# Is it possible to derive $2\times 2$ Lorentz transformation matrix from only eigenvectors?

As a preface, I am somewhat familiar with year 1 linear algebra but not too familiar with how one makes the connection to Lorentz transformation matrices so I apologize if the answer is obvious.

One of my friends recently showed me this video https://www.youtube.com/watch?v=Rh0pYtQG5wI which is part of a larger series the presenter Mark Rober/minutephysics did on special relativity.

In it he says that the Lorentz transformation he describes can be derived from two basic principles:

1) Velocity is relative and measured velocity depends on reference frame.

2) The speed of light $$c$$ is the same in all inertial frames.

From principle 2 I gathered that in that case the world lines for particles moving at speed $$c$$ must be eigenvectors to the Lorentz transformation matrix.

So from that I gathered that one could try to derive the transformation matrix from those eigenvectors.

$$v_1 = \left[\begin{matrix} -c \\ 1\end{matrix}\right]t, v_2 = \left[\begin{matrix} c \\ 1\end{matrix}\right]t$$

Using Matrix Eigendecomposition we get:

$$T_{\gamma} = \left[\begin{matrix} -c & c\\ 1 & 1\end{matrix}\right] \left[\begin{matrix} \lambda_1 & 0\\ 0 & \lambda_2\end{matrix}\right] \tfrac{1}{2}\left[\begin{matrix} -\tfrac{1}{c} & 1 \\ \tfrac{1}{c} & 1\end{matrix}\right]$$

$$T_{\gamma} = \left[\begin{matrix} \frac{\lambda_1 + \lambda_2}{2} & c\frac{\lambda_2 - \lambda_1}{2}\\ \frac{\lambda_2 - \lambda_1}{2c} & \frac{\lambda_1 + \lambda_2}{2}\end{matrix}\right]$$

as the transformation matrix.

I imagine that in order for the transformation to work it must make an object with velocity $$v$$ have zero displacement in the world line. Which means that:

$$T_{\gamma}*\left[\begin{matrix} v \\ 1 \end{matrix}\right] = \left[\begin{matrix} 0 \\ t'\end{matrix}\right]$$

From multiplying to get the top row we find out:

$$\frac{\lambda_1 + \lambda_2}{2}*v + c\frac{\lambda_2 - \lambda_1}{2} = 0$$

which leads to:

$$\frac{\lambda_2}{\lambda_1} = \frac{c-v}{c+v}$$

Now I think that for this matrix the eigenvalues should be the respective Doppler factors. Which means that I would have to arrive at:

$$\lambda_2 = \frac{1}{\lambda_1}$$

But I have no idea how to get that from our assumptions. I would appreciate any pointers.

• You can get $\lambda_1\lambda_2=\pm 1$ from the matrix equation $T_\gamma^T\eta T_\gamma=\eta$. – J.G. May 14 at 12:03