Doppler as an eigenvalue of the Lorentz transformation It is a known fact that
$$
\gamma (1\pm\beta) = \sqrt\frac{1\pm\beta}{1\mp\beta}
$$
is an eigenvalue of the Lorentz transformation (which is a linear transformation).
This is also (as stated in the linked SE question) the relativistic Doppler shift. I know how to show that this is both the eigenvalue and the relativistic Doppler shift, but I'm having a hard time understanding the meaning behind this (maybe some intuitive understanding) of what does this mean that the Doppler shift is an eigenvalue and why does this shift apply only for wave?
 A: Compute the eigenvectors and eigenvalues of a boost [expressed in terms of the rapidity $\theta$]
$$\left(
\begin{array}{cc}
\gamma & \gamma\beta\\
\gamma\beta & \gamma
\end{array}\right)
=
\left(
\begin{array}{cc}
\cosh\theta & \sinh\theta\\
\sinh\theta & \cosh\theta
\end{array}
\right)$$
where $\beta=\tanh\theta$ and $\gamma=\cosh\theta$. (What would $\exp\theta$ be?).
[UPDATE: The use of rapidity is enlightening, but not necessary,
for the calculations suggested below.]
Note the vector-type of the eigenvectors.
(What type of object is describable by such vectors?)
Imagine the "diamond" whose edges are formed by the two eigenvectors.
Under the boost, each eigenvector is scaled by its eigenvalue.
Note the product of the eigenvalues and note the area of the boosted diamond. 
(You might also note the "slopes" of the diagonals of that boosted diamond.)
I hope this helps.
(I can update with graphics later, if necessary.)

[UPDATE: @snatchysquid , here is a hint to get you started on the eigenvalue problem.
Given a matrix $B$, find its eigenvalues, the $\lambda$'s that make $ B\vec v =\lambda \vec v$ true for some special vectors $\vec v$ (called the eigenvectors associated with each eigenvalue). So, for the boost, we have
$$\left(
\begin{array}{ll}
\gamma -\lambda & \gamma\beta\\
\gamma\beta & \gamma -\lambda
\end{array}\right)
\left(\begin{array}{c}
v_x \\ v_y
\end{array}\right)
=
\left(\begin{array}{c}
0 \\ 0
\end{array}\right)
$$
Find the $\lambda$'s that satisfy
this equation. [You'll find that you need the determinant of the matrix $(B-\lambda I)$ must be zero.]
Then for each of the found $\lambda$'s, compute the specific form of its eigenvector
$
\left(\begin{array}{c}
v_x \\ v_y
\end{array}\right)
$, where the required $v_x$ and $v_y$ are related by the particular $\lambda$.]
