Exercise 5.3.15

Let $v$ be that nonnegative vector and say $d$ is the sum of all its entries. Thus we have $x=\frac{1}{d}v$ is a probability vector. Furthermore, if $y$ is a probability vector in $W$. They must be parallel, say $y=\mathit{kx}$. The sum of entries of $y$ is $k$ by the fact that $x$ is a probability vector. And the sum of entries of $y$ is $1$ since it itself is a probability vector. This means $k=1$. So the vector is unique.