Background - transmission line

$\newcommand{\ket}[1]{\left \lvert #1 \right \rangle}$ A transmission line can be modeled as an infinite sequence of inductors and capacitors:

     z   z+dz
     |   |
     v   v
  ---L---L--- <-- signal
   |   |   |
.. C   C   C ..
   |   |   |
  ----------- <-- ground

Each $L$ and $C$ is the inductance and capacitance per length. The current and voltage in this circuit is represented by the so-called telegrapher's equations \begin{align} V(z + dz) = V(z) - L \, dz \, \dot{I}(z) \quad \text{and} \quad I(z + dz) = I(z) - C \, dz \, \dot{V}(z) \end{align} which can be rearranged, in the limit $dz \rightarrow 0$, to $$ \frac{\partial V}{\partial z} = - L \frac{\partial I}{\partial t} \quad \text{and} \quad \frac{\partial I}{\partial z} = - C \frac{\partial{V}}{\partial t} \, . $$ If we take $\partial/\partial z$ on the left equation, and $\partial/\partial_t$ on the right equation, and substitude the right into the left, we get $$ \left( \frac{\partial V}{\partial z} \right)^2 - LC \left(\frac{\partial V}{\partial t} \right)^2 = 0 $$ which is the wave equation for waves moving with phase velocity $v^2 = 1 / LC$ (remember $L$ and $C$ are per-length quantities).

Context - coupled transmission lines

Now suppose we have coupled transmission lines

    z      z+dz
    |      |
    v      v
  ------------------- <-- ground
    |      |      |
..  Ca     Ca     Ca ..
    |      |      |
  -----La-----La----- <-- signal a
    |      |      |
..  Cm Lm  Cm Lm  Cm ..
    |      |      |
  -----Lb-----Lb----- <-- signal b
    |      |      |
..  Cb     Cb     Cb ..
    |      |      |
  ------------------- <-- ground

We have two lines $a$ and $b$, each with inductance per length $L_a$ or $L_b$, and capacitance per length $C_a$ or $C_b$. The lines are coupled through $C_m$ and $L_m$; the mutual inductance is hard to draw in ASCII art, so think of $L_m$ as a mutual inductance between the inductors $L_a$ and $L_b$. It's pretty easy to show that the equations for these coupled lines are \begin{align} \frac{\partial V_a}{\partial z} &= - L_a \frac{\partial I_a}{\partial t} - L_m \frac{\partial I_b}{\partial t} \\ \frac{\partial I_a}{\partial z} &= -(C_a + C_m) \frac{\partial V_a}{\partial t} + C_m \frac{\partial V_b}{\partial t} \end{align} and similarly for line $b$.

The equations for the coupled lines can be expressed in a matrix form \begin{align} \frac{\partial}{\partial z} \ket{V} &= - [L] \frac{\partial}{\partial t} \ket{I} \\ \frac{\partial}{\partial z} \ket{I} &= - [C] \frac{\partial}{\partial t} \ket{V} \end{align} where $\ket{V}$ is the vector of voltages, $\ket{I}$ is the vector of currents, and $[L]$ and $[C]$ are matrices. The matrices are $$ [L] = \begin{bmatrix} -L_a & -L_m \\ -L_m & -L_b \end{bmatrix} \quad \text{and} \quad [C] = \begin{bmatrix} -(C_a + C_m) & C_m \\ C_m & -(C_b + C_m) \end{bmatrix} \, . $$

Question

In the book Fundamentals of Microwave Transmission Lines by Jon C. Freeman, it is written

It turns out that the velocity of propagation is $$ [L][C] = \frac{1}{v^2} [I]$$ where $$ v = \frac{1}{\sqrt{\mu \epsilon}} $$ (here $[I]$ is the identity matrix)

Why is the product of the $L$ and $C$ matrices diagonal, and why are the diagonal entries equal to each other ($1/v^2$)? It seems like the author must be making an assumption and I haven't figured out what that assumption is.