I had a hard time proving fact 6 in this article:
If $A$ is a positive column-stochastic matrix, then there is a unique eigenvector corresponding to the eigenvalue $z = 1$ such that it has only positive entries the sum of its entries equals $1$.
Specifically, I know that by the Perron-Frobenius theorem, all entries of the eigenvector corresponding to the largest eigenvalue should be positive (at least in the case above where the matrix is positive), but I had a hard time proving that the sum of these entries is 1. Any thoughts on this would be appreciated.
Best Answer
You are fundamentally misunderstanding the statement, and your subject line contains conceptual errors.
You keep talking about "the eigenvector corresponding to $1$", as if there were only one such vector. That's not true.
If $V$ is any vector space, $T$ is any linear operator on $V$, and $x\in V$ is an eigenvector corresponding to $\lambda$, then for all scalars $a$ you have $T(ax) = aT(x) = a(\lambda x) = \lambda(ax)$, so $ax$ is also an eigenvector corresponding to $\lambda$ if $a\neq 0$. If all entries of $x$ are positive, and $a\gt 0$, then all entries of $ax$ are positive, and it is still an eigenvector corresponding to $\lambda$.
In particular, unless your field is $\mathbb{F}_2$ and your eigenspace is $1$-dimensional, you will necessarily have more than one eigenvector corresponding to that eigenvalue. So, no, the statement is not talking about "THE eigenvector corresponding to the largest eigenvalue" (from the comments), because there are multiple (in fact, infinitely many) eigenvectors corresponding to $1$.
The statement does not say that if $x$ is an eigenvector corresponding to $1$ then its entries will add up to $1$. It certainly does not say that "the" eigenvector corresponding to $1$ with positive entries have that property (because there isn't any one such vector; "the" is the singular definite article and can only be used when there is a singly determined object it is being applied to). What it says is that there exists an eigenvector corresponding to $1$ with the desired properties, and it will be the unique such eigenvector.
You already know that the properties of $A$ guarantee the existence of a vector $x$ with all entries positive (Perron-Frobenius doesn't say "the (sic) eigenvector corresponding to $1$ satisfies this condition"; it says there is an eigenvector corresponding to $1$ satisfying this condition). Say $x=(a_1,\ldots,a_n)$. Let $k=a_1+\cdots+a_n$. Because all the $a_i$ are positive, $k\gt 0$. Let $a=\frac{1}{k}$. Then $ax$ is also an eigenvector corresponding to $1$, also has all entries positive, but now we have $$ax = \left(\frac{a_1}{k},\ldots,\frac{a_n}{k}\right),\qquad \frac{a_1}{k}+\cdots + \frac{a_n}{k} = \frac{a_1+\cdots+a_n}{k} = 1.$$ So $ax$ is an eigenvector of $A$, corresponding to $1$, with all entries positive which add up to $1$. That proves existence.
To prove uniqueness, the Perron-Frobenius Theorem also tells you the eigenspace of $1$ is one-dimensional, so if $v$ is any eigenvector corresponding to $1$ whose entries add up to $1$, then $v$ is a multiple of our vector $ax$. If $v=\beta(ax)$, then the entries of $v$ add up to $\beta$ (since the entries of $ax$ add up to $1$), so $\beta=1$, hence $ax=v$, as desired.