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I have tried to prove that if a matrix $A\in\mathbb{C}^{n\times n}$ is such that there are no two sets $I,J\subseteq\{1,\dots,n\}$ that are disjoint, complementary, nonempty, and such that for all $(i,j)\in I\times J$ we have $a_{ij}=0$, $a_{ij}$ being the $ij$ entry of the matrix $A$, then the associate graph, as defined here, is strongly connected. The hypothesis will be called n-DISC (not Disjoint Index Set Condition) and the strong connectedness of the graph will be referred to as SCG in the proof. I have managed to give the following proof:

I have proved it in special cases, and then assumed the same line of thought can be used for any size, but the proof is anyway long and very unelegant, so any suggestion of a more elegant proof is very welcome indeed. Instead of proving n-SCG implies DISC, I tried to prove that n-DISC implies SCG, because trying the former didn't seem to lead me anywhere, while the latter gave me something that seemed to be OK for any case. Let's start with $n=2$. Then we have a $2\times2$ matrix and 2 nodes. Let's take one. it can't be cut off from the other one, otherwise we would have the DISC. So from this node, be it $A$, we can reach the other one, be it $B$. The same argument on $B$ proves we can reach $A$ from $B$, and therefore the graph is SC. Let's go on with $n=3$. We start at one node, be it $A$. From it, as said above, we can reach at least one other node, be it $B$. From $B$ we can reach at least another node. Now one is tempted to say it is the last node, $C$, but of course not. So we have to cases: either $A\to B\to C$, or $A\leftrightarrow B$. In the former case, from $C$ we can reach at least one node. If it is $A$, we have concluded. If it is $B$, i.e. $A\to B\leftrightarrow C$ then from either $B$ or $C$ we can reach $A$, otherwise the DISC is violated by $B,C$ opposed to $A$. If $A\to B\leftrightarrow C\to A$, we have concluded, and if $A\leftrightarrow B\leftrightarrow C$, we have also concluded. However, we still have the latter case: $A\leftrightarrow B$. Of course, from one of those we must reach $C$. So $A\leftrightarrow B\to C$ or $C\leftarrow A\leftrightarrow B$. In the former case, we anyway have $A\to B\to C$, and we have shown before that this case leads to a positive conclusion. In the latter case, from $C$ we can reach either $B$ or $A$, otherwise we violate the supposed DISC. If $C\to B$, then $A\to C\to B\to A$, which concludes the proof. If $C\to A$, then $C\leftrightarrow A\leftrightarrow B$, which also concludes the proof. I tried it for $n=4$, but got bored in the process, concluded a positive case and left many uninvestigated cases.

Now I fully well understand this proof is horribly complex and incredibly unelegant. This is why I'm asking if there is a simpler and more elegant proof for this statement. Naturally, this is only half of the theorem, which states the equivalence of n-DISC with SCG, i.e. the equivalence of the irreducibility of $A$ with the strong connectedness of its associated graph. Can someone help me here?

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marked as duplicate by ml0105, mau, apnorton, Davide Giraudo, mookid May 4 '14 at 18:27

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  • $\begingroup$ I suggest that what you have written out can be recast as the induction step of an induction argument, inducting on the size $n$ of the matrix. Perhaps casting it as an induction argument will also simplify the proof. $\endgroup$ – Lee Mosher May 4 '14 at 16:39
  • $\begingroup$ A trivial remark: you should add the condition that $I$ and $J$ are non-empty. $\endgroup$ – Andreas Caranti May 4 '14 at 16:54
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First translate your n-DISC criterion into a statement about directed graphs, then prove that condition is equivalent to being strongly connected.

In the language of directed graphs n-DISC just means that there is no partition of the vertices into sets $I,J$ such that no edges go from $I$ to $J$.

Clearly if such a partition exists then there is no path from any vertex in $I$ to a vertex in $J$, and the graph can't be strongly connected. Hence SCG $\implies $n-DISC.

Now suppose you have the n-disc property, choose a vertex $v$ in your directed graph. Let $I$ be the set of vertices that $v$ has a directed path to (including $v$), and let $J$ be the set of vertices which $v$ doesn't have a directed path to. By construction there can be no edges from $I$ to $J$. But then the n-DISC property tells us that this can only happen if one of $I$ or $J$ is empty, so $J$ must be empty as $I$ contains $v$. Hence there exists a directed path from $v$ to any vertex in the graph, this holds for all choices of $v$ so the graph is strongly connected. So n-DISC $\implies$ SCG.

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Suppose you cannot go from a vertex $x$ to a vertex $y$.

Consider the set $J$ of all vertices from which you can reach $y$, $$ J = \{ j \in \{1, 2, \dots, n\} : \text{there is a directed path from $j$ to $y$}\} $$ As $y \in J$, but $x \notin J$, you have that $J$ is a non-empty, proper subset.

Consider $I = \{1, 2, \dots, n\} \setminus J$, so that $I$ is also non-empty and proper.

Clearly the sets $I$ and $J$ are disjoint and complementary, and then the condition $a_{ij} = 0$ for $i \in I$ and $j \in J$ holds. This is because if $a_{ij} \ne 0$ for some $i \in I$, so that $i \notin J$, and $j \in J$, then there is a directed edge from $i \in I$ to $j \in J$, and thus there is a directed path from $i$ to $y$, so that $i \in J$ by the definition of $J$, a contradiction.

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