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Introduce the primal slack $s=Ax-b\in K$. The Lagrangian is $L(x,y)=c^{\top }x-⟨y,Ax-b⟩$, with $y\in K^{\ast }=K$. Minimizing over $x$ is finite only when $A^{\top }y=c$. Thus the conic dual problem isThe primal-dual gap at feasible points is $c^{\top }x-b^{\top }y=⟨s,y⟩\ge 0$. Self-duality specifies the multiplier cone; it does not alone guarantee feasible or bounded problems. The central-path discussion assumes primal and dual strict feasibility and a finite optimum. These additional existence conditions are not implied by the given full column rank of $A$.

Let $F$ be the canonical logarithmically homogeneous self-concordant barrier, with parameter $\nu$ and $F(ts)=F(s)-\nu \log t$. For each $\mu >0$, minimizing $c^{\top }x+\mu F(Ax-b)$ defines the primal central path. Its stationarity defines the dual path through $y=-\mu \nabla F(s)\in \mathrm{int}K$. The joint characterization isEuler's identity for logarithmic homogeneity gives $⟨s,\nabla F(s)⟩=-\nu$, so $\boxed{⟨s,y⟩=\nu \mu }$. The gap tends to zero as $\mu \downarrow 0$. If $F_{\ast }(y)={\sup }_{s\in \mathrm{int}K}[-⟨y,s⟩-F(s)]$ is the dual barrier, the equivalent dual relation is $s=-\mu \nabla F_{\ast }(y)$. A self-dual cone does not justify identifying two arbitrary primal and dual barrier functions without this relation.

For a target parameter $\bar{\mu }>0$, form residuals $r_p=Ax-b-s$, $r_d=A^{\top }y-c$ and $r_c=y+\bar{\mu }\nabla F(s)$. Linearization gives the central-path Newton systemFor a strictly feasible primal-dual iterate with $r_p=r_d=0$, elimination reduces it toThe barrier Hessian is positive definite and $A$ has full column rank, so the reduced matrix is positive definite. Alternatively a changing path parameter can be included as an additional linear term $\Delta \mu \nabla F(s)$; the displayed system instead fixes the new target parameter before solving.

This is an interior-point method because iterates stay in the cone interiors where the barrier and its gradient/Hessian are defined. A full Newton step need not do so. Use a fraction-to-boundary or backtracking step: decrease $\alpha >0$ until $s+\alpha \Delta s$ and $y+\alpha \Delta y$ lie in $\mathrm{int}K$, then enforce an appropriate barrier or residual decrease. Such a positive step exists because the current points are interior. Local barrier norms can also certify an interior step via the Dikin ellipsoid.

For a practical starting point, choose $e\in \mathrm{int}K$ and solve a conic phase-I problem, for example minimizing $t$ subject to $Ax-b+te\in K$ and $t\ge -1$. A large positive $t$ with an arbitrary $x$ gives a strictly feasible start for this auxiliary problem. A feasible point with $t<0$ certifies $Ax-b\in \mathrm{int}K$. If the original problem is strictly feasible, a small negative $t$ is feasible, so phase I can find such a certificate. A similar feasibility procedure handles the dual equality and interior. Alternatively an infeasible-start primal-dual method or homogeneous self-dual embedding starts with interior cone variables while allowing nonzero linear residuals, and can report infeasibility rather than presume an interior solution exists.