Let $A$ be a matrix of order $n$. Then $A$ is irreducible if and only if its digraph $D$ is strongly connected.
What I think is, any 'Symmetric q-Diagonal Matrix' must be irreducible.
Here I denote 'Symmetric q-Diagonal Matrix', as an example, a symmetric tri-diagonal matrix where it has 3-consecutive non-zero elements in each row centered at diagonal.
$\left[\begin{array}{ccc} +&a&0&0&0\\ a&+&b&0&0\\ 0&b&+&c&0\\ 0&0&c&+&d\\ 0&0&0&d&+ \end{array}\right] $
shows this example. {a, b, c, d} each of them are non-positive for the following discussion (otherwise we wouldn't need specific signs). Then, its digraph will be always strongly connected.
I was thinking of this because there is a very interesting case in Matrix Inverse (though I don't have proof for this. It appeared in a lecture).
If matrix B is
[1] irreducible, [2] symmetric, [3] strictly diagonally dominant
[4] $b_{ii} > 0$, [5] $b_{ij} (i \neq j) \leq 0$, then Its inverse Q has all the elements where $q_{ij}>0(\forall i,j)$.
[2], [3], [4] -> gives P.D. matrix, and it at least tells it is invertible. Then, how would you continue?
So please confirm the first argument about irreducibility, and if possible let's discuss the 2nd case where all the elements of inverse become strictly positive (magic).
According to this if one can show that a q-diagonal matrix is strongly connected, we are done.
A strongly connected graph is on which any node is reachable from any node.
Consider a tri-diagonal $n\times n$ matrix $A$, as long as all the entries in the diagonal and the two offsets are non-zero, every node $i$ except the first and the last are connected to $i-1$, $i$, and $i+1$. The first node is connected to the second, and the last node is connected to the one before.
Therefore to go to any node $j$ from node %i% such that $j\gt i$, take path $i$, $i+1$, ..., $j$. Similar logic works for $j\lt i$.
Hence A is strongly connected and therefore irreducible.
We showed irreducibility for a tri-diagonal graph, $q=3$. For any $q\gt 3$, there will be extra edges, but not fewer. This doesn't affect the property. So, any q-diagonal matrix is irreducible for $q\gt 2$.