(a) Saddle-node (fold)
A saddle-node occurs when an eigenvalue crosses through zero: =0\lambda=0=0 is a simple root of the characteristic polynomial. Algebraically this means
detJ(Xc)=0(i.e. c3=0),\det J(X_c) = 0 \quad\text{(i.e. } c_3 = 0\text{),}detJ(Xc)=0(i.e. c3=0),
with the zero eigenvalue being simple (algebraic multiplicity 1, geometric multiplicity 1). Equivalently:
c3=0c_3 = 0c3=0 and c20c_2 \neq 0c2=0 (generically).
Rank condition: rankJ=2\mathrm{rank}\,J = 2rankJ=2 (one dimensional nullspace).
Nondegeneracy / transversality for a saddle-node requires:
Let vvv span the right nullspace: Jv=0J v = 0Jv=0.
Let www span the left nullspace: wJ=0w^\top J = 0wJ=0.
Normalization wv=1w^\top v = 1wv=1.
Define
 a=12wDX2F(Xc)[v,v](quadratic coefficient),b=wEF(Xc),a \;=\; \tfrac{1}{2}\, w^\top D_X^2 F(X_c)[v,v] \quad\text{(quadratic coefficient)},\qquad b \;=\; w^\top \partial_{\mu_E} F(X_c),a=21wDX2F(Xc)[v,v](quadratic coefficient),b=wEF(Xc),
where DX2F[X][u,v]D_X^2 F[X][u,v]DX2F[X][u,v] is the bilinear form of second derivatives (Hessian applied to u,vu,vu,v).
A generic saddle-node requires a0a\neq 0a=0 and b0b\neq 0b=0. In our model EF=(0,1,0)\partial_{\mu_E}F=(0,1,0)^\topEF=(0,1,0) so b=w2b = w_2b=w2 (the second component of the left nullvector).
Interpretation: if a0a\neq0a=0 and b0b\neq0b=0 then locally near the fold the reduced scalar normal form is
y=ay2+b(Ec)+higher orders,\dot y = a y^2 + b (\mu_E-\mu_c) + \text{higher orders},y=ay2+b(Ec)+higher orders,
so the local scaling of the two merging equilibria and their behavior follow classical saddle-node normal form.
(b) Hopf
A Hopf bifurcation requires a pair of complex conjugate eigenvalues to cross the imaginary axis:
At Hopf, 1,2=i0\lambda_{1,2} = \pm i\omega_01,2=i0 (simple pair) and the third eigenvalue has (3)0\Re(\lambda_3) \neq 0(3)=0.
Algebraically at the Hopf point the cubic coefficients satisfy (Hurwitz borderline)
c1c2=c3,c1>0,c2>0,c3>0,c_1 c_2 = c_3,\qquad c_1>0,\quad c_2>0,\quad c_3>0,c1c2=c3,c1>0,c2>0,c3>0,
and the crossing is transversal:
ddE(1,2)E=h0.\left.\frac{d}{d\mu_E}\Re(\lambda_{1,2})\right|_{\mu_E=\mu_h} \neq 0.dEd(1,2)E=h=0.
