Exercise 1.7.3 (Action $r \cdot (x,y) = (x + ry, y)$ from $\mathbb{R}$ on $\mathbb{R}^2$)

Question

Show that the additive group $\mathbb{R}$ acts on the $x,y$ plane $\mathbb{R} 	imes \mathbb{R}$ by $r\cdot (x,y) = (x + ry, y)$.

Richard Ganaye · Accepted Answer

ewcommand{\R}{\mathbb{R}}

\begin{proof} 
For all $r,s \in \R$ and for all $(x,y) \in \R^2$,
\begin{enumerate}
\item[(i)]$ r \cdot (s \cdot (x,y)) = r \cdot (x + s y, y) = (x + sy + ry, y) = (x + (r+s) y, y) = (r+s) \cdot (x,y),$
\item[(ii)] $0 \cdot (x,y) = (x + 0 y, y) = (x,y).$
\end{enumerate}
Therefore the additive group $\mathbb{R}$ acts on $\R^2$ by $r\cdot (x,y) = (x + ry, y)$.
\end{proof}
Note: The map
$$
\varphi
\left\{
\begin{array}{ccl}
\R & 	o & \mathrm{GL}_2(\R)\
r & \mapsto & \begin{pmatrix} 1 & r\ 0 & 1 \end{pmatrix}
\end{array}
ight.
$$
is a group isomorphism, so a faithful matrix representation of the additive group $\R$, such that 
$$\varphi(r) \begin{pmatrix} x\ y \end{pmatrix} = \begin{pmatrix} 1 & r\ 0 & 1 \end{pmatrix} \begin{pmatrix} x \ y \end{pmatrix}  = \begin{pmatrix} x + ry \ y \end{pmatrix}.$$
 This explains the preceding action.

Exercise 1.7.3 (Action $r \cdot (x,y) = (x + ry, y)$ from $\mathbb{R}$ on $\mathbb{R}^2$)

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Exercise 1.7.3 (Action $r \cdot (x,y) = (x + ry, y)$ from $\mathbb{R}$ on $\mathbb{R}^2$)

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