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If a matrix is irreducible, based on the following definition

A matrix is reducible if there are two disjoint sets of indexes $I,J$ with $|I|=\mu$, $|J|=\nu$, $\mu+\nu=n$ such that for every $(i,j)\in I\times J$ we have $a_{ij}=0$

then is this equivalent to saying that the matrix graph is connected?

I found this result but it seems it is only one direction.

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jonem
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    ...that the graph is strongly connected (since we talk about directed graphs). – Artem Jul 11 '14 at 23:05
  • @Artem Thank you. If the matrix is symmetric does this simplify things? Does symmetry cause this to reduce to connectivity? – jonem Jul 11 '14 at 23:08
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    Yes, if the matrix is symmetric we can talk about simply about connectivity. – Artem Jul 11 '14 at 23:16

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For an undirected graph $G$: for each partition $\{I,J\}$ of the vertices of $G$ we check if there are no edges between vertices in $I$ and vertices in $J$.

If $G$ is disconnected, then $\{I,J\}$ can be found (choose $I$ as the vertices in one connected component). Conversely, if $\{I,J\}$ can be found, then there are no edges between $I$ and $J$ and $G$ is disconnected.

For the adjacency matrix $A=(a_{ij})$, if $\{I,J\}$ can be found then subgraph formed by the rows indexed by $I$ and the columns indexed by $J$ form an all-$0$ submatrix.

This means that, $A$ has the block structure $$\begin{array}{|c|c|} \hline A_I & 0 \\ \hline 0 & A_J \\ \hline \end{array}$$ where the rows and columns are indexed by the indices in $I$ then $J$.

In the directed case, we instead have no edges directed from a vertex in $I$ to a vertex in $J$. So we can find $\{I,J\}$ if $G$ is not strongly connected (take $I$ to be the vertices in one strongly connected component).

Rebecca J. Stones
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