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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.

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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 $SO(2)$.

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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.

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  • $\begingroup$ After some wikipedia-ing, matrices belonging to $SO(2, \mathbb{R})$ are those such that the product of such a matrix and its transpose is equal to the identity matrix? $\endgroup$ – Oiler Aug 2 '16 at 3:17
  • $\begingroup$ The determinant must be 1 too, consider $A=\pmatrix{0 &1 \cr 1 & 0}, A^T=A$ and $A^2=I$ but $A$ is not in $SO(2,R)$. $\endgroup$ – Tsemo Aristide Aug 2 '16 at 3:24
  • $\begingroup$ I forgot to add that bit in as well. Got it thanks. $\endgroup$ – Oiler Aug 2 '16 at 3:27

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