[proofplan] Fix $\mu>0$ and study the first-order optimality condition for the logarithmic-barrier objective on the affine space $Ax=b$. Since the feasible set is relatively open in that affine space, the derivative at the minimizer must vanish along every feasible direction $h\in\ker A$. Full row rank then converts this annihilation condition into a Lagrange multiplier equation. Defining the dual slack by the barrier gradient gives dual feasibility, coordinatewise complementarity $x_i(\mu)s_i(\mu)=\mu$, and positivity. [/proofplan] [step:Compute the derivative of the barrier objective along feasible directions] Fix $\mu>0$. Define the open feasible affine set \begin{align*} \mathcal{F}:=\{x\in\mathbb{R}^n:Ax=b,\ x_i>0\text{ for }i=1,\dots,n\}. \end{align*} Define the barrier objective $\varphi_\mu:\mathcal{F}\to\mathbb{R}$ by \begin{align*} \varphi_\mu(x):=c\cdot x-\mu\sum_{i=1}^n\log x_i. \end{align*} By hypothesis, $x(\mu)\in\mathcal{F}$ is the unique minimizer of $\varphi_\mu$. Let $h\in\ker A$. Since $A(x(\mu)+th)=b$ for every $t\in\mathbb{R}$ and since every coordinate of $x(\mu)$ is positive, there exists $\varepsilon>0$ such that $x_i(\mu)+th_i>0$ for all $i=1,\dots,n$ whenever $|t|<\varepsilon$. Define $\psi_h:(-\varepsilon,\varepsilon)\to\mathbb{R}$ by \begin{align*} \psi_h(t):=\varphi_\mu(x(\mu)+th). \end{align*} The function $\psi_h$ has a minimum at $t=0$, so $\psi_h'(0)=0$. Differentiating gives \begin{align*} 0=\psi_h'(0)=c\cdot h-\mu\sum_{i=1}^n\frac{h_i}{x_i(\mu)}. \end{align*} Define $w(\mu)\in\mathbb{R}^n$ by \begin{align*} w_i(\mu):=c_i-\frac{\mu}{x_i(\mu)},\qquad i=1,\dots,n. \end{align*} Then \begin{align*} w(\mu)\cdot h=0\qquad\text{for every }h\in\ker A. \end{align*} [guided] Fix $\mu>0$. The barrier subproblem is an optimization problem on the affine space cut out by $Ax=b$, but only inside the positive orthant. We name this feasible set: \begin{align*} \mathcal{F}:=\{x\in\mathbb{R}^n:Ax=b,\ x_i>0\text{ for }i=1,\dots,n\}. \end{align*} The objective is the map $\varphi_\mu:\mathcal{F}\to\mathbb{R}$ defined by \begin{align*} \varphi_\mu(x):=c\cdot x-\mu\sum_{i=1}^n\log x_i. \end{align*} By assumption, $x(\mu)$ is the unique minimizer of this map. The feasible directions at $x(\mu)$ are exactly the vectors $h\in\ker A$, because moving from $x(\mu)$ to $x(\mu)+th$ preserves the equality constraint precisely when $Ah=0$. Since $x_i(\mu)>0$ for all $i$, small enough positive and negative values of $t$ keep every coordinate $x_i(\mu)+th_i$ positive. Hence, for each $h\in\ker A$, there exists $\varepsilon>0$ such that the curve $t\mapsto x(\mu)+th$ lies in $\mathcal{F}$ for all $|t|<\varepsilon$. Define $\psi_h:(-\varepsilon,\varepsilon)\to\mathbb{R}$ by \begin{align*} \psi_h(t):=\varphi_\mu(x(\mu)+th). \end{align*} Because $x(\mu)$ minimizes $\varphi_\mu$ on $\mathcal{F}$, the one-variable function $\psi_h$ has a minimum at $t=0$. Therefore its derivative at $0$ vanishes. Differentiating the linear term and the logarithmic terms gives \begin{align*} 0=\psi_h'(0)=c\cdot h-\mu\sum_{i=1}^n\frac{h_i}{x_i(\mu)}. \end{align*} Now define $w(\mu)\in\mathbb{R}^n$ coordinatewise by \begin{align*} w_i(\mu):=c_i-\frac{\mu}{x_i(\mu)},\qquad i=1,\dots,n. \end{align*} The previous displayed identity is exactly \begin{align*} w(\mu)\cdot h=0\qquad\text{for every }h\in\ker A. \end{align*} This is the first-order condition written intrinsically on the equality-constrained feasible affine space. [/guided] [/step] [step:Convert the directional condition into a multiplier equation] We use the following finite-dimensional linear algebra fact. [claim:Annihilators of $\ker A$ are row-space vectors] If $A\in\mathbb{R}^{p\times n}$ has full row rank and $w\in\mathbb{R}^n$ satisfies $w\cdot h=0$ for every $h\in\ker A$, then there exists a unique $\lambda\in\mathbb{R}^p$ such that \begin{align*} w=A^\top\lambda. \end{align*} [/claim] [proof] The map $A^\top:\mathbb{R}^p\to\mathbb{R}^n$ is injective because $A$ has full row rank. Hence uniqueness follows once existence is proved. Let $\operatorname{Range}(A^\top):=\{A^\top z:z\in\mathbb{R}^p\}\subset\mathbb{R}^n$. For every $z\in\mathbb{R}^p$ and every $h\in\ker A$, \begin{align*} (A^\top z)\cdot h=z\cdot Ah=0. \end{align*} Thus $\operatorname{Range}(A^\top)\subset(\ker A)^\perp$. Since $A$ has rank $p$, the [Rank-Nullity Theorem](/theorems/385) gives $\dim\ker A=n-p$, so $\dim(\ker A)^\perp=p$. Also $\dim\operatorname{Range}(A^\top)=p$. Therefore $\operatorname{Range}(A^\top)=(\ker A)^\perp$. Since $w$ annihilates $\ker A$, we have $w\in(\ker A)^\perp$, and hence $w=A^\top\lambda$ for some $\lambda\in\mathbb{R}^p$. [/proof] Apply the claim to $w(\mu)$ from the previous step. There is a unique $\lambda(\mu)\in\mathbb{R}^p$ such that \begin{align*} c-\mu X(\mu)^{-1}e=A^\top\lambda(\mu), \end{align*} where $e\in\mathbb{R}^n$ denotes the vector with $e_i=1$ for all $i$, and $X(\mu)\in\mathbb{R}^{n\times n}$ denotes the diagonal matrix with diagonal entries $x_1(\mu),\dots,x_n(\mu)$. Equivalently, \begin{align*} c=A^\top\lambda(\mu)+\mu X(\mu)^{-1}e. \end{align*} [/step] [step:Define the dual variables and verify the central path equations] Define $y(\mu)\in\mathbb{R}^p$ by \begin{align*} y(\mu):=\lambda(\mu). \end{align*} Define $s(\mu)\in\mathbb{R}^n$ coordinatewise by \begin{align*} s_i(\mu):=\frac{\mu}{x_i(\mu)},\qquad i=1,\dots,n. \end{align*} Since $x_i(\mu)>0$ and $\mu>0$, each $s_i(\mu)>0$. The definition gives \begin{align*} x_i(\mu)s_i(\mu)=x_i(\mu)\frac{\mu}{x_i(\mu)}=\mu,\qquad i=1,\dots,n. \end{align*} The multiplier equation from the previous step becomes \begin{align*} A^\top y(\mu)+s(\mu)=c. \end{align*} Finally, $x(\mu)\in\mathcal{F}$ by definition of the barrier subproblem, so \begin{align*} Ax(\mu)=b \end{align*} and $x_i(\mu)>0$ for $i=1,\dots,n$. Thus the triple $(x(\mu),y(\mu),s(\mu))$ satisfies all four displayed central-path equations. [/step] [step:Prove uniqueness of the associated dual slack and multiplier] The slack vector is forced by complementarity and positivity of $x(\mu)$: if $\tilde{s}\in\mathbb{R}^n$ satisfies $x_i(\mu)\tilde{s}_i=\mu$ for every $i$, then \begin{align*} \tilde{s}_i=\frac{\mu}{x_i(\mu)}=s_i(\mu),\qquad i=1,\dots,n. \end{align*} With $s(\mu)$ fixed, dual feasibility forces \begin{align*} A^\top \tilde{y}=c-s(\mu)=A^\top y(\mu). \end{align*} Since $A$ has full row rank, $A^\top:\mathbb{R}^p\to\mathbb{R}^n$ is injective, and therefore $\tilde{y}=y(\mu)$. Hence the central-path triple associated to the unique barrier minimizer is unique. This completes the proof. [/step]