how to vertical align text before blockarray in blkarray package?

Question

I have to label row and column index around matrix. I know blkarray would help, so I write like this:

\documentclass[a4paper,12pt]{article}

\usepackage{amsmath}
\usepackage{blkarray}% http://ctan.org/pkg/blkarray
\newcommand{\matindex}[1]{\mbox{#1}}% Matrix index

\begin{document}

pmatrix way:
\[
J(i,j,\theta)=\begin{pmatrix}
1 & \cdots & 0 & \cdots & 0 & \cdots & 0 \\
\vdots & \ddots & \vdots & & \vdots & & \vdots \\
0 & \cdots & c & \cdots & -s & \cdots & 0 \\
\vdots & & \vdots & \ddots & \vdots & & \vdots \\
0 & \cdots & s & \cdots & c & \cdots & 0 \\
\vdots & & \vdots & & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & \cdots & 0 & \cdots & 1 
\end{pmatrix}
\]

blkarray way:
\[
  J(i,j,\theta)=\begin{blockarray}{cccccccc}
    \begin{block}{(ccccccc)c}
        1 & \cdots & 0 & \cdots & 0 & \cdots & 0 \\
        \vdots & \ddots & \vdots & & \vdots & & \vdots \\
        0 & \cdots & c & \cdots & -s & \cdots & 0 & \matindex{$i$} \\
        \vdots & & \vdots & \ddots & \vdots & & \vdots \\
        0 & \cdots & s & \cdots & c & \cdots & 0 & \matindex{$j$} \\
        \vdots & & \vdots & & \vdots & \ddots & \vdots \\
        0 & \cdots & 0 & \cdots & 0 & \cdots & 1 \\
    \end{block}
    & & \matindex{$i$} & & \matindex{$j$} & \\
  \end{blockarray}
\]

Why $J(i,j,\theta)$ isn't vertical aligned int blkarray way?
\end{document}

Here is the result: I want to know why isn't vertical aligned in blkarray way.

And how to vertical aligned it like pmatrix way.

Thanks.

Answer 1

The problem is that the block array is vertically centered with respect to its entire size, including the indices below.

The easiest way, in this case, is to put the whole thing in the blkarray.

\documentclass[a4paper,12pt]{article}

\usepackage{amsmath}
\usepackage{blkarray}% http://ctan.org/pkg/blkarray

\begin{document}

\[
\begin{blockarray}{r@{}cccccccc}
\begin{block}{r(ccccccc)c}
                 & 1 & \cdots & 0 & \cdots & 0 & \cdots & 0 \\
                 & \vdots & \ddots & \vdots & & \vdots & & \vdots \\
                 & 0 & \cdots & c & \cdots & -s & \cdots & 0 & i \\
J(i,j,\theta)={} & \vdots & & \vdots & \ddots & \vdots & & \vdots \\
                 & 0 & \cdots & s & \cdots & c & \cdots & 0 & j \\
                 & \vdots & & \vdots & & \vdots & \ddots & \vdots \\
                 & 0 & \cdots & 0 & \cdots & 0 & \cdots & 1 \\
\end{block}
                 & & & i & & j & \\
\end{blockarray}
\]

\end{document}

Answer 2

You can obtain this result either with pmatrixand pstricks, defining certain elements as nodes, and attaching labels to these nodes, or with blockarray, stackengine or makecell, and a \raisebox command:

\documentclass[a4paper,12pt]{article}

\usepackage{mathtools}
\usepackage{blkarray, makecell,  stackengine}
\usepackage{pst-node, auto-pst-pdf} %

\begin{document}

\verb|pmatrix way + pstricks way:|\bigskip

\[
 \begin{postscript}
 J(i,j,\theta)=%
 \begin{pmatrix}
1 & \cdots & 0 & \cdots & 0 & \cdots & 0 \\
\vdots & \ddots & \vdots & & \vdots & & \vdots\\
0 & \cdots & c & \cdots & -s & \cdots &\rnode{rowi}{0}\\%
\vdots & & \vdots & \ddots & \vdots & & \vdots \\
0 & \cdots & s & \cdots & c & \cdots & \rnode{rowj}{0}\\%
\vdots & & \vdots & & \vdots & \ddots & \vdots \\
0 & \cdots & \rnode{coli}{0} & \cdots & \rnode{colj}{0} & \cdots & 1
\end{pmatrix}
 \psset{labelsep=1.2em}
\nput{0}{rowi}{i}\nput{0}{rowj}{j}
 \psset{labelsep=1.2ex}
\nput{-90}{coli}{i}\nput{-90}{colj}{j}
\end{postscript}
\]

\verb|blkarray + makecell or stackengine way:  |
\[
  J(i,j,\theta)= \raisebox{-0.3\baselineskip}{$ \begin{blockarray}{(*{7}{c})c}
        1 & \cdots & 0 & \cdots & 0 & \cdots & 0 \\
        \vdots & \ddots & \vdots & & \vdots & & \vdots \\
        0 & \cdots & c & \cdots & -s & \cdots & 0 & i\\ 
        \vdots & & \vdots & \ddots & \vdots & & \vdots \\
        0 & \cdots & s & \cdots & c & \cdots & 0 & j \\
        \vdots & & \vdots & & \vdots & \ddots & \vdots \\
        0 & \cdots & \smash{\stackunder[1.2ex]{0}{$ i $}} & \cdots &\smash{\makecell[tc]{0\\j}} & \cdots & 1
  \end{blockarray} $} 
\]%

 \end{document} 

Answer 3

With {pNiceMatrix} of nicematrix, you will have directly the expected result.

\documentclass[a4paper,12pt]{article}

\usepackage{nicematrix}

\begin{document}

\[
J(i,j,\theta)=\begin{pNiceMatrix}[last-row=8,last-col=8]
1 & \cdots & 0 & \cdots & 0 & \cdots & 0 \\
\vdots & \ddots & \vdots & & \vdots & & \vdots \\
0 & \cdots & c & \cdots & -s & \cdots & 0 & i \\
\vdots & & \vdots & \ddots & \vdots & & \vdots \\
0 & \cdots & s & \cdots & c & \cdots & 0 & j \\
\vdots & & \vdots & & \vdots & \ddots & \vdots \\
0 & \cdots & 0 & \cdots & 0 & \cdots & 1 \\
  & & i & & j & \\
\end{pNiceMatrix}
\]

\end{document}