Prove that the kernel of $GL_2(\mathbb{R})$ acting on $\mathbb{R}^2$ defined by $M(v)=Mv$ is $\{I\}.$
Proof:
Let $\begin{pmatrix}a & b\\\ c & d\end{pmatrix}\in M$ and $\begin{pmatrix}x_1\\\ x_2\end{pmatrix}\in v.$ So, we want to find the kernel of $$\begin{pmatrix}a & b\\\ c & d\end{pmatrix}*\begin{pmatrix}x_1\\\ x_2\end{pmatrix}=\begin{pmatrix}ax_1+bx_2\\\ cx_1+dx_2\end{pmatrix}$$
Observe that if $x_1=x_2=0$ then,
$$\begin{pmatrix}a & b\\\ c & d\end{pmatrix}*\begin{pmatrix}0\\ 0\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$$
hence, $\begin{pmatrix}0\\ 0\end{pmatrix}\in ker.$
WLOG, suppose $x_2\neq0.$ Then we have
$$ax_1+bx_2=0$$
$$cx_1+dx_2=0$$
By multiplying $c$ tof the first equation and $a$ to the second we have,
$$cax_1+cbx_2=0$$
$$cax_1+dax_2=0$$
which gives us,
$$\begin{align}
cax_1+cbx_2&=cax_1+dax_2\\
cbx_2&=dax_2\\
cb&=da
\end{align}
$$
But this implies that the matrix is not invertible, which is a contradiction since $GL_2(\mathbb{R})$ is. Thus, $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$ is the only element in the kernel.
Is this correct? If so, how does this satisfy the definition of a kernel of an action i.e. in this case an element say $g\in GL_2(\mathbb{R})\times \mathbb{R}^2$ such that $g\cdot x=x, \forall x\in \mathbb{R}^2$?
Best Answer
I think you got confused with the definitions. Elements in the kernel are matrices in $GL_2(\mathbb{R})$ and not vectors in $\mathbb{R^2}$. Take a matrix $A=\begin{pmatrix}a & b\\\ c & d\end{pmatrix}$ in the kernel. Also let's write $e_1=\begin{pmatrix}1\\\ 0\end{pmatrix}$ and $e_2=\begin{pmatrix} 0\\\ 1\end{pmatrix}$. Then $Ae_1=e_1$ and $Ae_2=e_2$. From here you easily get that $a=d=1$ and $b=c=0$, so $A=I$.