Let $\mathscr{P}$ be the set of row vectors of size $d$ with non-negative coefficients whose coordinate sum equals 1: $$\mathscr{P} = \left\{ u \in \mathscr{M}_{1,d}\left(\mathbb{R}_{+}\right) : \sum_{j=1}^{d} u_j = 1 \right\}.$$ We consider a square matrix $P \in \mathscr{M}_d\left(\mathbb{R}_{+}\right)$ such that for all $i \in \{1,\ldots,d\}$, $$\sum_{j=1}^{d} P_{i,j} = 1$$ We further assume that there exist $\nu \in \mathscr{P}$ and $c > 0$ such that for all $i,j \in \{1,\ldots,d\}$, $$P_{i,j} \geqslant c\nu_j.$$ Show that there exists a unique element $\mu$ of $\mathscr{P}$ such that $\mu P = \mu$.
Let $\mathscr{P}$ be the set of row vectors of size $d$ with non-negative coefficients whose coordinate sum equals 1:
$$\mathscr{P} = \left\{ u \in \mathscr{M}_{1,d}\left(\mathbb{R}_{+}\right) : \sum_{j=1}^{d} u_j = 1 \right\}.$$
We consider a square matrix $P \in \mathscr{M}_d\left(\mathbb{R}_{+}\right)$ such that for all $i \in \{1,\ldots,d\}$,
$$\sum_{j=1}^{d} P_{i,j} = 1$$
We further assume that there exist $\nu \in \mathscr{P}$ and $c > 0$ such that for all $i,j \in \{1,\ldots,d\}$,
$$P_{i,j} \geqslant c\nu_j.$$
Show that there exists a unique element $\mu$ of $\mathscr{P}$ such that $\mu P = \mu$.