Nonnengative matrices sum and produt

Question

I can proof this situation on this way? If $A$ and $В$ are irreducible, then $AB$ is irreducible.

Proof: hyphothese: $A\geq0, B\geq 0$ are irredicible Thesis: $AB$ is irreducible.

If $A\geq0$ is irreducible, we have $\left(I+A\right)^{n-1}>0$, so $\left(I+A\right)^{n-1}A>0$

If $B\geq0$ is irreducible, we have $\left(I+B\right)^{n-1}>0$, so $\left(I+B\right)^{n-1}B>0$

So we have $\left(I+A\right)^{n-1}A\left(I+B\right)^{n-1}B>0$ if A and $\left(I+B\right)^{n-1}$ comute we have

$\left(I+A\right)^{n-1}\left(I+B\right)^{n-1}AB>0\Rightarrow \left(I+A+B+AB\right)^{n-1}AB>0$. How i conclude that $AB$ is irreducible?

Answer 1

The statement is false. For instance, if $A\ne I$ is a circulant permutation matrix of size $\ge2$, such as $$ A=\pmatrix{0&1\\ &\ddots&\ddots\\ &&0&1\\ 1&&&0} $$ and $B=A^T$, then $A$ and $B$ are irreducible but $AB=I$ is reducible.

In fact, $AB$ can be irreducible even if both $A$ and $B$ are primitive. The following rocket-shaped matrix was given as an example in my answer to the question Product of a primitive matrix and its transpose.

Let $$ A=B^T=\pmatrix{0&1&1&0\\ 0&1&1&0\\ 1&0&0&1\\ 1&0&0&1\\}. $$ Then $A$ and $B$ are primitive (hence irreducible) because $A^2$ is the all-one matrix. However, $$ AB=AA^T=\pmatrix{2&2&0&0\\ 2&2&0&0\\ 0&0&2&2\\ 0&0&2&2} $$ is reducible.

Answer 2

If $A$ and $B$ are irreducible then $AB$ is irreducible.

H: $A\ geq 0$, $B\geq 0$ are irreducible

T: $AB$ is irreducible

\fbox{Demo:} If $B$ is irreducible then there is a $r>0$ that is the maximum eigenvalue to which corresponds a $ x> 0 $ that a strictly positive maximum vector, that is:

$$Bx = rx$$, Multiplying right by $A\geq 0$ we have to

$ ABx =Arx$, but since $r$ is a number, then we can organize it as follows: $ABx=rAx$, $A\geq 0$ then multiplying by $x>0$ results in another positive vector which by the way is a maximum eigenvector of $AB$ corresponding to the maximum eigenvalue $r$, therefore $AB$ is irreducible.

I´m I wrong?