#math/linear-algebra
A linear map $T\in \mathcal{L}(V,W)$ is called invertible if there exists a linear map $S\in \mathcal{L}(W,V)$ such that $ST=I$ (the identity operator on $V$) and $TS=I$ (the identity operator on $W$). Then, $S$ is called the inverse of $T$ and it is also unique, $(S_{1}=S_{1}I=S_{1}(TS_{2})=(S_{1}T)S_{2}=S_{2})$.
---
Example: Suppose $T(x,y,z)=(-y,\,x,\,4z)$. Then $T$ is a counterclockwise rotation by $90^{\circ}$ in the $xy$-plane and a stretch by a factor of $4$ in the $z\text{-axis}$. Hence, $T^{-1}=\left( y,-x, \frac{1}{4}z \right)$ is a clockwise rotation by $90^{\circ}$ in the $xy\text{-plane}$ and a stretch by a factor of $\frac{1}{4}$ in the $z\text{-axis}$.
```tikz
\usepackage{pgfplots}
\pgfplotsset{compat=1.16}
\begin{document}
\begin{tikzpicture}
\begin{axis}[
hide axis,
view={25}{20},
axis equal image,
xmin=-1.6, xmax=2.2, ymin=-0.4, ymax=2.2, zmin=0, zmax=4.7,
width=15cm,
clip=false,
]
% ---- axes ----
\draw[->,thick,black!65] (axis cs:-1.4,0,0) -- (axis cs:2.1,0,0) node[anchor=north,black]{$x$};
\draw[->,thick,black!65] (axis cs:0,-0.3,0) -- (axis cs:0,2.1,0) node[anchor=west,black]{$y$};
\draw[->,thick,black!65] (axis cs:0,0,0) -- (axis cs:0,0,4.6) node[anchor=south,black]{$z$};
% ---- input vector v = (1,1,1) ----
\draw[dashed,blue!55] (axis cs:1,1,1)--(axis cs:1,1,0);
\draw[dashed,blue!55] (axis cs:0,0,0)--(axis cs:1,1,0);
\node[blue,fill=blue,circle,inner sep=1pt] at (axis cs:1,1,0){};
\draw[->,blue,very thick] (axis cs:0,0,0)--(axis cs:1,1,1);
\node[blue,anchor=south west] at (axis cs:1,1,1){$(1,1,1)$};
% ---- image vector T(v) = (-1,1,4) ----
\draw[dashed,red!55] (axis cs:-1,1,4)--(axis cs:-1,1,0);
\draw[dashed,red!55] (axis cs:0,0,0)--(axis cs:-1,1,0);
\node[red,fill=red,circle,inner sep=1pt] at (axis cs:-1,1,0){};
\draw[->,red,very thick] (axis cs:0,0,0)--(axis cs:-1,1,4);
\node[red,anchor=south east] at (axis cs:-1,1,4){$T(1,1,1)=(-1,1,4)$};
\end{axis}
\end{tikzpicture}
\end{document}
```