Numbering Last Equation in List of Equations

Question

I am using the following code to create a list of equations,

\begin{equation}
  \begin{array}{rcll}
  \| \hat{\textbf{v}}_{i}^{*} \|^{2} & = & \left\| \hat{\textbf{v}}_{i}^{*} + \sum_{j=1}^{i-1} \mu_{i,j}\hat{\textbf{v}}_{j}^{*} \right\|^{2} & \mbox{from (\ref{eqn:gram-schmit-orthogonalization})}, \notag \\
    & = & \| \hat{\textbf{v}}_{i}^{*} \|^{2} + \sum_{j = 1}^{i-1} \mu_{i,j}^{2} \| \hat{\textbf{v}}_{j}^{*} \|^{2} & \mbox{since~} \hat{\textbf{v}}_{1}^{*}, \ldots, \hat{\textbf{v}}_{n}^{*} \mbox{~are orthogonal}, \notag \\
    & \leq & \| \hat{\textbf{v}}_{i}^{*} \|^{2} + \sum_{j = 1}^{i-1} \frac{1}{4} \| \hat{\textbf{v}}_{j}^{*} \|^{2} & \mbox{since~} | \mu_{i,j} | \leq \frac{1}{2}, \notag \\
    & = & \| \hat{\textbf{v}}_{i}^{*} \|^{2} + \sum_{j = 1}^{i-1} 2^{i-j-2} \| \hat{\textbf{v}}_{i}^{*} \|^{2} & \mbox{from (\ref{eqn:lovasz-useful-estimate})}, \notag \\
    & = & \frac{1+2^{i-1}}{2} \| \hat{\textbf{v}}_{i}^{*} \|^{2} & \notag \\
    & \leq & 2^{i-1} \| \hat{\textbf{v}}_{i}^{*} \|^{2} & \mbox{since~} 1 \leq 2^{i-1} \mbox{~for all~} i \geq 1. \label{eqn:lovasz-simplification} \\
  \end{array}
\end{equation}

What I want to do is put a label (or number) against the last equation in the list. I am not quite sure how to do this. I have tried searching online, and the closest that I have been able to find is using the \tag{...} command of the align environment. Unfortunately, this environment only allows the equation to be split into two. I want more control over splitting the equation, which is why I chose the array environment to split the equations into four segments. Any help would be appreciated.

Answer 1

I know you wrote that you don't like align, but I believe that the following code does exactly what you want. Notice that I use \text and \mathbf, which is more proper than \mbox and \textbf. The \text command does better font handling and \mathbf should be used for upright bold math.

\documentclass{article}
\usepackage{amsmath}
\begin{document}

\begin{align}
  \| \hat{\mathbf{v}}_{i}^{*} \|^{2} & = \left\| \hat{\mathbf{v}}_{i}^{*} + \sum_{j=1}^{i-1} \mu_{i,j}\hat{\mathbf{v}}_{j}^{*} \right\|^{2} && \text{from (\ref{eqn:gram-schmit-orthogonalization})}, \notag \\
    & = \| \hat{\mathbf{v}}_{i}^{*} \|^{2} + \sum_{j = 1}^{i-1} \mu_{i,j}^{2} \| \hat{\mathbf{v}}_{j}^{*} \|^{2} && \text{since~} \hat{\mathbf{v}}_{1}^{*}, \ldots, \hat{\mathbf{v}}_{n}^{*} \text{~are orthogonal}, \notag \\
    & \leq \| \hat{\mathbf{v}}_{i}^{*} \|^{2} + \sum_{j = 1}^{i-1} \frac{1}{4} \| \hat{\mathbf{v}}_{j}^{*} \|^{2} && \text{since~} | \mu_{i,j} | \leq \frac{1}{2}, \notag \\
    & = \| \hat{\mathbf{v}}_{i}^{*} \|^{2} + \sum_{j = 1}^{i-1} 2^{i-j-2} \| \hat{\mathbf{v}}_{i}^{*} \|^{2} && \text{from (\ref{eqn:lovasz-useful-estimate})}, \notag \\
    & = \frac{1+2^{i-1}}{2} \| \hat{\mathbf{v}}_{i}^{*} \|^{2} \notag \\
    & \leq 2^{i-1} \| \hat{\mathbf{v}}_{i}^{*} \|^{2} && \text{since~} 1 \leq 2^{i-1} \text{~for all~} i \geq 1. \label{eqn:lovasz-simplification}
\end{align}

\end{document}

Answer 2

Instead of using \notag you can define \eqnumtag taht inserts the number in {align*}.

\documentclass{article}
\usepackage{amsmath}

\newcommand{\eqnumtag}{%
   % step the counter and make it lable-able
   \refstepcounter{equation}%
   % print the counter as tag
   \tag{\theequation}%
}

\begin{document}
Reference to eq.~(\ref{eqn:test}) and (\ref{eqn:test2}).
\begin{equation}
1+1=2
\end{equation}
Text
\begin{align*}
   a &= b \\
   0 &= 0 \\
   1 &= 1 \eqnumtag\label{eqn:test}
\end{align*}
Text
\begin{equation}
2^2=4
\end{equation}
Text
\begin{subequations}
\begin{align*}
   a &= b \\
   0 &= 0 \\
   1 &= 1 \eqnumtag\label{eqn:test2}
\end{align*}
Text
\begin{equation}
2^2=4
\end{equation}
Text
\end{subequations}
\end{document}

Note that this solution also works with {subequations}.