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The Lorentz-transformation can be deduced without any assumption on the speed of light, see https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#From_group_postulates.

This calculation leaves an undetermined constant $\kappa$, giving $L(v)=\frac{1}{\sqrt{1 + \kappa v^2}} \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix}.$ Now let's calculate the eigenvectors. We don't need the eigenvalues, so we can ignore any constant factor. We can assume that an eigenvector has the form $\begin{bmatrix}1\\ u\end{bmatrix}$.

Now we have to solve $\begin{bmatrix}1\\ u\end{bmatrix}= \lambda \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix} \begin{bmatrix}1\\ u\end{bmatrix}=\lambda\begin{bmatrix}1+uv\kappa\\-v+ u\end{bmatrix}=\frac{\lambda}{1+uv\kappa}\begin{bmatrix}1\\\frac{u-v}{1+uv\kappa}\end{bmatrix}$.

Thus $u=\frac{u-v}{1+uv\kappa}\Rightarrow u+u^2v\kappa=u-v\Rightarrow u=\pm\frac 1{\sqrt{ -\kappa}} $.

Now we can interpret \begin{bmatrix}1\\ u\end{bmatrix} as an object moving with constant speed $u$. This special $u=\frac 1{\sqrt{ -\kappa}}$ is independent of the observer.

Experimentation and observation then shows that $u=c=\frac 1{\sqrt{ -\kappa}}$ is the speed of light.

Edit (inspired by @Timaeus): note that if for a matrix $A$ that has a positive real eigenvalue $\lambda_0$ with all other eigenvalues $|\lambda| <\lambda_0$ we can define $\vec x_{n+1} := \frac{1}{||x_n||}A\vec x_n$, Then $\lim_{n\to\infty}\vec x_n$ will be the eigenvector belonging to $\lambda_0$ (except for some very unlucky starting points.)

In other words, giving an object a series of Lorentz-boosts will bring it arbitrarily close to the speed given by the eigenvector.

The Lorentz-transformation can be deduced without any assumption on the speed of light, see https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#From_group_postulates.

This calculation leaves an undetermined constant $\kappa$, giving $L(v)=\frac{1}{\sqrt{1 + \kappa v^2}} \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix}.$ Now let's calculate the eigenvectors. We don't need the eigenvalues, so we can ignore any scalar factor. We can assume that an eigenvector has the form $\begin{bmatrix}1\\ u\end{bmatrix}$.

Now we have to solve $\begin{bmatrix}1\\ u\end{bmatrix}= \lambda \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix} \begin{bmatrix}1\\ u\end{bmatrix}=\lambda\begin{bmatrix}1+uv\kappa\\-v+ u\end{bmatrix}=\lambda\cdot(1+uv\kappa)\begin{bmatrix}1\\\frac{u-v}{1+uv\kappa}\end{bmatrix}$.

Thus $u=\frac{u-v}{1+uv\kappa}\Rightarrow u+u^2v\kappa=u-v\Rightarrow u=\pm\frac 1{\sqrt{ -\kappa}} $.

Now we can interpret \begin{bmatrix}1\\ u\end{bmatrix} as an object moving with constant speed $u$. This special $u=\frac 1{\sqrt{ -\kappa}}$ is independent of the observer.

Experimentation and observation then shows that $u=c=\frac 1{\sqrt{ -\kappa}}$ is the speed of light.

Edit (inspired by @Timaeus): note that if for a matrix $A$ that has a positive real eigenvalue $\lambda_0$ with all other eigenvalues $|\lambda| <\lambda_0$ we can define $\vec x_{n+1} := \frac{1}{||x_n||}A\vec x_n$, Then $\lim_{n\to\infty}\vec x_n$ will be the eigenvector belonging to $\lambda_0$ (except for some very unlucky starting points.)

In other words, giving an object a series of Lorentz-boosts will bring it arbitrarily close to the speed given by the eigenvector.

The Lorentz-transformation can be deduced without any assumption on the speed of light, see https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#From_group_postulates.

This calculation leaves an undetermined constant $\kappa$, giving $L(v)=\frac{1}{\sqrt{1 + \kappa v^2}} \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix}.$ Now let's calculate the eigenvectors. We don't need the eigenvalues, so we can ignore any constant factor. We can assume that an eigenvector has the form $\begin{bmatrix}1\\ u\end{bmatrix}$.

Now we have to solve $\begin{bmatrix}1\\ u\end{bmatrix}= \lambda \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix} \begin{bmatrix}1\\ u\end{bmatrix}=\lambda\begin{bmatrix}1+uv\kappa\\-v+ u\end{bmatrix}=\frac{\lambda}{1+uv\kappa}\begin{bmatrix}1\\\frac{u-v}{1+uv\kappa}\end{bmatrix}$.

Thus $u=\frac{u-v}{1+uv\kappa}\Rightarrow u+u^2v\kappa=u-v\Rightarrow u=\pm\frac 1{\sqrt -\kappa} $.

Now we can interpret \begin{bmatrix}1\\ u\end{bmatrix} as an object moving with constant speed $u$. This special $u=\frac 1{\sqrt -\kappa}$ is independent of the observer.

Experimentation and observation then shows that $u=c=\frac 1{\sqrt -\kappa}$ is the speed of light.

Edit (inspired by @Timaeus): note that if for a matrix $A$ that has a positive real eigenvalue $\lambda_0$ with all other eigenvalues $|\lambda| <\lambda_0$ we can define $\vec x_{n+1} := \frac{1}{||x_n||}A\vec x_n$, Then $\lim_{n\to\infty}\vec x_n$ will be the eigenvector belonging to $\lambda_0$ (except for some very unluck starting points.)

In other words, giving an object a series of Lorentz-boosts will bring it arbitrarily close to the speed given by the eigenvector.

The Lorentz-transformation can be deduced without any assumption on the speed of light, see https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#From_group_postulates.

This calculation leaves an undetermined constant $\kappa$, giving $L(v)=\frac{1}{\sqrt{1 + \kappa v^2}} \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix}.$ Now let's calculate the eigenvectors. We don't need the eigenvalues, so we can ignore any constant factor. We can assume that an eigenvector has the form $\begin{bmatrix}1\\ u\end{bmatrix}$.

Now we have to solve $\begin{bmatrix}1\\ u\end{bmatrix}= \lambda \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix} \begin{bmatrix}1\\ u\end{bmatrix}=\lambda\begin{bmatrix}1+uv\kappa\\-v+ u\end{bmatrix}=\frac{\lambda}{1+uv\kappa}\begin{bmatrix}1\\\frac{u-v}{1+uv\kappa}\end{bmatrix}$.

Thus $u=\frac{u-v}{1+uv\kappa}\Rightarrow u+u^2v\kappa=u-v\Rightarrow u=\pm\frac 1{\sqrt{ -\kappa}} $.

Now we can interpret \begin{bmatrix}1\\ u\end{bmatrix} as an object moving with constant speed $u$. This special $u=\frac 1{\sqrt{ -\kappa}}$ is independent of the observer.

Experimentation and observation then shows that $u=c=\frac 1{\sqrt{ -\kappa}}$ is the speed of light.

Edit (inspired by @Timaeus): note that if for a matrix $A$ that has a positive real eigenvalue $\lambda_0$ with all other eigenvalues $|\lambda| <\lambda_0$ we can define $\vec x_{n+1} := \frac{1}{||x_n||}A\vec x_n$, Then $\lim_{n\to\infty}\vec x_n$ will be the eigenvector belonging to $\lambda_0$ (except for some very unlucky starting points.)

In other words, giving an object a series of Lorentz-boosts will bring it arbitrarily close to the speed given by the eigenvector.

The Lorentz-transformation can be deduced without any assumption on the speed of light, see https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#From_group_postulates.

This calculation leaves an undetermined constant $\kappa$, giving $L(v)=\frac{1}{\sqrt{1 + \kappa v^2}} \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix}.$ Now let's calculate the eigenvectors. We don't need the eigenvalues, so we can ignore any constant factor. We can assume that an eigenvector has the form $\begin{bmatrix}1\\ u\end{bmatrix}$.

Now we have to solve $\begin{bmatrix}1\\ u\end{bmatrix}= \lambda \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix} \begin{bmatrix}1\\ u\end{bmatrix}=\lambda\begin{bmatrix}1+uv\kappa\\-v+ u\end{bmatrix}=\frac{\lambda}{1+uv\kappa}\begin{bmatrix}1\\\frac{u-v}{1+uv\kappa}\end{bmatrix}$.

Thus $u=\frac{u-v}{1+uv\kappa}\Rightarrow u+u^2v\kappa=u-v\Rightarrow u=\pm\frac 1{\sqrt -\kappa} $.

Now we can interpret \begin{bmatrix}1\\ u\end{bmatrix} as an object moving with constant speed $u$. This special $u=\frac 1{\sqrt -\kappa}$ is independent of the observer.

Experimentation and observation then shows that $u=c=\frac 1{\sqrt -\kappa}$ is the speed of light.

The Lorentz-transformation can be deduced without any assumption on the speed of light, see https://en.wikipedia.org/wiki/Derivations_of_the_Lorentz_transformations#From_group_postulates.

This calculation leaves an undetermined constant $\kappa$, giving $L(v)=\frac{1}{\sqrt{1 + \kappa v^2}} \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix}.$ Now let's calculate the eigenvectors. We don't need the eigenvalues, so we can ignore any constant factor. We can assume that an eigenvector has the form $\begin{bmatrix}1\\ u\end{bmatrix}$.

Now we have to solve $\begin{bmatrix}1\\ u\end{bmatrix}= \lambda \begin{bmatrix} 1 & \kappa v \\ -v & 1 \end{bmatrix} \begin{bmatrix}1\\ u\end{bmatrix}=\lambda\begin{bmatrix}1+uv\kappa\\-v+ u\end{bmatrix}=\frac{\lambda}{1+uv\kappa}\begin{bmatrix}1\\\frac{u-v}{1+uv\kappa}\end{bmatrix}$.

Thus $u=\frac{u-v}{1+uv\kappa}\Rightarrow u+u^2v\kappa=u-v\Rightarrow u=\pm\frac 1{\sqrt -\kappa} $.

Now we can interpret \begin{bmatrix}1\\ u\end{bmatrix} as an object moving with constant speed $u$. This special $u=\frac 1{\sqrt -\kappa}$ is independent of the observer.

Experimentation and observation then shows that $u=c=\frac 1{\sqrt -\kappa}$ is the speed of light.

Edit (inspired by @Timaeus): note that if for a matrix $A$ that has a positive real eigenvalue $\lambda_0$ with all other eigenvalues $|\lambda| <\lambda_0$ we can define $\vec x_{n+1} := \frac{1}{||x_n||}A\vec x_n$, Then $\lim_{n\to\infty}\vec x_n$ will be the eigenvector belonging to $\lambda_0$ (except for some very unluck starting points.)

In other words, giving an object a series of Lorentz-boosts will bring it arbitrarily close to the speed given by the eigenvector.