In a positive Markov matrix, why is the eigenvector corresponding to λ = 1 non-zero? This is a theorem I've seen in multiple places, but haven't been able to find a proof of. Is it because all of the eigenvectors are non-zero? What's the intuition behind it?
 A: You seem to be confusing a nonzero eigenvector with an entrywise nonzero eigenvector.
Every eigenvector is nonzero by definition, but not every eigenvector is entrywise nonzero (counterexample: consider any diagonal matrix of size $>1$ with distinct diagonal entries).
A positive Markov matrix does possess an entrywise nonzero eigenvector, however. Actually we can do better: every positive Markov matrix $M$ has left and right entrywise positive eigenvectors corresponding to the eigenvalue $1$.
Without loss of generality, suppose all column sums of $M$ are equal to $1$. Then $M$ of course possess an entrywise positive left eigenvector $e^T=(1,1,\ldots,1)$. The remaining question is whether $M$ has an entrywise positive right eigenvector for the eigenvalue $1$. One simple classical proof (as mentioned by Calvin Lin in another answer) of this fact is to note that the function $f(x)=Mx/\|Mx\|_1$ maps the simplex $S=\{x\ge0: \sum_ix_i=1\}$ into itself, and hence by Brouwer's fixed-point theorem, it has a fixed point in $S$. As $x\in S$, it is entrywise nonnegative. However, since $M$ is positive, $f(x)=x$ can only occur only when $x$ is entrywise positive. Finally, as
$$
1=e^Tx=(e^TM)x=e^T(Mx)=e^T(\|Mx\|_1x)=\|Mx\|_1,
$$
we see that $x$ is an eigenvector for the eigenvalue $1$.
There is actually an even stronger result, namely, the celebrated Perron-Frobenius theorem, which implies that the eigenvalue $1$ is also simple, so that $x$ is unique up to scaling. I shall not go into details here, as the complete statement of Perron-Frobenius theorem is quite involved, and you can find it easily in any good reference books with chapters on nonnegative matrices.
A: An eigenvector is non-zero by definition. 
