Question about image of homomorphism from $\mathbb{R}$ to $GL(2,\mathbb{R})$ An exercise from Gallian's Contemporary Abstract Algebra tasks the reader to prove that the mapping $\phi: \mathbb{R} \rightarrow GL(2, \mathbb{R})$ defined by $$x \mapsto \left[\begin{array}{c c }
\cos x & \sin x \\
-\sin x & \cos x
\end{array} \right]$$
is a homomorphism and identify the kernel (here $\mathbb{R}$ is regarded as a group under ordinary addition). I managed to prove that $\phi$ is indeed a homomorphism and that $\ker \phi = \{ 2 k \pi \in \mathbb{R} \mid k \in \mathbb{Z} \}$. 
My question is what is the image of $\phi$? I can see that $$ \det \left( \left[\begin{array}{c c }
\cos x & \sin x \\
-\sin x & \cos x
\end{array} \right]\right) = \cos^2 x + \sin^2 x = 1$$ for all $x \in \mathbb{R}$, which leads me to believe that $Im \, \phi = SL(2, \mathbb{R})$, but I am unsure if this correct. 
 A: The image of $\phi$ is $SO(2,R)$ the space of orthogonal maps whose determinant is 1. It cannot be $SL(2,R)$ since its dimension is 3.
A: Matrices in $SO(2)$ are by definition those satisfying 
$$AA^T=I$$
and $$det(A)=1.$$
Spelling this out for a matrix $$A=\left(\begin{array}{cc}a&b\\
c&d\end{array}\right)$$
yields
$$a^2+b^2=1$$
$$c^2+d^2=1$$
$$ac+bd=0$$
$$ad-bc=1$$
The first equation implies that $(a,b)$ lies on the unit circle in ${\mathbb R}^2$ and thus there is some $\phi$ with $a=cos\phi, b=sin\phi$.
Similarly the second equation implies there is some $\psi$ with $d=cos\psi, c=sin\psi$.
The third equation then yields $cos\phi sin\psi+sin\phi cos\psi=0$, thus $sin(\phi+\psi)=0$. The fourth equation yields $cos\phi cos\psi -sin\phi sin\psi=1$, thus $cos(\phi+\psi)=1$. From $sin=0$ and $cos=1$ we have $\phi+\psi=0$, i.e. $c=-b, d=a$.
So all matrices in $SO(2)$ are of the form $\left(\begin{array}{cc}cos\phi&sin\phi\\
-sin\phi&cos\phi\end{array}\right)$ and thus in the image of your homomorphism. One easily checks that also conversely all matrices in the image are in $SO(2)$, so the image is exactly $SO(2)$.
