Thus Jacobian simplifies to
J=(0RsRbR/(1+P)2PsPbP/(1+R)20).(6)J = \begin{pmatrix} 0 & R^* s_R b_R /(1+\kappa P^*)^2\\[6pt] P^* s_P b_P /(1+\kappa R^*)^2 & 0 \end{pmatrix}. \tag{6}J=(0PsPbP/(1+R)2RsRbR/(1+P)20).(6)
This symmetric off-diagonal form is convenient: the characteristic equation is 2=0\lambda^2 - \Delta =02=0 with determinant
=detJ=(RsRbR(1+P)2)(PsPbP(1+R)2).(7)\Delta = \det J = \bigg( R^* s_R \frac{b_R}{(1+\kappa P^*)^2}\bigg)\bigg( P^* s_P \frac{b_P}{(1+\kappa R^*)^2}\bigg). \tag{7}=detJ=(RsR(1+P)2bR)(PsP(1+R)2bP).(7)
Eigenvalues are =\lambda = \pm\sqrt{\Delta}=. Because the trace is zero (here), eigenvalues are real if >0\Delta>0>0 (one positive, one negative saddle) or purely imaginary if <0\Delta<0<0 (not possible since 0\Delta\ge00 with positive parameters). However, if we include additional negative self-derivative terms (costs leading to negative FR,RF_{R,R}FR,R or FP,PF_{P,P}FP,P), the trace will no longer be zero and complex conjugate eigenvalues can arise.
To account for realistic self-regulation (for example, costs that make FR,R<0F_{R,R}<0FR,R<0 or include logistic saturation in growth), augment FRF_RFR and FPF_PFP with explicit negative self-feedback terms:
FR=aR+bRP1+PcRR,FP=aP+bPR1+RcPP,(8)F_R = a_R + b_R\frac{P}{1+\kappa P} - c_R R,\qquad F_P = a_P + b_P\frac{R}{1+\kappa R} - c_P P, \tag{8}FR=aR+bR1+PPcRR,FP=aP+bP1+RRcPP,(8)
with cR,cP>0c_R,c_P>0cR,cP>0. Then
FR,R=cR,FP,P=cP,F_{R,R} = -c_R,\quad F_{P,P}=-c_P,FR,R=cR,FP,P=cP,
and the Jacobian becomes
J=(RsRcRRsRbR/(1+P)2PsPbP/(1+R)2PsPcP).(9)J = \begin{pmatrix} - R^* s_R c_R & R^* s_R b_R /(1+\kappa P^*)^2\\[6pt] P^* s_P b_P /(1+\kappa R^*)^2 & - P^* s_P c_P \end{pmatrix}. \tag{9}J=(RsRcRPsPbP/(1+R)2RsRbR/(1+P)2PsPcP).(9)
