Why was is thought necessary to quote my entire post to reply without addressing any specific parts of my post?
amazing work rpenner truly... amazing...
I am sure, anyone with a degree in physics would call it elementary, since the main lines of argumentation were laid out 1905-1910. (Einstein, Lorentz, Minkowski, von Ignatowsky).
but can I ask?
What presumptions/assumptions/assumed to be true premises are you making?
I assumed that the transform with velocity v would convert motionlessness into motion of v in the x direction, and convert motion of -v in the x direction to motionlessness. I assumed the structure of space-time was invariant with respect to translations in time and/or space such that it was only necessary to discuss coordinate differences. I also assumed isometry in x,y and z directions and that the transform was self-consistent.
More useless math and that has nothing to do with the assumption that the speed of light is invariant.
This math is just a trick, to distract your attention from the main issue, the speed of light is invariant.
That was not one of my assumptions. However, if you ask what the eigenvectors of $$\Lambda_K(v)$$ are, the results (for non-zero v) depend on K.
The eigenvectors are $$\begin{pmatrix} \sqrt{K} \\ -1 \\ 0 \\ 0 \end{pmatrix}, \quad\begin{pmatrix} \sqrt{K} \\ +1 \\ 0 \\ 0 \end{pmatrix}, \quad\begin{pmatrix}0 \\ 0 \\ 1 \\ 0 \end{pmatrix}, \quad\begin{pmatrix}0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$$
Any scalar multiple of an eigenvector is also an eigenvector.
So if, as nature prescribes, $$K>0$$, then it follows that $$\Delta S = \begin{pmatrix} \sqrt{K} \\ \pm 1 \\ 0 \\ 0 \end{pmatrix} \sqrt{K^{-1}} \Delta t = \begin{pmatrix} \Delta t \\ \pm \sqrt{K^{-1}} \Delta t \\ 0 \\ 0 \end{pmatrix}$$ represents motion of $$\pm \sqrt{K^{-1}} $$ in the x direction which transforms into
$${\huge \Delta S'} = \begin{pmatrix} \frac{1}{\sqrt{1 - K v^2}} & \frac{K v}{\sqrt{1 - K v^2}} & 0 & 0 \\ \frac{v}{\sqrt{1 - K v^2}} & \frac{1}{\sqrt{1 - K v^2}} & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} \sqrt{K} \\ \pm 1 \\ 0 \\ 0 \end{pmatrix} \sqrt{K^{-1}} \Delta t = \begin{pmatrix} \frac{\sqrt{K} \pm K v}{\sqrt{1-Kv^2}} \\ \frac{\pm 1 + v \sqrt{K}}{\sqrt{1-Kv^2}} \\ 0 \\ 0 \end{pmatrix} \sqrt{K^{-1}} \Delta t = \frac{\sqrt{K^{-1}} \pm v}{\sqrt{1-Kv^2}} \Delta t \begin{pmatrix} \sqrt{K} \\ \pm 1 \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix} \Delta t' \\ \pm \sqrt{K^{-1}} \Delta t' \\ 0 \\ 0 \end{pmatrix}$$
So K > 0 implies that there is an invariant speed $$\sqrt{K^{-1}}$$ such that motion at this speed in either direction in the x-direction is left unchanged by the generalized Galilean transform. Since we (science-minded folk) let evidence from nature decide that $$K=c^{\tiny -2}$$ then it follows that this invariant speed is c, but that was not one of my assumptions.
$$K=c^{\tiny -2}$$ is supported by research based on the color of gold, and experiments with flowing water, airplanes, starlight, electron beams, cosmic ray showers, antimatter, and isotope weights. The math I have shown is 100% compatible with the assumption that K=0, but such an assumption is not supported by evidence from nature.