What is the essence behind representing Möbius transformations as matrices? I know that Möbius transformations generally maps lines/circles to line/circles using a function $f\left( z \right) = \frac{{az + b}}{{cz + d}}$ defined over $\mathbb C$. However, what I do not understand is that they usually are represented via a matrix and often written as
$$\left[ {\begin{array}{*{20}{c}}
z&1
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
a&c\\
b&d
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{az + b}&{cz + d}
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
{\frac{{az + b}}{{cz + d}}}&1
\end{array}} \right] = f\left( z \right)$$
I took the last equation from this  Wikipedia article. I cannot understand the meaning of this equation and how could you equate $f(z)$ to the row vector $\left[ {\begin{array}{*{20}{c}}
{\frac{{az + b}}{{cz + d}}}&1
\end{array}} \right]$. Could someone put these in simple terms so that I see how things work. Thanks in advance.
 A: Geometrically, the set of lines through the origin in $\mathbf{C}^{2}$ corresponds to an equivalence relation on non-zero vectors:
$$
(Z_{0}, Z_{1}) \sim (tZ_{0}, tZ_{1}),\quad t \neq 0.
$$
Particularly, if $Z_{1} \neq 0$ and we define $z_{0} = Z_{0}/Z_{1}$, then $(Z_{0}, Z_{1}) \sim (Z_{0}/Z_{1}, 1) = (z_{0}, 1)$. Similarly, if $Z_{0} \neq 0$ and we define $z_{1} = Z_{1}/Z_{0}$, then $(Z_{0}, Z_{1}) \sim (1, Z_{1}/Z_{0}) = (1, z_{1})$. (The diagram comes from a question about visualizing the projective line. Over the complex numbers, the projective line is identified with the Riemann sphere.)

If $ad - bc \neq 0$, multiplication by the matrix $\left[\begin{array}{@{}cc@{}}a & c \\ b & d \\ \end{array}\right]$ induces a bijection on the set of non-zero vectors, and maps lines through the origin to lines through the origin. Via the interpretation of the preceding paragraph, this matrix induces a (holomorphic) bijection of the Riemann sphere to itself. Finally to answer the question, if $cz + d \neq 0$, then
$$
(az + b, cz + d) \sim \Bigl(\frac{az + b}{cz + d}, 1\Bigr).
$$
