 Solution

Let $a=B_{12}$, $b=B_{13}$, $c=B_{21}$, $d=B_{23}$, $e=B_{31}$, and $f=B_{32}$. Zero row sums put $B$ in the form used in the spectrum of a three-dimensional matrix with zero row sums. In particular, $B(1,1,1{)}^T=0$. DefineExpanding the characteristic polynomial gives $\det (\lambda I-B)=\lambda ({\lambda }^2+S\lambda +T)$, soFor the physical coefficients, $L_j{B}_{jk}=L_k{B}_{kj}$ with $L_j=M_j\sqrt{G{M}_{\star }{a}_j}$. Thus $W=\mathrm{diag}(L_j)$ gives positive diagonal symmetrization of a matrix, andConsequently the eigenvalues are real and nonpositive, and $B$ has a complete eigenvector basis. This justifies the mode expansion in part (i).