If $A \in M\_n(\mathbb{C})$, then $A$ is called *reducible* if there is a permuation matrix $P$ such that 
$$ 
P^\top A P =
\begin{bmatrix}
A\_{11} & A\_{12} \\
0 & A\_{22}
\end{bmatrix}, $$
in which $A\_{11}$ and $A\_{22}$ are square matrices of order at least one. If $A$ is not reducible, then $A$ is called *irreducible*. Notice that, with this definition, every one-by-one matrix is irreducible. A matrix $A$ is irreducible if and only if its *directed graph* or *digraph* is *strongly connected*. 


It is known (see, e.g., Brualdi and Ryser [Theorem 3.2.4; MR1130611]) that if $A \in M\_n(\mathbb{C})$, then $A$ is irreducible or there is a permutation matrix $P$ such that 
\begin{equation}
P^\top A P =
\begin{bmatrix}
A\_{11} & \cdots & A\_{1k} \\
 & \ddots & \vdots \\
 & & A\_{kk}
\end{bmatrix}, \tag{1} \label{fnf}
\end{equation}
in which the matrices $A\_{11},\dots, A\_{kk}$ are square, irreducible matrices. The matrix in \eqref{fnf} is called a *Frobenius* or *irreducible normal form (of $A$)* (FNF) and it is not unique. However, the blocks $A\_{11},\dots, A\_{kk}$ are unique up to permutation similarity. 


A matrix $A$ is nonnegative ($A \ge 0$) if $a\_{ij} \ge 0$, $1 \le i,j \le n$. A nonnegative matrix is *stochastic* if $Ae = e$ and *doubly stochastic* if, in addition, $A^\top e = e$ (i.e., $A$ has row and column sums equal to one). 


In a 1965 paper, Perfect and Mirsky [MR0175917] state, without proof, that if $A$ is a doubly stochastic matrix, then every FNF of $A$ is of the form 
$$
\begin{bmatrix}
A\_{11} & & \\
 & \ddots & \\
 & & A\_{kk}
\end{bmatrix},
$$
i.e., every doubly stochastic matrix is either irreducible or is permutationally similar to a direct sum of irreducible, doubly stochastic matrices. After giving the result Perfect and Mirsky state: 



> 
> "This result is almost certainly well-known. As, furthermore, it
>  follows very easily from the definitions, we omit the details of the
>  proof." – p. 38. 
> 
> 
> 


This result is easy to establish for matrices of order less than or equal to four, but does not seem obvious or seem to follow simply from the definitions above. 


**Question 1:** Is this result obvious or is there a simple proof of this fact?


**Question 2:** Does anybody know of a reference? 



Frobenius normal form of a doubly stochastic matrix
The matrix $P^\top AP$ is doubly stochastic. Suppose that $A\_{11}$ is $m\_{1} \times m\_{1}.$ Note that the sum of the entries of each column of $A\_{11}$ is $1$,and the sum of entries in each row of $A\_{11}$ is at most $1$ (using non-negativity of all entries of $P^\top AP)$. Notice that the sum of all entries of $A\_{11}$ is $m\_{1}$. Hence, the sum of the entries of each row of $A\_{11}$ is in fact $1$ (if any of the row sums were strictly less than $1$, then the sum of the row-sums, which equals the sum of the entries of $A\_{11}$, would be strictly less than $m\_1$, a contradiction), i.e., $A\_{11}$ is doubly stochastic. Thus, $A\_{12},A\_{13},\ldots,A\_{1k}$ are all zero. Likewise, $A\_{22}$ must be doubly stochastic and $A\_{23}, \ldots , A\_{2k}$ must all be zero, and so on.


