If T(z)\in PSL(2,\mathbb{R}), then the connection between fixed points of T(z) and lines of eigenvectors for the matrix T corresponding to T(z) is explained by Theorem 1. Now we consider the transformations of the group G which belong to .
Let be any transformation with the corresponding matrix T=\pm \left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\in GL(2,\mathbb{R}). The characteristic polynomial for this matrix is
{\lambda}^{2}(trT)\lambda 1=0.
(3.1)
We use the eigenvector representation \left(\begin{array}{c}{k}_{1}\\ {k}_{2}\end{array}\right) for the matrix T. First we begin with the glide reflections.
3.1 Glide reflections
We will show that the fixed points of a glide reflection T(z) correspond to the two lines of eigenvectors for the matrix T corresponding to T(z). In the following two lemmas, we determine the eigenvalues and eigenvectors of the matrices which correspond to the glide reflections.
Lemma 1 Let the matrix T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) correspond to any glide reflection. The eigenvalues of T are
and the eigenvectors of
T
are
\left(\genfrac{}{}{0ex}{}{{k}_{1}}{{k}_{2}}\right)=\left(\genfrac{}{}{0ex}{}{\frac{ad\pm \sqrt{{(a+d)}^{2}+4}}{2c}r}{r}\right).
(3.3)
Proof It is easy to compute the eigenvalues by the condition a+d\ne 0 and (3.1). For an eigenvalue λ, we obtain the eigenvector by the following equation
\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\left(\genfrac{}{}{0ex}{}{{k}_{1}}{{k}_{2}}\right)=\lambda \left(\genfrac{}{}{0ex}{}{{k}_{1}}{{k}_{2}}\right).
Thus we have (a\lambda ){k}_{1}+b{k}_{2}=0 and c{k}_{1}+(d\lambda ){k}_{2}=0. If we choose {k}_{2}=r as a parameter, we find the eigenvector as \left(\begin{array}{c}\frac{\lambda d}{c}r\\ r\end{array}\right). Therefore we obtain the eigenvectors as
\left(\genfrac{}{}{0ex}{}{{k}_{1}}{{k}_{2}}\right)=\left(\genfrac{}{}{0ex}{}{\frac{ad\pm \sqrt{{(a+d)}^{2}+4}}{2c}r}{r}\right).
□
Theorem 2 Let T(z) be a glide reflection map in the group G with corresponding matrix T. Then T(w)=w if and only if L(w) is a line of eigenvectors of T.
Proof Let T(z) be a glide reflection map in the group G with a corresponding matrix T. For glide reflections, the lines with slope w, where w=\frac{ad\pm \sqrt{{(a+d)}^{2}+4}}{2c} is a fixed point of T(z), are
L(w)=\{r\left(\genfrac{}{}{0ex}{}{\frac{ad\pm \sqrt{{(a+d)}^{2}+4}}{2c}}{1}\right):r\in \mathbb{C}\}.
(3.4)
Then T maps L(w) to L({w}^{\mathrm{\prime}}) if and only if T(w)={w}^{\mathrm{\prime}}. Thus, w is a fixed point of T(z) if and only if T maps L(w) to itself, and so if and only if each nonzero point on L(w) is an eigenvector of T. □
Example 1 By (2.3) we find the fixed points of the glide reflection T=\left(\begin{array}{cc}1& 2\\ 1& 1\end{array}\right) as x=\pm \sqrt{2}. By (3.2) we find the eigenvalues as . Hence, by Lemma 1, we obtain the following eigenvectors
\left(\genfrac{}{}{0ex}{}{\sqrt{2}r}{r}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0ex}{}{\sqrt{2}r}{r}\right)
respectively. We have the slopes {w}_{1}=\sqrt{2} and {w}_{2}=\sqrt{2}.
3.2 Reflections
Recall that we have tr(T)=0 for any reflection transformation.
Lemma 2 Let the matrix T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) correspond to any reflection. The eigenvalues of T are
Proof By (3.1), if we use the condition tr(T)=0, the result is obtained. □
First we begin the case c=0. For this case, the set of fixed points is a circle with radius ∞ (that is, a line on the complex plane).
Lemma 3 Let the matrix T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) correspond to any reflection with c=0. We have
T=\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{or}}\phantom{\rule{1em}{0ex}}T=\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right).
Proof The proof is easy by the facts that adbc=1 (a,b,c,d\in \mathbb{R}) and tr(T)=0. □
Lemma 4

(i)
For the matrix T=\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right), we have the eigenvalues \lambda =1 and \lambda =1 and the eigenvectors
\left(\genfrac{}{}{0ex}{}{r}{0}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0ex}{}{\frac{b}{2}r}{r}\right),
(3.5)
respectively.

(ii)
For the matrix T=\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right), we have the eigenvalues \lambda =1 and \lambda =1 and the eigenvectors
\left(\genfrac{}{}{0ex}{}{\frac{b}{2}r}{r}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0ex}{}{r}{0}\right),
(3.6)
respectively.
Proof It is easy to compute the eigenvalues \lambda =1 and \lambda =1 by the condition a+d=0 and (3.1). For these eigenvalues, we obtain the eigenvectors by the following equation
\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right)\left(\genfrac{}{}{0ex}{}{{k}_{1}}{{k}_{2}}\right)=\lambda \left(\genfrac{}{}{0ex}{}{{k}_{1}}{{k}_{2}}\right).
If we choose {k}_{2}=r as a parameter, we find the eigenvectors as \left(\begin{array}{c}r\\ 0\end{array}\right) and \left(\begin{array}{c}\frac{b}{2}r\\ r\end{array}\right). The second part of the proof can be obtained similarly. □
In the first part of Lemma 4, we have the slopes as {w}_{1}=\mathrm{\infty} and {w}_{2}=\frac{b}{2}. In the second part, we have {w}_{1}=\frac{b}{2} and {w}_{2}=\mathrm{\infty}.
Lemma 5

(i)
The matrix T=\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right) represents the reflection T(z)=\frac{\overline{z}+b}{1}. The set of the fixed points of this reflection is a circle with radius ∞, that is, the line x=\frac{b}{2}.

(ii)
The matrix T=\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right) represents the reflection T(z)=\overline{z}+b. The set of the fixed points of this reflection is the circle x=\frac{b}{2}.
Proof The proof follows by straightforward computations. □
In the following theorem, we explain the relationship between fixed points of the reflections with c=0 and eigenvectors of the matrices corresponding to those reflections.
Theorem 3 Let T(z) be a reflection map in the group G with c=0 and let T be the matrix corresponding to T(z). Then L(\mathrm{\infty}) and L(\pm \frac{b}{2}) are the lines of the eigenvectors of the matrix T and the set of the fixed points of the reflection T(z) is the line x=\pm \frac{b}{2}.
Proof The proof follows by Lemma 3, Lemma 4 and Lemma 5. □
Finally, we consider the reflections with c\ne 0. Lemma 6 can be proven in a similar way as Lemma 4.
Lemma 6 Let the matrix T=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right) correspond to any reflection with c\ne 0. We have the following eigenvectors for the eigenvalues \lambda =1 and \lambda =1
\left(\genfrac{}{}{0ex}{}{\frac{1d}{c}r}{r}\right)\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0ex}{}{\frac{d+1}{c}r}{r}\right),
(3.7)
respectively.
In Lemma 6, we have the slopes as {w}_{1}=\frac{1d}{c} and {w}_{2}=\frac{d+1}{c}. In the following theorem, we explain the relationship between the set of fixed points of the reflections with c\ne 0 and eigenvectors of matrices corresponding to those reflections.
Theorem 4 Let T(z) be a reflection map in the group G with c\ne 0 and let T be the corresponding matrix of T(z). If L({w}_{1}) and L({w}_{2}) are the lines of the eigenvectors of the matrix T, then the set of the fixed points of the reflection T(z) is the circle centered at M(\frac{{w}_{1}+{w}_{2}}{2},0) and of radius \frac{{w}_{1}{w}_{2}}{2}.
Proof For the slopes {w}_{1}=\frac{1d}{c} and {w}_{2}=\frac{d+1}{c}, we have
\frac{{w}_{1}+{w}_{2}}{2}=\frac{a}{c}
and
\frac{{w}_{1}{w}_{2}}{2}=\frac{1}{c}.
Then the proof follows by Lemma 6. □
Example 2 The fixed point set of the reflection T=\left(\begin{array}{cc}2& 3\\ 1& 2\end{array}\right) is a circle. By Theorem 4, we find the equation of this circle. By Lemma 6, eigenvectors of the matrix T are
\left(\genfrac{}{}{0ex}{}{3r}{r}\right)\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\left(\genfrac{}{}{0ex}{}{r}{r}\right).
Then we have {w}_{1}=3 and {w}_{2}=1. Thus the fixed point set is a circle centered at M(\frac{{w}_{1}+{w}_{2}}{2},0)=M(2,0) and of radius \frac{{w}_{1}{w}_{2}}{2}=\frac{31}{2}=1.