Proposition

Suppose

\[\begin{align*} A = \begin{bmatrix} a & b \\ c & d \end{bmatrix} \end{align*}\]

is a $2 \times 2$ matrix of complex numbers whose determinant $ad - bc$ is nonzero. Then we can define a corresponding Mobius transformation on $\hat{\mathbb{C}}$ by $T_A(z) = \frac{az + b}{cz + d}$. Show that $T_A \circ T_B = T_{A \cdot B}$, where $\circ$ denotes composition and $\cdot$ denotes matrix multiplication.

Solution

Let

\[\begin{align*} A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}, B = \begin{bmatrix} e & f \\ g & h \end{bmatrix}. \end{align*}\]

Then

\[\begin{align*} T_A \circ T_B &= \frac{aT_B(z) + b}{cT_B(z) + d} \\ &= \frac{a(ez + f) + b(gz + h)}{c(ez + f) + d(gz + h)} \\ &= \frac{(ae + bg)z + (af + bh)}{(ce + dg)z + (cf + dh)}. \end{align*}\]

Since

\[\begin{align*} A \cdot B &= \begin{bmatrix} ae + bg & af + bh \\ ce + dg & cf + dh \end{bmatrix}, \end{align*}\]

$T_A \circ T_B = T_{A \cdot B}$.