Suppose map a 2-dimensional point $(x,y)$ to its image point $(x',y')$ by a projective transformation $H$. So we derive that $$x'(h_{31}x+h_{32}y+h_{33})=h_{11}x+h_{12}y+h_{13}$$ $$y'(h_{31}x+h_{32}y+h_{33})=h_{21}x+h_{22}y+h_{23}$$ So a point correspondence lead to two linear equations which are linear in the elements in $H$. Now suppose there are 4 point correspondences: $$(x^{(1)},y^{(1)}) \leftrightarrow (x'^{(1)},y'^{(1)}) $$ $$(x^{(2)},y^{(2)}) \leftrightarrow (x'^{(2)},y'^{(2)}) $$ $$(x^{(3)},y^{(3)}) \leftrightarrow (x'^{(3)},y'^{(3)}) $$ $$(x^{(4)},y^{(4)}) \leftrightarrow (x'^{(4)},y'^{(4)}) $$ From above 4 point correspondences we derive 8 equations about elements of $H$, and write it in matrix form: $$\begin{bmatrix} x^{(1)} & y^{(1)} & 1 & 0 & 0 & 0 & x^{(1)}x'^{(1)} & y^{(1)}x'^{(1)} & x'^{(1)} \\ 0 & 0 & 0 & x^{(1)} & y^{(1)} & 1 &x^{(1)}y'^{(1)} & y^{(1)}y'^{(1)} & y'^{(1)} \\ x^{(2)} & y^{(2)} & 1 & 0 & 0 & 0 & x^{(2)}x'^{(2)} & y^{(2)}x'^{(2)} & x'^{(2)} \\ 0 & 0 & 0 & x^{(2)} & y^{(2)} & 1 &x^{(2)}y'^{(2)} & y^{(2)}y'^{(2)} & y'^{(2)} \\ x^{(3)} & y^{(3)} & 1 & 0 & 0 & 0 & x^{(3)}x'^{(3)} & y^{(3)}x'^{(3)} & x'^{(3)} \\ 0 & 0 & 0 & x^{(3)} & y^{(3)} & 1 &x^{(3)}y'^{(3)} & y^{(3)}y'^{(3)} & y'^{(3)} \\ x^{(4)} & y^{(4)} & 1 & 0 & 0 & 0 & x^{(4)}x'^{(4)} & y^{(4)}x'^{(4)} & x'^{(4)} \\ 0 & 0 & 0 & x^{(4)} & y^{(4)} & 1 &x^{(4)}y'^{(4)} & y^{(4)}y'^{(4)} & y'^{(4)} \\ \end{bmatrix} \begin{bmatrix} h_{11} \\ h_{12} \\ h_{13} \\ h_{21} \\ h_{22} \\ h_{23} \\ h_{31} \\ h_{32} \\ h_{33} \\\end{bmatrix}=0$$ Denoting above system as $A \mathbf h=0$ where $A \in R^{8 \times 9} $ and $\mathbf h \in R^9$.
To my understanding, dimension of A's nullspace: $$dim(N(A))=9-Rank(A) \ge 1 $$ because $Rank(A) \le 8$. So there must exist a nonzero solution to system $A \mathbf h=0$.
My question: why qualify that three points cannot be collinear?
Original problem comes from R.Hartley & A.Zisserman Multiple View Geometry in Computer Vision at page 35: