Uniqueness in the Perron–Frobenius Theorem I'm working through proving the basics of Perron's theorem, but I'm stuck on uniqueness for the positive eigenvector.
Given a positive square matrix $A$, I see how to use Brouwer's fixed point theorem to show the existence of a positive eigenvector $\mathbf{v}$. Call its eigenvalue $\lambda$.
Clearly any constant multiple $c \mathbf{v}$ is also an eigenvector with eigenvalue $\lambda$. And following these notes I think I see why any other positive eigenvector $\mathbf{w}$ must have the same eigenvalue $\lambda$. But how do we know $\mathbf{w} = c \mathbf{v}$? Why can't $\mathbf{w}$ be independent of $\mathbf{v}$ despite having the same eigenvalue?
This seems to be Corollary 1 in the linked notes, but the proof there has me stumped. (The definition of what gives us $\mathbf{u} = (1, 1, \ldots, 1)$? How could we possibly know this about $\mathbf{u}$? And how does that tell us $\mathbf{w} = c \mathbf{v}$?)
 A: I found a nice solution here.
Let $\mathbf{v}$ and $\mathbf{w}$ be positive eigenvectors of a positive matrix $A$, both associated with eigenvalue $\lambda$. And suppose for reductio that they are linearly independent.
Because they're independent, we can find a constant $c$ so that $\mathbf{v}-c\mathbf{w}$ is nonnegative and nonzero, but has at least one zero entry. We just start with $c=0$, and increase until $cw_i = v_i$ for some $i$.
Now observe that
$$
  \begin{aligned}
    \mathbf{v}-c\mathbf{w} 
      &= \frac{A}{\lambda} \mathbf{v} - \frac{A}{\lambda} c\mathbf{w}\\
      &= \frac{A}{\lambda} \left(\mathbf{v}- c\mathbf{w}\right).
  \end{aligned}
$$
But $A$ is positive, so $A/\lambda$ is positive, which means $(A/\lambda) (\mathbf{v}- c\mathbf{w})$ must be positive too (recall $\mathbf{v}- c\mathbf{w}$ was nonnegative and nonzero). This contradicts our assumption that $\mathbf{v}- c\mathbf{w}$ had at least one zero. So $\mathbf{v}$ and $\mathbf{w}$ could not have been linearly independent.
(Now I'm just stuck on showing that $\lambda$ must be the leading eigenvalue, i.e. $\lambda > |\lambda'|$ for every other eigenvalue $\lambda'$ of $A$. There's a short proof in Section 10.3 of Strang's Introduction to Linear Algebra, but I can't follow it. If anybody can explain it, or has another nice solution, feel free to combine it with the above solution to this problem and I'll accept their answer as the solution.)
