DURABLE CONSUMPTION AS A STATUS GOOD: A STUDY OF NEOCLASSICAL CASES

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 + δ}X^{1 + β + δ}) ^— [^cμM ^(Fkk ^{+ F}kk^lk^{) —} lμβFkk (^у/Г)] } ^ω^
+β {(1 + δ)(1 + β + δ) + [c_μμ (F_kk + F_kkl_k) — l_μμF_kk (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) + [c_μμ (Fkk + Fkklk) — l_μβFkk (r∕^r)] } < 0,

¹¹In calculating tr(J) = 2β, we substitute for l_μ and Z⅛ in the expression for tr(J). In deriving det J > 0,

we use the fact that

1 -I J ∙ J J 2     2 T~<       T~∣2       1∕T-∣O T-I T-Ih         T-I ∕~∕τh

and substitute for FkkFii = F_m and (Fkiβ — FkkFi) = —F_fefe(y∕l).

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