Suppose in a complex projective plane CP2, I use homogeneous coordinates
$$(x, y, z)$$
and the following transformation:
$$A =\begin{pmatrix} \cos{\alpha} & -\sin{\alpha} & 0 \\ \sin{\alpha} & \cos{\alpha} & 0 \\ 0 & 0 & 1 \end{pmatrix} $$
The point $I=(1, i, 0)$ is transformed in $(e^{-i\alpha}, i e^{-i\alpha}, 0)$.
So $A(I) = e^{-i\alpha} I$ and $I$ is invariant.
Likewise for $J=(1,-i,0)$: $A(J) = e^{i\alpha} J$ and $J$ is invariant.
The point $P = I + J = (2,0,0)$ transforms into $P' = (2\cos{\alpha},2\sin{\alpha},0)$.
The matrix is linear and indeed:
$$A(P) = A(I+J) = A(I) + A(J) = e^{-i\alpha} I + e^{i\alpha} J = P'.$$
Now I call the projective transformation itself $B$, so:
$$B(I) = I, \quad B(J) = J \quad\mbox{and}\quad B(P) = P'.$$
I thougth the projective transformation would be linear as well. This would mean:
$$B(P) = B(I+J) = B(I) + B(J) = I + J =P.$$
Which is clearly false.
What error am I making here precisely?
Projective transformations take (projective) lines to (projective) lines but they don't preserve the "midpoints of lines" or "sums of points". In fact, midpoints and "sums" are not even defined in the projective geometry.
You can just consider $\mathbb{RP}^1$, homogenous coordinates and the identity. "$(1,0)+(0,1)$" is mapped to $(1,1)$ but "(1,0)+(0,2)" is not mapped to a multiple of $(1,1)$. There is no $+$ in the projective geometry.