= Solution
In the unnormalized embedding, take
$$
x=s=e,\qquad \tau=\kappa=\theta=\rho=1,\qquad\lambda=0.
$$
Then the first equation is $Ae-b+\bar b=0$, the second is $c-e-\bar c=0$, and the third is $-c^Te-1+\bar z=0$. The fourth becomes
$$
\bar c^Te-\bar z+N
=(c^Te-n)-(c^Te+1)+(n+1)=0.
$$
The multiplier elimination preserves this point because $q=c-e-\bar c=0$ and hence $Hq=0$. Every coordinate of $y$ is one. Dividing by $n^*=2n+4$ therefore gives
$$
\boxed{y^{(0)}=(1/n^*,\ldots,1/n^*)^T,}
$$
which satisfies $A^*y^{(0)}=0$, $\mathbf1^Ty^{(0)}=1$, and strict positivity. Its existence is independent of feasibility of the original input problem.
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