Eigenvector corresponding to eigenvalue $ 1 $ of a stochastic matrix

Question

I am trying to justify fact $ 5 $ in this link which states that if $ A $ is a column stochastic matrix, then $ A $ has eigenvalue $ 1 $ and a unique eigenvector such that all entries are either negative or positive. I have successfully proved that $ A $ has an eigenvalue $ 1 $ but still stuck on the second part of the fact. It seems that this is the Perron Frobenius theorem, but the proof for this theorem requires materials that are beyond a first course in linear algebra that I haven't learned about, so is there any way to prove this fact without using the PF theorem?

Answer 1

For one thing, the eigenvector is unique up to scalar. That's easier to prove for the transpose $A^T$: the only eigenvectors for $1$ are multiples of $(1,...,1)^T.$ If $v$ is any eigenvector, then let $k$ be an index such that $v_k$ is maximal; then $$v = A^T v, \; \mathrm{so} \; v_k = \sum_{i=1}^n a_{ik} v_i \le \sum_{i=1}^n a_{ik} v_k = v_k,$$ and equality implies that all $v_i$ equal $v_k.$

Now that we know this, let $v$ be an eigenvector of $A$. It's enough to show that $|v|$, whose components are the absolute values $|v_k|$, is also an eigenvector.

I will use the notation $v \le w$ and $v < w$ to mean that $v_k \le w_k$ or $v_k < w_k$ for all $k$.

Define $y := A|v| - |v| \ge |Av| - |v| = 0.$ Assume that $y \ne 0$; then $Ay$ and $A|v|$ both have strictly positive elements, so there is $\varepsilon > 0$ with $Ay > \varepsilon A |v|.$ Then $$Ay = A (A|v| - |v|) > \varepsilon A|v|$$ implies $$A^2 |v| = A (A|v| - |v|) + A|v| > (1+\varepsilon) A |v|.$$ By taking the $\ell^1$-norm (i.e. sum of elements), it follows that $$\|A |v|\|_1 = \|A\|_1 \cdot \|A |v|\|_1 \ge \|A^2 |v|\|_1 > \|(1+\varepsilon) A |v|\|_1 = (1+\varepsilon) \|A|v|\|_1.$$ This is a contradiction. So $y = 0$ and $|v|$ was already our eigenvector for $1$.