# Two-Wire Power Line Microwave Communication Using Low-Loss Modes
## 2 Problem Definition and Exact Solution
Since the matrix in (9) is symmetric, it can be written as follows:  
\[ \begin{bmatrix}A&B\\ B&A\end{bmatrix}  \tag{11}\]The eigenvalues of this matrix (\(\lambda\)) (not to be confused with \(\lambda\), which is used elsewhere to designate \"wavelength\") are defined by  
\[ \begin{bmatrix}A&B\\ B&A\end{bmatrix}\begin{bmatrix}q_{a}\\ q_{b}\end{bmatrix} =\lambda \begin{bmatrix}q_{a}\\ q_{b}\end{bmatrix}  \tag{12}\]  
where the vectors \(\overline{q}=[q_{a},q_{b}]^{T}\) are its eigenvectors. (not to be confused with \(q\), which is used elsewhere to represent charge). Since (12) represents a homogeneous set of equations, it only has a solution if  
\[\left| \begin{array}{cc}A-\lambda&B\\ B&A-\lambda\end{array} \right|=0 \tag{13}\]  
This occurs when the quadratic  
\[\lambda^{2}-2A\lambda+\left(A^{2}-B^{2}\right)=0 \tag{14}\]  
Hence, the eigenvalues are  
\[\lambda_{1/2}=A\pm B=\widetilde{G}_{ez}^{11}-z_{iw}\pm\widetilde{G}_{ez}^{12} \tag{15}\]  
The eigenvectors can be found by inserting the eigenvalues into (12) as  
\[ \begin{bmatrix}\mp B&B\\ B&\mp B\end{bmatrix}\begin{bmatrix}q_{a}\\ q_{b}\end{bmatrix} =0 \tag{16}\]  
Given that the matrix in (16) is symmetric, its eigenvectors are simply  
\[\overline{q}_{1}=a_{1} \begin{bmatrix}1\\ 1\end{bmatrix} \text{ and }\overline{q}_{2}=a_{2} \begin{bmatrix}1\\ -1\end{bmatrix}  \tag{17}\]  
where \(a_{1}\) and \(a_{2}\) are arbitrary constants. \(\overline{q}_{1}\) represents the common component (again often called the \"common mode\"), which has equal currents on each conductor, and \(\overline{q}_{2}\) represents the differential component (again often called the \"differential mode\"). These eigenvectors can be written as a matrix of eigenvectors (by columns) that are normalized to a magnitude of 1, as follows:  
\[|\eta|=\frac{1}{\sqrt{2}} \begin{bmatrix}1&1\\ 1&-1\end{bmatrix}  \tag{18}\]  
If the column matrix of conductor currents \(\hat{\tilde{I}}_{1}\) and \(\hat{\tilde{I}}_{2}\) is expanded in the eigenvectors of the symmetric matrix, then  
\[ \begin{bmatrix}\hat{\tilde{I}}_{1}\\ \hat{\tilde{I}}_{2}\end{bmatrix} =[\eta] \begin{bmatrix}\hat{\tilde{I}}_{gm1}^{ \prime}\\ \hat{\tilde{I}}_{gm2}\end{bmatrix}  \tag{19}\]  
where \(\hat{\tilde{I}}_{gm1}\) and \(\hat{\tilde{I}}_{gm2}\) are the \"geometric mode\" amplitudes. If (19) is substituted into (9), the entire equation is pre-multiplied by the inverse matrix \([\eta]^{-1}\), which (in this case) is equal to \([\eta]\). \[\frac{1}{2} \begin{bmatrix}1&1\\ 1&-1\end{bmatrix}\begin{bmatrix}\widetilde{G}_{ez}^{11}-z_{iw}&\widetilde{G}_ {ez}^{12}\\ \widetilde{G}_{ez}^{21}&\widetilde{G}_{ez}^{22}-z_{iw}\end{bmatrix}\begin{bmatrix} 1&1\\ 1&-1\end{bmatrix}\begin{bmatrix}\hat{\tilde{I}}_{gm1}\\ \hat{\tilde{I}}_{gm2}\end{bmatrix} =\frac{1}{\sqrt{2}} \begin{bmatrix}1&1\\ 1&-1\end{bmatrix}\begin{bmatrix}-\hat{\tilde{V}}_{1}-\hat{\tilde{\tilde{ \tilde{\tilde{\tilde{\tilde{\chi}}}}}}}^{1}_{2v}\\ -\hat{\tilde{V}}_{2}-\hat{\tilde{\tilde{\chi}}}^{2}_{2v}\end{bmatrix}  \tag{20}\]However, pre-multiplying and post-multiplying a matrix by a matrix of its eigenvectors results in a diagonalized matrix of eigenvalues, as follows:  
\[[\eta]^{-1}\left[ \begin{matrix}\widetilde{G}_{ez}^{11}-z_{iw}&\widetilde{G}_{ez}^ {12}\\ \widetilde{G}_{ez}^{21}&\widetilde{G}_{ez}^{22}-z_{iw}\end{matrix} \right][\eta]= \left[ \begin{matrix}\widetilde{G}_{ez}^{11}-z_{iw}+\widetilde{G}_{ez}^{12}&0\\ 0&\widetilde{G}_{ez}^{22}-z_{iw}-\widetilde{G}_{ez}^{12}\end{matrix} \right] \tag{21}\]  
In addition, the right-hand side of (20) becomes  
\[\frac{1}{\sqrt{2}} \begin{bmatrix}1&1\\ 1&-1\end{bmatrix}\begin{bmatrix}-\hat{V}_{1}-\hat{\tilde{E}}_{ez}\\ -\hat{V}_{2}-\hat{\tilde{E}}_{ez}\end{bmatrix} =-\frac{1}{\sqrt{2}} \begin{bmatrix} \hat{V}_{1}+\hat{V}_{2}+\hat{\tilde{E}}_{ez}^{1}+\hat{\tilde{E}}_{ez}^{2}\\ \hat{V}_{1}-\hat{V}_{2}+\hat{\tilde{E}}_{ez}^{1}-\hat{\tilde{E}}_{ez}^{2}\end{bmatrix}  \tag{22}\]  
So that  
\[\frac{1}{2}