Relationship between total support and fully indecomposability. I have to prove that if a nonnegative matrix $A$ has total support then there exist $P,Q$ permutation matrix such that $$PAQ=\bigoplus_{i=1}^k A_i$$ where $A_i$ is fully indecomposable $\forall i=1,\dots, k$ and the $\bigoplus$ symbol means "direct sum" of matrices.
I found many reference on this theorem but no one is complete.
Could you help me?
Some papers I found cite  , but I cannot find a link between the two ones.
 A: Let $A$ be a non-negative $n\times n$ matrix. 
If $\sigma$ is a permutation of $\{1,\ldots,n\}$ then the entries $a_{1\sigma(1)},\ldots,a_{n,\sigma(n)}$ are called a diagonal of $A$.
A diagonal of $A$ is positive if it does not contain any zero entry.
A non-negative matrix $A$ has total support if every $a_{ij}\neq0$ belongs to some positive diagonal of A. 
So $A$ does not have total support if and only if there is $a_{ij}\neq0$ such that the submatrix obtained by deleting row $i$ and column $j$ has no positive diagonal.
Frobenius-König theorem says that $A$ does not contain a positive diagonal (its permanent is zero) if and only if there is a submatrix of A occupying s rows and t columns, which is identically zero, such that $s+t>n$. 
Now, assume that $A$ is not fully indecomposable, but has total support. Then there are permutation matrices $P,Q$ such that $PAQ=\begin{pmatrix}X_{s\times s} &Y_{s\times n-s}\\
0_{n-s\times s}&Z_{n-s\times n-s}\end{pmatrix}$.
If $Y_{ij}\neq 0$ then deleting its row and column we obtain a submatrix $M$ of $PAQ$ of order $n-1$. Notice that $O_{n-s\times s}$ still is a submatrix of $M$ and $n-s+s=n>n-1$. Hence, $M$ does not contain a positive diagonal and $PAQ$ does not have total support. Absurd! So $Y_{s\times n-s}$ must be identically zero.
Now, use induction on $X$ and $Z$. 
