# Are tridiagonal stochastic matrices irreducible?

According to Wikipedia, a tridiagonal matrix is a band matrix that has nonzero elements only on the main diagonal, the subdiagonal and the supradiagonal. To my understanding, in a tridiagonal stochastic matrix, from state $$i$$, we could always reach another state $$j$$ (imagining transitioning to neighboring states in each step). Therefore, I thought that a tridiagonal stochastic matrix is always irreducible. However, in this paper, it is stated that

A tridiagonal matrix ($$a_{ij}$$) is irreducible if and only if $$a_{ij}a_{ji} \neq 0$$ for all $$j = i + 1$$.

This statement confused me because I thought that $$a_{ij} a_{ji} \neq 0$$ is always true for a tridiagonal matrix. I'm wondering if I'm missing anything here.

• Most of the time, tridiagonal means that the only non-zero entries are in those three diagonals but not that the entries of those diagonals are all non-zero. Oct 30, 2022 at 7:29
• Ah I see. I misread the definition in Wikipedia. Thanks for pointing that out!
– Jack
Oct 30, 2022 at 7:30

As @Mariano Suárez-Álvarez pointed out in the comment section, it is not required in a tridiagonal matrix that all entries of the main diagonal, subdiagonal and the supradiagonal be non-zero, which necessitates the condition that $$a_{ij}a_{ji}\neq 0$$ for a tridiagonal stochastic matrix to be irreducible.