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Consider the cubic feedback systemAt the origin its eigenvalues are $0,-2,-1$. With $v=(x+y)/2$ and $w=(x-y)/2$, the center subspace is the $v$ axis and the two remaining directions are strictly stable. The equations becomeThe symmetry $(v,w,z)\mapsto (-v,-w,z)$ allows a center manifold graph with $w(v)$ odd and $z(v)$ even. Write $w=c_3{v}^3+O(v^5)$ and $z=d_2{v}^2+d_4{v}^4+O(v^6)$. The invariance equation for a graph requires $w^{\prime }(v)\dot{v}=\dot{w}$ and $z^{\prime }(v)\dot{v}=\dot{z}$, rather than setting the stable-coordinate derivatives to zero.

The quadratic $z$ equation gives $d_2=1$; the cubic $w$ equation gives $-2c_3+(a+1)/2=0$. At fourth order the $z$ invariance equation is $(a-1)v^4=-d_4{v}^4$, soSubstitution into the $v$ equation yieldsIn particular, the quintic coefficient is $[(3a+1)c_3-d_4]/2$, which records both the cubic stable graph and the quartic lag in $z$. For $a<1$, the leading odd term points toward zero on both sides; for $a>1$ it points outward. At $a=1$ the cubic vanishes but the quintic coefficient is one and still points outward. Since the transverse directions are strictly stable, the centre manifold theorem gives local asymptotic stability exactly when $a<1$.

For a finite-dimensional linearization $\dot{v}=Av$, the center subspace is the real invariant subspace associated with all generalized eigenvectors whose eigenvalues have zero real part. It includes zero eigenvalues and purely imaginary conjugate pairs. Generalized eigenspaces are needed when $A$ is not diagonalizable.

For $A=\mathrm{diag}(-1,0,2)$, it is the second coordinate axis. For a planar rotation matrix generator, it is the whole plane. The linear system alone gives no exponential contraction or expansion on this subspace; nonlinear terms can still cause growth or decay. A local center manifold is tangent to it at the equilibrium.