where ω = 1 — 4 are the eigenvalues of J.11 The condition tr (J) > 0 rules out the case of
four negative eigenvalues, while det (J) > 0 implies that the following cases do not obtain:
i) one negative and three positive eigenvalues; and ii) three negative and one positive
eigenvalue. This leaves the cases that are not ruled-out by tr (J) > 0 and det (J) > 0: i)
two negative and two positive eigenvalues; and ii). four positive eigenvalues. To determine
which of the two cases holds, we use the fact that the characteristic equation, denoted by
det (J—ωl) = 0, can be factored as:
det (J—ωl) = (ω — ωι)(ω — ω2)(ω — ω3)(ω — ω4)
= ω4 — tr (J) ω3 + (ω1ω2 + ω1ω3 + ω1ω4 + ω2ω3 + ω2ω4 + ω3ω4) ω2
— (ω1ω2ω3 + ω1ω2ω4 + ω1ω3ω4 + ω2ω3ω4) ω + det (J) = 0. (2.16a)
Calculating det (J—ωl) from (2.15), we hnd:
det (J—ωl) =
ω"' — tr (J) ω3 + {β2 — (1 + δX1 + β + δ) — [cμM (Fkk + Fkklk) — lμβFkk (у/Г)] } ω^
+β {(1 + δ)(1 + β + δ) + [cμμ (Fkk + Fkklk) — lμμFkk (у/l)] } ω + det (J) = 0.
(2.16b)
Matching the coefficients of (2.16a, b), we observe that:
ω1ω2ω3 + ω1ω2ω4 + ω1ω3ω4 + ω2ω3ω4
= —β {(1 + δ)(1 + β + δ) + [cμμ (Fkk + Fkklk) — lμβFkk (r∕r)] } < 0,
11In calculating tr(J) = 2β, we substitute for lμ and Z⅛ in the expression for tr(J). In deriving det J > 0,
we use the fact that
Fkk + Fkllk = Fkk -
FkkV "
V '' + μFu
< 0,
1 -I J ∙ J J 2 2 T~< T~∣2 1∕T-∣O T-I T-Ih T-I ∕~∕τh
and substitute for FkkFii = Fm and (Fkiβ — FkkFi) = —Ffefe(y∕l).
10