编辑

Question 1

据我所知，你不能写类似的东西

\draw<overlay specification>

或者

\node<overlay specification>

并让事情可靠地正确运行。如果它有时有效，那很大程度上是运气问题，当然不能指望，我想。至少，它似乎在这里造成了问题；它对我来说不能可靠地工作；而且有一堆专业代码在流传，如果它真的有效，那将完全毫无意义。由于人们很少能想出这样的解决方案来解决不存在的问题，因此声称这不是可行的方法的说法得到了证据的合理支持，尽管是间接的。

它是可以在 s\onslide<overlay>中使用规范tikzpicture，这（通常）似乎可靠地工作。对于更微妙的叠加效果，需要抓取一些为此目的设计的专业 TikZ 样式。但是，在这种情况下，不需要专业抓取，我们只需使用即可实现所需的结果\onslide<>。

我已经删除了\onslide<1->{}周围的所有内容，因为无论如何这都是默认设置。

\documentclass{beamer}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{arrows, shapes, calc, decorations.pathmorphing, tikzmark, decorations.pathreplacing, scopes, fit}
\newcommand{\Bmat}{\mathbf{B}}
\newcommand{\zvec}{\mathbf{z}}
\newcommand{\trace}[1]{\text{Tr}\left[#1\right]}
\begin{document}
\begin{frame}{The Wahba Problem can be solved through an \textbf{eigenvalue--eigenvector} problem.}
  The attitude quaternion is the eigenvector corresponding to the largest eigenvalue in the system
  \begin{center}
    \begin{tikzpicture}[remember picture]
      \node (dport) {$\subnode[inner sep=0pt]{dmat}{$\textbf{K}$} \bar{\textbf{q}} = \lambda \subnode[inner sep=0pt]{qbar}{$\bar{\textbf{q}}$}$};
      {[draw=blue, ultra thick, <-]
        \onslide<2->
        \draw  (dmat.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (dav) {$\textbf{K} = \left[\begin{array}{c c}
            \subnode[inner sep=0pt]{smat}{$\textbf{S}$} - \mu \textbf{I}_{3 \times 3} & \textbf{z} \\
            \subnode[inner sep=0pt]{zvec}{$\textbf{z}^T$} & \subnode[inner sep=0pt]{mu}{$\mu$}
          \end{array}\right]$};
        \draw  (qbar.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (quat) {\small attitude quaternion};
        \onslide<3->
        \draw  (smat.north)--+(-10pt,10pt) node[inner sep=0pt, anchor=south east, align=center] (sexp) {\small $\mathbf{S}=\Bmat+\Bmat^T$};
        \draw  (zvec.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (zexp) {\small $\zvec=\sum_{i=1}^m\frac{1}{\sigma_i^2}\left(\widetilde{\mathbf{b}_i}\times\mathbf{a}_i\right)$};
        \draw (mu.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (muexp) {\small $\mu=\trace{\subnode[inner sep=0pt]{bmat}{$\Bmat$}}$};
        \onslide<4->
        \draw  (bmat.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (bexp) {\small $\Bmat=\sum_{i=1}^m\frac{1}{\sigma_i^2}\widetilde{\mathbf{b}_i}\mathbf{a}_i^T$};
      }
%       \foreach \i in {qbar,dmat,smat,zvec,mu,bmat} \draw [fill,red] (\i) circle (1pt) node [red, font=\tiny] {\i};
    \end{tikzpicture}
  \end{center}
\end{frame}
\end{document}

编辑

正如安德鲁，此代码会抹去您的主题可能使用的任何脚注。如果您使用这样的主题，那么在以下问题的答案中演示的解决方法这个问题可用于确保预期的输出。例如，使用Joachim Breitner 的方法：

\documentclass{beamer}
\usepackage{amsmath}
\usetheme{Madrid}
\usepackage{tikz}
\usetikzlibrary{arrows, shapes, calc, decorations.pathmorphing, tikzmark, decorations.pathreplacing, scopes, fit}
\newcommand{\Bmat}{\mathbf{B}}
\newcommand{\zvec}{\mathbf{z}}
\newcommand{\trace}[1]{\text{Tr}\left[#1\right]}
\begin{document}
\begin{frame}{The Wahba Problem can be solved through an \textbf{eigenvalue--eigenvector} problem.}
  The attitude quaternion is the eigenvector corresponding to the largest eigenvalue in the system
  \begin{center}
    \begin{tikzpicture}[remember picture]
      \node (dport) {$\subnode[inner sep=0pt]{dmat}{$\textbf{K}$} \bar{\textbf{q}} = \lambda \subnode[inner sep=0pt]{qbar}{$\bar{\textbf{q}}$}$};
      {[draw=blue, ultra thick, <-]
        \onslide<2->
        \draw  (dmat.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (dav) {$\textbf{K} = \left[\begin{array}{c c}
            \subnode[inner sep=0pt]{smat}{$\textbf{S}$} - \mu \textbf{I}_{3 \times 3} & \textbf{z} \\
            \subnode[inner sep=0pt]{zvec}{$\textbf{z}^T$} & \subnode[inner sep=0pt]{mu}{$\mu$}
          \end{array}\right]$};
        \draw  (qbar.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (quat) {\small attitude quaternion};
        \onslide<3->
        \draw  (smat.north)--+(-10pt,10pt) node[inner sep=0pt, anchor=south east, align=center] (sexp) {\small $\mathbf{S}=\Bmat+\Bmat^T$};
        \draw  (zvec.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (zexp) {\small $\zvec=\sum_{i=1}^m\frac{1}{\sigma_i^2}\left(\widetilde{\mathbf{b}_i}\times\mathbf{a}_i\right)$};
        \draw (mu.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (muexp) {\small $\mu=\trace{\subnode[inner sep=0pt]{bmat}{$\Bmat$}}$};
        \onslide<4->
        \draw  (bmat.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (bexp) {\small $\Bmat=\sum_{i=1}^m\frac{1}{\sigma_i^2}\widetilde{\mathbf{b}_i}\mathbf{a}_i^T$};
      }
%       \foreach \i in {qbar,dmat,smat,zvec,mu,bmat} \draw [fill,red] (\i) circle (1pt) node [red, font=\tiny] {\i};
      \onslide<1->
    \end{tikzpicture}
  \end{center}
\end{frame}
\end{document}

生产

Answer

据我所知，你不能写类似的东西

\draw<overlay specification>

或者

\node<overlay specification>

并让事情可靠地正确运行。如果它有时有效，那很大程度上是运气问题，当然不能指望，我想。至少，它似乎在这里造成了问题；它对我来说不能可靠地工作；而且有一堆专业代码在流传，如果它真的有效，那将完全毫无意义。由于人们很少能想出这样的解决方案来解决不存在的问题，因此声称这不是可行的方法的说法得到了证据的合理支持，尽管是间接的。

它是可以在 s\onslide<overlay>中使用规范tikzpicture，这（通常）似乎可靠地工作。对于更微妙的叠加效果，需要抓取一些为此目的设计的专业 TikZ 样式。但是，在这种情况下，不需要专业抓取，我们只需使用即可实现所需的结果\onslide<>。

我已经删除了\onslide<1->{}周围的所有内容，因为无论如何这都是默认设置。

\documentclass{beamer}
\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{arrows, shapes, calc, decorations.pathmorphing, tikzmark, decorations.pathreplacing, scopes, fit}
\newcommand{\Bmat}{\mathbf{B}}
\newcommand{\zvec}{\mathbf{z}}
\newcommand{\trace}[1]{\text{Tr}\left[#1\right]}
\begin{document}
\begin{frame}{The Wahba Problem can be solved through an \textbf{eigenvalue--eigenvector} problem.}
  The attitude quaternion is the eigenvector corresponding to the largest eigenvalue in the system
  \begin{center}
    \begin{tikzpicture}[remember picture]
      \node (dport) {$\subnode[inner sep=0pt]{dmat}{$\textbf{K}$} \bar{\textbf{q}} = \lambda \subnode[inner sep=0pt]{qbar}{$\bar{\textbf{q}}$}$};
      {[draw=blue, ultra thick, <-]
        \onslide<2->
        \draw  (dmat.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (dav) {$\textbf{K} = \left[\begin{array}{c c}
            \subnode[inner sep=0pt]{smat}{$\textbf{S}$} - \mu \textbf{I}_{3 \times 3} & \textbf{z} \\
            \subnode[inner sep=0pt]{zvec}{$\textbf{z}^T$} & \subnode[inner sep=0pt]{mu}{$\mu$}
          \end{array}\right]$};
        \draw  (qbar.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (quat) {\small attitude quaternion};
        \onslide<3->
        \draw  (smat.north)--+(-10pt,10pt) node[inner sep=0pt, anchor=south east, align=center] (sexp) {\small $\mathbf{S}=\Bmat+\Bmat^T$};
        \draw  (zvec.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (zexp) {\small $\zvec=\sum_{i=1}^m\frac{1}{\sigma_i^2}\left(\widetilde{\mathbf{b}_i}\times\mathbf{a}_i\right)$};
        \draw (mu.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (muexp) {\small $\mu=\trace{\subnode[inner sep=0pt]{bmat}{$\Bmat$}}$};
        \onslide<4->
        \draw  (bmat.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (bexp) {\small $\Bmat=\sum_{i=1}^m\frac{1}{\sigma_i^2}\widetilde{\mathbf{b}_i}\mathbf{a}_i^T$};
      }
%       \foreach \i in {qbar,dmat,smat,zvec,mu,bmat} \draw [fill,red] (\i) circle (1pt) node [red, font=\tiny] {\i};
    \end{tikzpicture}
  \end{center}
\end{frame}
\end{document}

编辑

正如安德鲁，此代码会抹去您的主题可能使用的任何脚注。如果您使用这样的主题，那么在以下问题的答案中演示的解决方法这个问题可用于确保预期的输出。例如，使用Joachim Breitner 的方法：

\documentclass{beamer}
\usepackage{amsmath}
\usetheme{Madrid}
\usepackage{tikz}
\usetikzlibrary{arrows, shapes, calc, decorations.pathmorphing, tikzmark, decorations.pathreplacing, scopes, fit}
\newcommand{\Bmat}{\mathbf{B}}
\newcommand{\zvec}{\mathbf{z}}
\newcommand{\trace}[1]{\text{Tr}\left[#1\right]}
\begin{document}
\begin{frame}{The Wahba Problem can be solved through an \textbf{eigenvalue--eigenvector} problem.}
  The attitude quaternion is the eigenvector corresponding to the largest eigenvalue in the system
  \begin{center}
    \begin{tikzpicture}[remember picture]
      \node (dport) {$\subnode[inner sep=0pt]{dmat}{$\textbf{K}$} \bar{\textbf{q}} = \lambda \subnode[inner sep=0pt]{qbar}{$\bar{\textbf{q}}$}$};
      {[draw=blue, ultra thick, <-]
        \onslide<2->
        \draw  (dmat.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (dav) {$\textbf{K} = \left[\begin{array}{c c}
            \subnode[inner sep=0pt]{smat}{$\textbf{S}$} - \mu \textbf{I}_{3 \times 3} & \textbf{z} \\
            \subnode[inner sep=0pt]{zvec}{$\textbf{z}^T$} & \subnode[inner sep=0pt]{mu}{$\mu$}
          \end{array}\right]$};
        \draw  (qbar.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (quat) {\small attitude quaternion};
        \onslide<3->
        \draw  (smat.north)--+(-10pt,10pt) node[inner sep=0pt, anchor=south east, align=center] (sexp) {\small $\mathbf{S}=\Bmat+\Bmat^T$};
        \draw  (zvec.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (zexp) {\small $\zvec=\sum_{i=1}^m\frac{1}{\sigma_i^2}\left(\widetilde{\mathbf{b}_i}\times\mathbf{a}_i\right)$};
        \draw (mu.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (muexp) {\small $\mu=\trace{\subnode[inner sep=0pt]{bmat}{$\Bmat$}}$};
        \onslide<4->
        \draw  (bmat.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (bexp) {\small $\Bmat=\sum_{i=1}^m\frac{1}{\sigma_i^2}\widetilde{\mathbf{b}_i}\mathbf{a}_i^T$};
      }
%       \foreach \i in {qbar,dmat,smat,zvec,mu,bmat} \draw [fill,red] (\i) circle (1pt) node [red, font=\tiny] {\i};
      \onslide<1->
    \end{tikzpicture}
  \end{center}
\end{frame}
\end{document}

生产

Question 2

更新日期：2016-04-06

我添加了为 tikzmark 名称添加前缀和后缀的功能，并在加载 beamer 时自动添加后缀。这使原始代码能够正常工作，因为它使子节点名称在每个帧上都是唯一的。

这需要最新版本的 tikzmark，可从github。

首先，你能写入\draw<overlay specification>后它就会按照您的意愿运行：\draw只有满足覆盖规范时才会出现。

问题是，该\subnode命令的工作原理是将内容写入辅助文件，然后在后续运行时将其读回。但是，这需要使用唯一的名称，因为每个名称都会覆盖先前具有相同名称的名称。如果所有具有相同名称的子节点都位于同一位置，则不会出现问题。但事实并非如此。您的图片跳来跳去，这意味着在每一帧中子节点都位于不同的位置。

一种解决方案是修改名称，使其包含唯一部分，帧号就是一个合适的选择。一个更好的解决方案（在我看来）是修复跳跃。这也创造了更好的视觉效果，每次焦点都集中在新的部分上，观众不必每一步都四处寻找额外的内容。

幸运的是，这个问题已经解决了。添加来自我怎样才能修复投影机中跳跃的 TikZ 图片？似乎有效。

\documentclass{beamer}
%\url{https://tex.stackexchange.com/q/302517/86}



\usepackage{amsmath}
\usepackage{tikz}
\usetikzlibrary{arrows, shapes, calc, decorations.pathmorphing, tikzmark, decorations.pathreplacing, scopes, fit}

\newcommand{\Bmat}{\mathbf{B}}
\newcommand{\zvec}{\mathbf{z}}
\newcommand{\trace}[1]{\text{Tr}\left[#1\right]}

\makeatletter
\newcounter{jumping}
\resetcounteronoverlays{jumping}

\def\jump@setbb#1#2#3{%
  \@ifundefined{jump@#1@maxbb}{%
    \expandafter\gdef\csname jump@#1@maxbb\endcsname{#3}%
  }{%
    \csname jump@#1@maxbb\endcsname
    \pgf@xa=\pgf@x
    \pgf@ya=\pgf@y
    #3
    \pgfmathsetlength\pgf@x{max(\pgf@x,\pgf@xa)}%
    \pgfmathsetlength\pgf@y{max(\pgf@y,\pgf@ya)}%
    \expandafter\xdef\csname jump@#1@maxbb\endcsname{\noexpand\pgfpoint{\the\pgf@x}{\the\pgf@y}}%
  }
  \@ifundefined{jump@#1@minbb}{%
    \expandafter\gdef\csname jump@#1@minbb\endcsname{#2}%
  }{%
    \csname jump@#1@minbb\endcsname
    \pgf@xa=\pgf@x
    \pgf@ya=\pgf@y
    #2
    \pgfmathsetlength\pgf@x{min(\pgf@x,\pgf@xa)}%
    \pgfmathsetlength\pgf@y{min(\pgf@y,\pgf@ya)}%
    \expandafter\xdef\csname jump@#1@minbb\endcsname{\noexpand\pgfpoint{\the\pgf@x}{\the\pgf@y}}%
  }
}

\tikzset{
  stop jumping/.style={
    execute at end picture={%
      \stepcounter{jumping}%
      \immediate\write\pgfutil@auxout{%
        \noexpand\jump@setbb{\the\value{jumping}}{\noexpand\pgfpoint{\the\pgf@picminx}{\the\pgf@picminy}}{\noexpand\pgfpoint{\the\pgf@picmaxx}{\the\pgf@picmaxy}}
      },
      \csname jump@\the\value{jumping}@maxbb\endcsname
      \path (\the\pgf@x,\the\pgf@y);
      \csname jump@\the\value{jumping}@minbb\endcsname
      \path (\the\pgf@x,\the\pgf@y);
    },
  }
}

\makeatother

\begin{document}

\begin{frame}{The Wahba Problem can be solved through an \textbf{eigenvalue--eigenvector} problem.}
    The attitude quaternion is the eigenvector corresponding to the largest eigenvalue in the system
    \begin{center}
%        \only<1->{
        \begin{tikzpicture}[remember picture,stop jumping]
            \node<1-> (dport) {$\subnode[inner sep=0pt]{dmat}{$\textbf{K}$} \bar{\textbf{q}} = \lambda \subnode[inner sep=0pt]{qbar}{$\bar{\textbf{q}}$}$};
            {[draw=blue, ultra thick, <-]
                \draw<2-> (dmat.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (dav) {$\textbf{K} = \left[\begin{array}{c c}
                        \subnode[inner sep=0pt]{smat}{$\textbf{S}$} - \mu \textbf{I}_{3 \times 3} & \textbf{z} \\
                        \subnode[inner sep=0pt]{zvec}{$\textbf{z}^T$} & \subnode[inner sep=0pt]{mu}{$\mu$}
                    \end{array}\right]$};

                 \draw<2-> (qbar.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (quat) {\small attitude quaternion};

                 \draw<3-> (smat.north)--+(-10pt,10pt) node[inner sep=0pt, anchor=south east, align=center] (sexp) {\small $\mathbf{S}=\Bmat+\Bmat^T$};
                 \draw<3-> (zvec.south)--+(-10pt,-10pt) node[inner sep=0pt, anchor=north east, align=center] (zexp) {\small $\zvec=\sum_{i=1}^m\frac{1}{\sigma_i^2}\left(\widetilde{\mathbf{b}_i}\times\mathbf{a}_i\right)$};
                 \draw<3-> (mu.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (muexp) {\small $\mu=\trace{\subnode[inner sep=0pt]{bmat}{$\Bmat$}}$};
                 \draw<4> (bmat.south)--+(10pt,-10pt) node[inner sep=0pt, anchor=north west, align=center] (bexp) {\small $\Bmat=\sum_{i=1}^m\frac{1}{\sigma_i^2}\widetilde{\mathbf{b}_i}\mathbf{a}_i^T$};

            }


        \end{tikzpicture}
%    }
    \end{center}

\end{frame}

\end{document}

Answer