Nondegeneracy of dominant singular value and positivity of dominant singular vector of connected nonnegative matrix

Question

Call a (not necessarily square) nonnegative matrix $M$ connected if there do not exist permutation matrices $P$ and $Q$ such that $PMQ=\begin{pmatrix}A&0\\0&B\end{pmatrix}$ for some $A$ and $B$. [Perhaps this property already has a standard name.]

I'm looking for a proof of the following (assuming it's true, as I believe):

If a matrix is (nonnegative and) connected, then

1. its dominant (greatest) singular value $s$ is not degenerate (i.e. it has unique left and right singular vectors),
2. the left and right singular vectors of $s$ are positive (not just nonnegative).

My guess is that this is probably relatively straightforward, or even follows directly from some standard results. However, I'm a combinatorialist with limited linear algebra, so would appreciate some help.

The context is this paper by Albert & Vatter.

Answer 1

One can check (details on this to be given later) that your condition on $M$ implies that the symmetric nonnegative matrix $MM^T$ is not of the form $P^T\begin{pmatrix} N_1&0\\ 0&N_2\end{pmatrix}P$ for any permutation matrix $P$. So, the matrix $MM^T$ is primitive: it is non-negative and, for some natural number $m$, all entries of the $m$th power of $MM^T$ are $>0$.

So, your desired result for $MM^T$ follows by the Perron–Frobenius theorem. Similarly, for the matrix $M^T M$.

Answer 2

Consider $M$ to be the weighted biadjacency matrix of a bipartite graph $G$. If $M$ is connected then so is $G$. (This was the motivation for the terminology.)

$MM^T$ and $M^TM$ are then symmetric weighted adjacency matrices of the projections $G_1$ and $G_2$ of $G$ onto its two vertex sets (in which nodes are adjacent if they share a neighbour in $G$). If $G$ is connected, then so are $G_1$ and $G_2$. Thus $MM^T$ and $M^TM$ are both irreducible.