Every primitive matrix is irreducible? $A$ 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?
 A: 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. 
A: Your definition of primitive matrix is wrong. A primitive matrix is a nonnegative matrix $A$ such that $A^k>0$ for some natural integer $k$. You cannot remove the nonnegativity requirement on $A$ and the positivity requirement on $k$.
If $A$ is reducible, then it is, by definition, permutation-similar to a block upper triangular matrix $M$. Since $M^k$ is block upper triangular for all $k\in\mathbb{N}$, it always contains a zero block. Yet $M^k$ is permutation-similar to $A^k$. So, $A^k$ always contains a zero entry and hence it is not primitive.
A: For n ≥ 2, an n × n non-negative matrix $A$ is
irreducible if $\sum_{k=0}^{n-1}A_k$ is positive matrix.
For n ≥ 2, an n × n non-negative matrix A is
primitive if there exists $k\in N$ such that  $A_k$ is positive matrix.
According to the definition, we can get example:
$A=[0,1;1,0]$, with $spec(A)=\{1,-1\}$ and eigenvector are $[1,0]^T, [0,1]^T$. So diagonalization is :
$\left[ \begin{matrix} 1&0\\0&-1 \end{matrix} \right]=D=Q^{-1}AQ$
Look at this! the sum of A^k (from k=1 to n-1) is positive definitely. But does it primitive? No$\lim_{k\rightarrow \infty}A^k=\lim_{k\rightarrow \infty}QD^kQ^{-1}$, but A is primitive if the following holds: $∃k,∀(i,j):A_{ij}^k>0$ . Obviously can we say A is primitive when k approaching infty?
So the irreducible matrix may not be the primitive matrix.
