TITLE: Every primitive matrix is irreducible?
QUESTION [6 upvotes]: $A$ is reducible if there is some permutation matrix $P$ such that
$$ PAP^T = 
 \begin{bmatrix}
B & C \\ 
O & D \\
 \end{bmatrix}
$$ 
And, if $A^k > O$ for some k, then $A$ is called primitive.
Then, how can I show that every primitive matrix is irreducible?

REPLY [8 votes]: An easier way (IMO) to understand that every non-negative primitive matrix $A$ is also irreducible is by simply examining the definitions (some of them might be considered as alternative - but equivalent - definitions):
Let $A$ be a non-negative matrix.

$A$ is primitive if the following holds:
$\exists k \forall (i,j) : A^k_{ij} > 0$
$A$ is irreducible if the following holds:
$\forall (i,j) \exists k : A^k_{ij} > 0$

Notice the slight difference between the 2 definitions - for (2) it means that every pair $(i,j)$ has its own value of $k$ (or perhaps it's better to denote it as a function: $k(i,j)$), while in (1) it's the same value of $k$ for all the $(i,j)$ pairs. Thus, (1) is clearly a particular case of (2), and therefore, if a non-negative matrix is primitive, then it's also irreducible.