From telegraph equation
\begin{align*}
- & \frac{\partial \bf{v}(z,\omega) }{\partial z}=[\bf{R}(\omega )+j\omega \bf{L}(\omega)]\bf{i}(z,\omega )=\bf{Z}(\omega) \bf{i}(z,\omega) \\
- & \frac{\partial \bf{i}(z,\omega) }{\partial z}=[\bf{G}(\omega )+j\omega \bf{C}(\omega)]\bf{v}(z,\omega )= \bf{Y}(\omega ) \bf{v} (z, \omega )\\
\end{align*}
we have the following wave equation
\begin{align*}
- \frac{\partial ^2 \bf{v}(z)}{\partial z} = [Z][Y]\bf{v}(z) \\
- \frac{\partial ^2 \bf{i}(z)}{\partial z} = [Y][Z]\bf{i}(z)
\end{align*}
The modal solution to the wave equation will be of the form
\begin{align*}
\bf{v}(z) & = \bf{V}^{\pm}e ^{\mp \gamma _0 z} \\
\bf{i}(z) & = \bf{I}^{\pm}e ^{\mp \gamma _1 z }\\
\end{align*}
Substitute above equation into wave equation, we have
\begin{align*}
( \gamma _0 ^2 \bf{U} & -[Z][Y]) \bf{V} ^{\pm} =0 \\
( \gamma _1 ^2 \bf{U} & -[Y][Z]) \bf{I} ^{\pm} =0 ....(4)\\
\end{align*}
For non-trivial solution, the determinant of
\begin{align*}
| \gamma _0 ^2 \bf{U} & -[Z][Y] | =0 \\
( \gamma _1 ^2 \bf{U} & -[Y][Z]| =0 ....(5)\\
\end{align*}
It can be shown r0=r1=r, due to symmetric of [Z],[Y] and ([Z][Y])T=[YT][ZT]=[Y][Z].
The solution to (4) will be called modes, while the eigenvalues will be called spectrum.
Assume(5) (4) has n distinct eigenvalues and eigenvectors, combine all eigenvectors to form matrix, it will be
\begin{align*}
[\bf{V}_1 e ^ {- \gamma _1 z}, \bf{V}_2 e ^ {- \gamma _1 z},...\bf{V}_N e ^ {- \gamma _N z} ] = \bf{M} _V diag(e ^ {- \gamma _1 z},e ^ {- \gamma _2 z},...,e ^ {- \gamma _N z})=\bf{M}_v \bf{D} ....(6)\\
\end{align*}
where Mv=[V1,V2,...VN] are eigenvectors of voltage wave, Mv is non-singular and span a N-dimensional space. Similarity, we have
\begin{align*}
[\bf{I}_1 e ^ {- \gamma _1 z}, \bf{I}_2 e ^ {- \gamma _1 z},...\bf{I}_N e ^ {- \gamma _N z} ] = \bf{M} _I diag(e ^ {- \gamma _1 z},e ^ {- \gamma _2 z},...,e ^ {- \gamma _N z})=\bf{M}_I \bf{D}
\end{align*}
Substitute (6) into (1), we have
\begin{align*}
\bf{M}_V =[Z]\bf{M}_I diag(\gamma_1 ^ {-1},\gamma_2 ^ {-1},...\gamma_N^ {-1})=\bf{Z}\bf{M}_I \bf{S}
\end{align*}
Similarity, we have
\begin{align*}
\bf{M}_I =[Y]\bf{M}_V diag(\gamma_1 ^ {-1},\gamma_2 ^ {-1},...\gamma_N^ {-1})=\bf{Y}\bf{M}_V \bf{S}
\end{align*}
Now, we have the N engenvectors spanning N-dimensional space, any voltage V(z) can be represented as linear combination of these eighenvectors.
\bf{V}(z)=\bf{V}_i(z)+\bf{V}_r(z)=\bf{M_v(DV_0^++D^{-1}V_0^-)}
where
\begin{align*}
\bf{V_0^+}= &(V_{01}^+,V_{02}^+,....,V_{0N}^+)\\
\bf{V_0^-}= & (V_{01}^-,V_{02}^-,....,V_{0N}^-)
\end{align*}
Same thing for I(z),where
\bf{I}(z)=\bf{I}_i(z)+\bf{I}_r(z)=\bf{M_I(DI_0^+-D^{-1}I_0^-)}
Note: minus (-1) sign is due to the directions of current Ir is opposite of Ii.
The characteristic matrix Zc is defined as
\begin{align*}
\bf{V_i}(z) & =\bf{Z}_C \bf{I}_i(z) \\
\bf{V_r}(z) & =-\bf{Z}_C \bf{I}_r(z) \\
w&here\\
\bf{Z}_C & =\bf{M}_V \bf{M}_I ^{-1}
\end{align*}
