100 绘制如下形式的「抽象」矩阵的最佳实践？

这个问题本质上和 matrix 没有丝毫关系

\documentclass{article}
\usepackage{tikz, bm, lipsum}
\usetikzlibrary{scopes}
\begin{document}

\lipsum[1][1-4]

\begin{center}
\begin{tikzpicture}

  \node[font=\large] {$\bm{R_0} = $};

  \draw[thick] (1.2,2)--(1,2)--(1,-2)--(1.2,-2);

  {[]
    \draw (2,0.9)--(2,-1.8) (2.5,0.9)--(2.5,-1.8) (3,0.9)--(3,-1.8);
    \node[font=\scriptsize] at (2.5,1.15){$K$};
  \node[font=\large] at (3.15,-2.15){\bm{$S$}}; }

  {[shift={(4.7,0.2)}]
    \node[draw,minimum size=1.7cm,label={[above=0.5,font=\scriptsize]$K$},
    label={[right=1.35cm,below=1.25cm,font=\scriptsize]$K$}] (m) {};

    \node[font=\large] at(0.3,-2.4) {$E(\bm{aa}^H)$};
  }

  {[shift={(7,0.2)}]
    \draw (0,0)--(2.8,0) (0,0.62)--(2.8,0.62) (0,-0.62)--(2.8,-0.62);
    \node[font=\scriptsize] at(1.35,1.3){$M$};
    \node[font=\scriptsize] at(3.3,0.1){$K$};

    \node[font=\large] at(1.7,-2.4) {$\bm{S}^H$};
  }

  \draw[thick] (10.9,2)--(11.1,2)--(11.1,-2)--(10.9,-2);

  \node[font=\large,align=center] at(6,-3.1) {\scalebox{0.9}[1.5]{$\uparrow$} \\[1ex] non-singular};
\end{tikzpicture}
\end{center}

\lipsum[1][1-4]
\end{document}

虽然有一些神秘的baseline微调...

\documentclass{article}
\usepackage{fourier}
\usepackage{amsmath}
\usepackage{nicematrix}
\usepackage{tikz}
\usepackage{lipsum}
\begin{document}

\lipsum[2]

\[
R_o = 
\begin{bNiceArray}{ccc}[margin,last-row]
    \begin{tikzpicture}
    \clip (-.75,-.25) rectangle (1,3);
    \draw 
    (-.45,0) -- ++(0,2.25)
    (0.0,0) -- ++(0,2.25)
    node[above] {$K$}
    (+.45,0) -- ++(0,2.25);
    \end{tikzpicture}
    & 
    \begin{tikzpicture}[baseline=-1.25cm]
    \draw 
    (0,0) rectangle (1.5,1.5)
    (.75,1.5) node[above] {$K$}
    (1.5,.75) node[right] {$K$}
    ;
    \end{tikzpicture}
    & 
    \begin{tikzpicture}[baseline=-2cm]
    \draw 
    (0,-.5) --  ++(2,0)
    (0,0) -- ++(2,0) node[right] {$K$}
    (0,.5) -- node[midway,above] {$M$} ++(2,0);
    \end{tikzpicture} \\
    \mathbf{S} 
    & 
    % \mathbb{E}(\mathbf{a}\mathbf{a}^H) 
    \tikz[baseline=5.5ex]{
        \node (O) {non-singular};
        \node at (0,1) {$\mathbb{E}(\mathbf{a}\mathbf{a}^H) $}edge[latex-] (O.north);
    }
    &
    \mathbf{S}^H
\end{bNiceArray}
\]

\lipsum[2]

\end{document}