The pinhole model

A pinhole camera is a plane with a hole in it and a sensor behind. Light from a scene point travels in a straight line through the hole and lands somewhere on the sensor. That is the entire physical model, and everything else is bookkeeping about coordinates.

Put the hole at the origin of the camera frame, with $+Z$ pointing into the scene. Put the image plane at $Z = f$ in front of the hole rather than behind it — the real image is inverted, the virtual one is not, and since flipping twice is tedious, everyone uses the virtual plane and never mentions it again. Similar triangles then give

$$x = f\,\frac{X}{Z}, \qquad y = f\,\frac{Y}{Z}.$$

Now write it in homogeneous coordinates. With $\mathbf{X} = (X, Y, Z, 1)^\top$ and the image point $\mathbf{x} = (u, v, 1)^\top$ known only up to scale:

$$\lambda \begin{pmatrix} u \\ v \\ 1 \end{pmatrix} = \begin{bmatrix} f & 0 & 0 & 0 \\ 0 & f & 0 & 0 \\ 0 & 0 & 1 & 0 \end{bmatrix} \begin{pmatrix} X \\ Y \\ Z \\ 1 \end{pmatrix}.$$

The matrix is linear. The division has not vanished — it hides in $\lambda = Z$ , and it reappears the moment you insist on a Euclidean answer. That is the trade homogeneous coordinates offer: the algebra becomes linear, and the nonlinearity is quarantined in one normalisation step at the end.