Applying a Möbius map to a circle or line

We now suppose we have a Möbius map $m=\begin{bmatrix}\alpha&\beta\\ \gamma&\delta\end{bmatrix}$ and a circle or line $C$ given by the Hermitian matrix $H=\begin{bmatrix}a & b \\ \bar{b} & c \end{bmatrix}$. We wish to calculate the image $m(C)$ of $C$ under the Möbius map $m$.

It will turn out to simply be matrix multiplication, but let's reason out why this is so. First, $C$ consists of all $z$ satisfying $$0 = \begin{bmatrix}\bar{z} & 1\end{bmatrix} H \begin{bmatrix}z\\ 1\end{bmatrix}.$$ Then $m(C)$ consists of all points $z$ such that the inverse image $m^{-1}(z)$ belongs to $C$. The inverse image is also a linear fractional transformation, namely, $$\frac{\delta z-\beta}{-\gamma z + \alpha}$$ Therefore, $m(C)$ consists of all $z$ satisfying $$0 = \begin{bmatrix}\bar{\frac{\delta z-\beta}{-\gamma z + \a... ...ix}\frac{\delta z-\beta}{-\gamma z + \alpha}\\ 1\end{bmatrix}.$$ Multiply this equation by $\bar{(-\gamma z+\alpha)}{(-\gamma z+\alpha)}$ to obtain
$$\begin{align*} 0 &= \begin{bmatrix}\bar{\delta z-\beta}&\bar{-\gamma z + \alpha}... ...d{bmatrix} \bar{m^{-1}}^T H m^{-1} \begin{bmatrix}z\\ 1\end{bmatrix}\end{align*}$$
The matrix in the middle $\bar{m^{-1}}^T H m^{-1}$ is again a Hermitian matrix of negative determinant (in fact the same determinant as $H$ since we assume $m$ has determinant 1).

This proves $m(C)$ is also a circle or straight line, and a Hermitian matrix corresponding to $C$ is just obtained by simple matrix multiplication $$\fbox{ $\displaystyle \bar{m^{-1}}^T H m^{-1} $}$$