I wonder what is really a reducible matrix?

Question

Let A = \begin{bmatrix} 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0 \end{bmatrix} According to Matrix Analysis 2nd edition Definition 6.2.21, this is a reducible matrix if I exchange column 1 and 3.
However, obviously $A$ is irreducible as $(A+I)^{n-1}>0$, and $A$ represents a strongly connected graph too.
I read this question: Reducible matrices and strongly connected graphs. In the answer @thanasissdr mentioned that $B$ and $D$ must be square blocks. But my $A$ above is a $4\times4$ matrix and the result $B$ and $D$ are exactly square.
So this is what I'm confused: whether my $A$ is reducible or not?

Answer 1

A matrix $M$ is reducible if there is a permutation matrix $P$ so that

$$PMP^T=\begin{bmatrix}B&C\\0&D\end{bmatrix}$$

where $B$ and $D$ are square.

On the other hand, $M$ is partly decomposable if there are permutation matrices $P$ and $Q$ so that

$$PMQ=\begin{bmatrix}B&C\\0&D\end{bmatrix}$$

where $B$ and $D$ are square.

The matrix

$$A = \begin{bmatrix} 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\\ 1& 0& 0& 0 \end{bmatrix}$$

happens to be partly decomposable, but not reducible.

You give a permutation $P^T$ of the columns so that $AP^T$ is of the form $\begin{bmatrix}B&C\\0&D\end{bmatrix}$, but for reducibility we need a simultaneous permutation of the rows and columns; $PAP^T$ must be of this form.