Nonnegative inverse matrices preserve extremal basis rays

Suppose $n>0$. For each row $i$, the diagonal equation $1=(AB{)}_{ii}={\sum }_k{a}_{ik}{b}_{ki}$ gives an index $k$ with $a_{ik}>0$ and $b_{ki}>0$. If also $a_{i\mathrm{\ell }}>0$ for some $\mathrm{\ell }\mathrm{\ne }k$, then the off-diagonal equation $0=(BA{)}_{k\mathrm{\ell }}={\sum }_r{b}_{kr}{a}_{r\mathrm{\ell }}$ has the strictly positive summand $b_{ki}{a}_{i\mathrm{\ell }}$, a contradiction. Thus each row of $A$ has exactly one positive entry and all its other entries are zero.

Let $k(i)$ be the column of the positive entry in row $i$. If distinct rows $i,j$ had $k(i)=k(j)=k$, then 1.1 supplies $b_{ki}>0$, while $0=(AB{)}_{ji}={\sum }_r{a}_{jr}{b}_{ri}$ contains the positive summand $a_{jk}{b}_{ki}$, a contradiction. Hence the $n$ indices $k(i)$ are distinct and cover all $n$ columns. Define $P$ to have entry $1$ at each $(i,k(i))$ and zero elsewhere, and let $D_{kk}$ be the unique positive entry of $A$ in column $k$. Then $P$ is a permutation matrix, $D$ is positive diagonal, and $A=PD$. For $n=0$, the unique empty matrices satisfy the same conclusion with $P=D=I_0$.

Under the basis hypotheses, column $j$ of $A=PD$ gives $b_j=d_j{c}_{\pi (j)}$ for a permutation $\pi$ and $d_j=D_{jj}>0$. By bilinearity and the unit root identities, $1=Q(b_j,b_j)=d_j^{2}Q(c_{\pi (j)},c_{\pi (j)})=d_j^2$. Positivity gives $d_j=1$ for every $j$; this is also vacuous for $n=0$. Therefore $D=I_n$ and $A=P$.