# Replicate the Singular Value Decomposition figure in LaTeX

I have tried to replicate a portion of this matrix decomposition system in LaTeX, but I am unable to draw the first and the last matrices with rectangular boxes.

This is what I could do until now, which of course, is easy to write. I can use LatexDraw to draw this figure, but the figure will not look very elegant as this one:

 $$\underbrace{\mathbf{A}}_{W \times D} = \underbrace{\mathbf{U}}_{W \times W} \times \underbrace{\mathbf{\Sigma}}_{W\times D} \times \underbrace{\mathbf{V}^{\text{T}}}_{D \times D} = \left( \begin{array}{ccccc} \sigma_1\\ & . & & \text{\huge0}\\ & & .\\ & \text{\huge0} & & \sigma_r\\ & & & & 0 \end{array} \right)$$

• It would be welcomed that the code you provided started with \documentclass and ended with \end{document}. Jun 4 '14 at 15:54
• @Manuel: I am sorry about this. It has been two times since I have been reminded about it. Next time there won't be any complains, I promise. Jun 5 '14 at 0:13

Using TikZ and some matrix of math nodes:

The code:

\documentclass{article}
\usepackage[margin=2cm]{geometry}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{matrix,positioning,decorations.pathreplacing}

\DeclareMathOperator{\Mcol}{col}
\DeclareMathOperator{\Mrow}{row}
\DeclareMathOperator{\Mnull}{null}

\begin{document}

\begin{align*}
A &= U\Sigma V^{T} \\
&=
\begin{tikzpicture}[
baseline,
mymat/.style={
matrix of math nodes,
ampersand replacement=\&,
left delimiter=(,
right delimiter=),
nodes in empty cells,
nodes={outer sep=-\pgflinewidth,text depth=0.5ex,text height=2ex,text width=1.2em}
}
]
\begin{scope}[every right delimiter/.style={xshift=-3ex}]
\matrix[mymat] (matu)
{
\& \& \& \& \& \\
\& \& \& \& \& \\
\& \& \& \& \& \\
\& \& \& \& \& \\
\& \& \& \& \& \\
\& \& \& \& \& \\
};
\node
at ([shift={(3pt,-7pt)}]matu-3-2.west)
{$\cdots$};
\node
at ([shift={(3pt,-7pt)}]matu-3-5.west)
{$\cdots$};
\foreach \Columna/\Valor in {1/1,3/r,4/{r+1},6/m}
{
\draw
(matu-1-\Columna.north west)
rectangle
([xshift=4pt]matu-6-\Columna.south west);
\node[above]
at ([xshift=2pt]matu-1-\Columna.north west)
{$u_{\Valor}$};
}
\draw[decorate,decoration={brace,mirror,raise=3pt}]
(matu-6-1.south west) --
node[below=4pt] {$\Mcol(A)$}
([xshift=4pt]matu-6-3.south west);
\draw[decorate,decoration={brace,mirror,raise=3pt}]
(matu-6-4.south west) --
node[below=4pt] {$\Mnull(A)$}
([xshift=4pt]matu-6-6.south west);
\end{scope}
\matrix[mymat,right=10pt of matu] (matsigma)
{
\sigma_{1} \& \& \& \& \& \\
\& \ddots \& \& \& \& \\
\& \& \sigma_{r} \& \& \& \\
\& \& \& 0 \& \& \\
\& \& \& \& \ddots \& \\
\& \& \& \& \& 0 \\
};
%\begin{scope}[every right delimiter/.style={xshift=-3ex}]
\matrix[mymat,right=25pt of matsigma] (matv)
{
\& \& \& \& \& \\
\& \& \& \& \& \\
\& \& \& \& \& \\
\& \& \& \& \& \\
\& \& \& \& \& \\
\& \& \& \& \& \\
};
\foreach \Fila/\Valor in {1/1,3/r,4/{r+1},6/n}
{
\draw
([yshift=-6pt]matv-\Fila-1.north west)
rectangle
([yshift=-10pt]matv-\Fila-6.north east);
\node[right=12pt]
at ([yshift=-8pt]matv-\Fila-6.north east)
{$v^{T}_{\Valor}$};
}
\draw[decorate,decoration={brace,raise=37pt}]
([yshift=-6pt]matv-1-6.north east) --
node[right=38pt] {$\Mrow(A)$}
([yshift=-10pt]matv-3-6.north east);
\draw[decorate,decoration={brace,raise=37pt}]
([yshift=-6pt]matv-4-6.north east) --
node[right=38pt] {$\Mnull(A)$}
([yshift=-10pt]matv-6-6.north east);
\end{tikzpicture}
\end{align*}

\end{document}


The few missing elements are easy to add.

Not exactly your scheme, but it approximates...

\documentclass{article}
\usepackage{amsmath}

\begin{document}

\newcommand{\vect}{\mathbf}
\newcommand{\nul}{\operatorname{Nul}}
\newcommand{\col}{\operatorname{Col}}
\newcommand{\row}{\operatorname{Row}}

$A= U\Sigma V^T= \begin{matrix} \underbrace{\left[\begin{matrix}\vect u_1 & \vect u_2 & \dots & \vect u_r\end{matrix}\right.}& \underbrace{\left.\begin{matrix}\vect u_{r+1} & \dots & \vect u_m\end{matrix}\right]}\\ \col A & \nul A^T \end{matrix} \begin{bmatrix} \sigma_1 & 0 & \dots & 0 & 0 & \dots & 0 \\ 0 & \sigma_2 & \dots & 0 & 0 & \dots & 0 \\ \dots& & & & & \\ 0 & 0 & \dots & \sigma_r & 0 & \dots & 0 \\ 0 & 0 & \dots & 0 & 0 & \dots & 0 \\ \dots& & & & & \\ 0 & 0 & \dots & 0 & 0 & \dots & 0 \end{bmatrix} \begin{bmatrix} \vect v_1^T \\ \vect v_2^T \\ \dots \\ \vect v_r^T \\ \vect v_{r+1}^T \\ \dots \\ \vect v_n^T \end{bmatrix} \begin{matrix} \left.\vphantom{\begin{bmatrix} \vect v_1^T \\ \vect v_2^T \\ \dots \\ \vect v_r^T \end{bmatrix}}\right\}\row A \\ \left.\vphantom{\begin{bmatrix} \vect v_{r+1}^T \\ \dots \\ \vect v_n^T \end{bmatrix}}\right\}\nul A \end{matrix}$

\end{document}


• Your solution seems very good as well. Thanks a lot. But unfortunately, I am forced to accept only one of the two. Jun 5 '14 at 0:15