The item linked to, proof number 10. It is a bit terse. I would add that the matrix $A$ is determinant $1.$ If there were any nonzero lattice points on the indicated line $L,$ there would be a lattice point $(c,d)$ with $0 < d < c,$ $c^2 = 2 d^2$ and minimal $c^2 + d^2 = 3 d^2 \neq 0.$ However, the part about the contraction says that the image under the matrix $A,$ namely $(-c+2d,c-d)$ is still nonzero, still in the lattice, still in the line, but of strictly smaller norm $$(-c+2d)^2 + (c-d)^2 = 2 c^2 - 6 cd + 5 d^2 = -6 cd + 9 d^2 = 3 d(3d-2c) < 3 d^2, $$
as $0 < d < c$ and $3d - 2 c < d.$ This contradicts the assumption of existence of the lattice point $(c,d)$ on the line, therefore of any lattice point on the line.
This proof too is by D. Kalman et al (Variations on an Irrational Theme-Geometry, Dynamics, Algebra, Mathematics Magazine, Vol. 70, No. 2 (Apr., 1997), pp. 93-104).
Let A be the matrix
A = \begin{pmatrix} -1 & \space 2\\ \space 1 & -1\\ \end{pmatrix}
By the definition,
\begin{pmatrix} -1 & \space 2\\ \space 1 & -1\\ \end{pmatrix} \begin{pmatrix} a \\ b \end{pmatrix} = \begin{pmatrix} 2b-a \\ a-b \end{pmatrix}
Two facts are worth noting: (a) matrix A maps integer lattice onto itself, (b) the line with the equation $a = \sqrt{2}b$ is an eigenspace L, say, corresponding to the eigenvalue $\sqrt{2} - 1:$
\begin{pmatrix} -1 & \space 2\\ \space 1 & -1\\ \end{pmatrix} \begin{pmatrix} \sqrt{2} \\ 1 \end{pmatrix} = \begin{pmatrix} 2-\sqrt{2} \\ \sqrt{2}-1 \end{pmatrix} $=(\sqrt{2}-1)$ \begin{pmatrix} \sqrt{2} \\ 1 \end{pmatrix}.
Since $0 \lt \sqrt{2} - 1 \lt 1,$ the effect of A on L is that of a contraction operator. So that the repeated applications of matrix A to a point on L remain on L and approach the origin. On the other hand, if the starting point was on the lattice, the successive iteration would all remain on the lattice, meaning that there are no lattice points on L.
