pgfdeclaredecoration{penciline}{initial}{
\state{initial}[width=+\pgfdecoratedinputsegmentremainingdistance,auto corner on length=1mm,]{
\pgfpathcurveto%
{% From
\pgfqpoint{\pgfdecoratedinputsegmentremainingdistance}
{\pgfdecorationsegmentamplitude}
}
{%  Control 1
\pgfmathparse{0.1*rand}
{\pgfqpoint{-\pgfdecorationsegmentaspect\pgfdecoratedinputsegmentremainingdistance}%
{\pgfmathresult\pgfdecorationsegmentamplitude}
}
}
{%TO
}
}
\state{final}{}
}
\makeatother

\begin{document}\ECFTallPaul

\begin{tikzpicture}
[zeros/.style={},scale=0.8,transform shape,
xes/.style={outer sep=1.5pt,decoration=penciline,decorate}]
\foreach \y in {1,...,4}{
\foreach \x in {1,...,5}{
\ifnum\x>\numexpr\y+1
\node[zeros] (mynode-\x-\y) at (\y,-\x) {0};
\else
\pgfmathparse{\y-\x<1?"draw":""}
\node[xes,\pgfmathresult](mynode-\x-\y) at (\y,-\x) {X};
\ifnum\x<5
\ifnum\x<\y\draw[<-,xes] (mynode-\x-\y) -- +(0,-0.7);\fi
\ifnum\x=\y\draw[<-,xes] (mynode-\x-\y) -- +(0,-0.7);\fi
\ifnum\y<4\draw[<-,xes] (mynode-\x-\y) -- +(0.7,-0.7);\fi
\fi
\fi
}
}
\node[fit=(mynode-1-1)(mynode-5-4),draw,thick,xes,red,label=90:{\Large dependencies}] {};

\begin{scope}[xshift=4.7cm]
\foreach \x[remember=\x as \lastx] in {1,...,5}{
\node (myvecs1-\x) at (0,-\x) {X};
\node (myvecs2-\x) at (1,-\x) {X};
\node (myvecsn-\x) at (5,-\x) {X};
\ifnum\x>1
\draw[->,xes] (myvecs1-\x) -- (myvecs1-\lastx);
\draw[->,xes] (myvecs2-\x) -- (myvecs2-\lastx);
\draw[->,xes] (myvecsn-\x) -- (myvecsn-\lastx);
\fi
}
\node[fit=(myvecs1-1)(myvecsn-5),draw,thick,xes,green!50!blue,label=90:{\Large Free of diagonal dep.}] {};
\end{scope}
\end{tikzpicture}

\end{document} Mostly stolen from

Simulating hand-drawn lines

TikZ marking several blocks in a matrix

and for the font

CTAN font catalogue

• I'm speechless. It's amazing, my friend! :) Sep 4 '12 at 0:59
• The horizontal dots inside the green box are missing ;-) (I'm just kidding). Sep 4 '12 at 1:16
• lol amazing! howcan I make it look not handwritten ?? lol Sep 4 '12 at 3:47
• @GiovanniAzua Just remove the decorations here and there and remove the font related commands for the base drawing. Note that the arrows are not from node-to-node so it might need some tweaks. Let me know if it doesn't solve your problem. Sep 4 '12 at 7:59