The definition I am thinking of is
A nonnegative matrix $A$ is said to be reducible if there exists a aprtition of the index set $\{1,2,\dots,n\}$ into nonempty disjoint sets $I_1,I_2$ such that $a_{ij} =0$ whenever $i\in I_1$ and $j\in I_2$
I am wondering if anyone has some intuition on what such matrices look like. Or perhaps some intuition on the structure this puts on matrices.
To me it seems like a reducible matrix is one that has to have a certain number of zeros (at least $(n-1)$ zeros?) and is such that these zeros are, in perhaps a loose sense, asymmetrically located in the matrix. (because if $a_{ij}=0$ then it cant be that $a_{ji}=0$ because then $j\in I_2$ and $j\in I_i$ violating disjointness)
But the above way of thinking about reducible seems too literal.
I found this blog post about reducibility That talks about reducibility in terms of having an invariant non-trivial subspace. I see why this is useful but I don't see how to link this definition to the one I provided
It means the submatrix $A(I_1,I_2)$ is zero. Alternatively, $A$ is permutationally similar to a block triangular matrix, i.e. $P^{-1}AP=\pmatrix{B&\ast\\ 0&C}$, where $B,C$ are non-empty square matrices of possibly different sizes. (More specifically, $P$ is the permutation matrix such that $P(I_2,\{1,2,\ldots,|I_2|\})$ and $P(I_1,\{(|I_2|+1),\ldots,n\})$ are identity matrices of sizes $|I_2|$ and $|I_1|$ respectively.)
As for the blog entry in the link, at the time of writing, I don't think the author's definition in terms of invariant subspace is correct. For example, every positive matrix $A$ has a positive eigenvector $v$ corresponding to the eigenvalue $\rho(A)$. Hence the linear span of $v$ is always a proper invariant subspace of $\mathbb R^n$. But $A$ is clearly not reducible because it hasn't any zero entry. Since reducibility is a basis-dependent concept (it is not preserved under similarity transform), the author's definition just does not sound right.
By the way, if I remember correctly, $A$ is not required to be nonnegative in Frobenius' original definition, but nowadays, in the context of Perron-Frobenius theorem, nonnegativeness of $A$ is commonly accepted as part of the definition.