4

Here what I have done so far. I want the middle space to reduce a bit.

\documentclass[10pt]{article}
\usepackage{amsmath, amsthm, amssymb}
\begin{document}
\begin{align}
\Vert\mathbf{R}\Vert_F^2  & \le  \Vert\mathbf{R}\Vert_F^2 +\lambda\Vert 
\mathbf{B}^{'}\Vert_F^2-\sum_{k=1}^{m}\Vert \mathbf{B}^{'}\mathbf{A}\Vert_F^2 & & = 
\Vert\mathbf{R}\Vert_F^2 +\lambda\Vert \mathbf{B}^{'}\Vert_F^2-\sum_{k=1}^{m}\Vert
\mathbf{b}_k\mathbf{a}_{[k]}\Vert_F^2 \nonumber \\
& = \Vert\mathbf{R}\Vert_F^2 +\lambda\sum_{k=1}^{m}\Vert \mathbf{b}_k\Vert_2^2 -
\sum_{k=1}^{m}\Vert \mathbf{W}_k\Vert_F^2 & &= \Vert\mathbf{R}\Vert_F^2 -
\sum_{k=1}^{m}\Vert    \mathbf{b}_k\Vert_2^2\Vert\mathbf{a}_{[k]}\Vert_2^2 +\lambda m 
\label{eq6} 
\end{align}
\end{document}    
1
  • 1
    What does ^{'} signify? It it signifies the transpose operation, it should probably be written as ', i.e., not raised and made smaller by the ^{...} "wrapper".
    – Mico
    Dec 18, 2014 at 6:45

2 Answers 2

7

You can add some negative space, for example \hspace*{-1em} before the last =

MWE

\documentclass[10pt]{article}
\usepackage{amsmath, amsthm, amssymb}
\begin{document}
\begin{align}
\Vert\mathbf{R}\Vert_F^2  & \le  \Vert\mathbf{R}\Vert_F^2 +\lambda\Vert
\mathbf{B}^{'}\Vert_F^2-\sum_{k=1}^{m}\Vert \mathbf{B}^{'}\mathbf{A}\Vert_F^2 & \hspace*{-1em} & =
\Vert\mathbf{R}\Vert_F^2 +\lambda\Vert \mathbf{B}^{'}\Vert_F^2-\sum_{k=1}^{m}\Vert
\mathbf{b}_k\mathbf{a}_{[k]}\Vert_F^2 \nonumber \\
& = \Vert\mathbf{R}\Vert_F^2 +\lambda\sum_{k=1}^{m}\Vert \mathbf{b}_k\Vert_2^2 -
\sum_{k=1}^{m}\Vert \mathbf{W}_k\Vert_F^2 & \hspace*{-1em} &= \Vert\mathbf{R}\Vert_F^2 -
\sum_{k=1}^{m}\Vert    \mathbf{b}_k\Vert_2^2\Vert\mathbf{a}_{[k]}\Vert_2^2 +\lambda m
\label{eq6}
\end{align}
\end{document}  

Output

enter image description here

Anyway, as you can see, the equation is still too long to fit in the lines. I'd suggest to insert some more breaks:

\documentclass[10pt]{article}
\usepackage{amsmath, amsthm, amssymb}
\begin{document}
\begin{equation}
\begin{aligned}
\Vert\mathbf{R}\Vert_F^2  & \le  \Vert\mathbf{R}\Vert_F^2 +\lambda\Vert
\mathbf{B}^{'}\Vert_F^2-\sum_{k=1}^{m}\Vert \mathbf{B}^{'}\mathbf{A}\Vert_F^2 \\
& = \Vert\mathbf{R}\Vert_F^2 +\lambda\Vert \mathbf{B}^{'}\Vert_F^2-\sum_{k=1}^{m}\Vert
\mathbf{b}_k\mathbf{a}_{[k]}\Vert_F^2 \\
& = \Vert\mathbf{R}\Vert_F^2 +\lambda\sum_{k=1}^{m}\Vert \mathbf{b}_k\Vert_2^2 -
\sum_{k=1}^{m}\Vert \mathbf{W}_k\Vert_F^2 \\
& = \Vert\mathbf{R}\Vert_F^2 -\sum_{k=1}^{m}\Vert
\mathbf{b}_k\Vert_2^2\Vert\mathbf{a}_{[k]}\Vert_2^2 +\lambda m
\label{eq6}
\end{aligned}
\end{equation}
\end{document} 

Output

enter image description here

Notice the use of equation and aligned instead of align to avoid using \nonumber.

2
  • @VinayakAbrol you're welcome Dec 18, 2014 at 6:30
  • An aligned environment with a single alignment point is equivalent to the split environment. Using \lVert and \rVert, as needed, instead of the plain \Vert might produce better spacing in some cases. It certainly makes the code easier to read and debug...
    – Mico
    Dec 18, 2014 at 6:48
3

Here is how I would do it:

\documentclass{article}

\usepackage[margin = 4cm]{geometry}
\usepackage{mathtools}

\newcommand*\Vector[1]{\mathbf{#1}}
\DeclarePairedDelimiter{\norm}{\lVert}{\rVert}
\DeclarePairedDelimiter{\something}{\lbrack}{\rbrack}

\begin{document}

\noindent Either
\begin{alignat}{2}
  \norm{\Vector{R}}_{F}^{2}
  &\le \norm{\Vector{R}}_{F}^{2} + \lambda \norm{\Vector{B}'}_{F}^{2}
       -\sum_{k = 1}^{m} \norm{\Vector{B}'{\mkern -2mu}\Vector{A}}_{F}^{2}
  &&= \norm{\Vector{R}}_{F}^{2} + \lambda \norm{\Vector{B}'}_{F}^{2}
      -\sum_{k = 1}^{m} \norm{\Vector{b}_{k}\Vector{a}_{\something{k}}}_{F}^{2} \nonumber \\
  &= \norm{\Vector{R}}_{F}^{2} + \lambda \sum_{k = 1}^{m} \norm{\Vector{b}_{k}}_{2}^{2}
     -\sum_{k = 1}^{m} \norm{\Vector{W}_{k}}_{F}^{2}
  &&= \norm{\Vector{R}}_{F}^{2} - \sum_{k = 1}^{m} \norm{\Vector{b}_{k}}_{2}^{2}
      \,\norm{\Vector{a}_{\something{k}}}_{2}^{2} + \lambda m 
\end{alignat}
or
\begin{align}
  \norm{\Vector{R}}_{F}^{2}
  &\le \norm{\Vector{R}}_{F}^{2} + \lambda \norm{\Vector{B}'}_{F}^{2}
       -\sum_{k = 1}^{m} \norm{\Vector{B}'{\mkern -2mu}\Vector{A}}_{F}^{2} \nonumber \\
  &= \norm{\Vector{R}}_{F}^{2} + \lambda \norm{\Vector{B}'}_{F}^{2}
     -\sum_{k = 1}^{m} \norm{\Vector{b}_{k}\Vector{a}_{\something{k}}}_{F}^{2} \nonumber \\
  &= \norm{\Vector{R}}_{F}^{2} + \lambda \sum_{k = 1}^{m} \norm{\Vector{b}_{k}}_{2}^{2}
     -\sum_{k = 1}^{m} \norm{\Vector{W}_{k}}_{F}^{2} \nonumber \\
  &= \norm{\Vector{R}}_{F}^{2} - \sum_{k = 1}^{m} \norm{\Vector{b}_{k}}_{2}^{2}
     \,\norm{\Vector{a}_{\something{k}}}_{2}^{2} + \lambda m 
\end{align}

\end{document}

output

Notice the use of \DeclarePairedDelimiter from the mathtools package.

5
  • 1
    In the pursuit of perfection, I'd be tempted to add \, (thinspace) between the two adjacent norm expressions in the final row. I'd also replace all instances of ^{'} with ', or I'd write ^{\prime}.
    – Mico
    Dec 18, 2014 at 6:53
  • 2
    following Mico in pursuit of perfection, i'd be tempted to make all the delimiters uniform in size -- the sub/sup on the version with \Vector{B}^{\prime} look more nicely spaced than those on the shorter ones. however, i agree with karlkoeller that aligned is preferable to align and lots of \nonumbers Dec 18, 2014 at 14:44
  • @Mico The \, is already used. :) I've now implemented ^{\prime}. Dec 18, 2014 at 15:40
  • @barbarabeeton ^{\prime} inplemented. The align is because I guess the reference is only to the last expression. Dec 18, 2014 at 15:41
  • 2
    @SvendTveskæg Using ' is easier than ^{\prime} and completely equivalent; also '' is equivalent to ^{\prime\prime} and so on. Instead ^{'} is wrong.
    – egreg
    Dec 18, 2014 at 16:03

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