# Matrix graph and irreducibility

How do I prove that if $A\in\mathbb C^{n\times n}$ is a matrix then it is irreducible if and only if its associated graph (defined as at Graph of a matrix) is strongly connected?

Update:

Seeing as no-one answered for over a week, I tried to do it by myself.

1. The first thing I did was try to show column or row permutation didn't change the strong connectedness of the graph. I didn't manage, and actually proved the opposite. On the way, though, I managed to show transposition doesn't. The argument is that transposition affects the graph in that it inverts all the arrows, but if there is a loop through all nodes then inverting the arrows means you go through it the other way round, so it's still there, and the graph stays strongly connectedness, and if there isn't, well, transposing can't make one appear, as otherwise transposing back would make it disappear, which we have proved impossible. This result may not be useful, but since I've done it I thought I might well write it down.
2. Then I tried to think of what permuting both rows and columns does to the graph. Why that? Let's recall the notion of irreducible matrix:

A matrix $A\in\mathbb{C}^{n\times n}$ is said to be reducible if there exists a permutation matrix $\Pi$ for which $\Pi\cdot A\cdot > \Pi^T$ is a block upper triangular matrix, i.e. has a block of zeroes in the bottom-left corner.

So if this operation does not alter the graph's strong connectedness, then I can work on the reduced matrix to show its graph is not strongly connected and prove one implication. Now such multiplications as in the definition of a reducible matrix, with $\Pi$ a matrix that swaps line $i$ with line $j$ - what do they do to the graph? Swapping the lines makes all arrows that go out of $i$ go out of $j$ and viceversa; swapping the columns does the same for arrows leaving $i$ (or $j$). So imagine we have a loop. Say it starts from a node other than $i$ and $j$. At a certain point it reaches, say, $i$. Before that, everything is unchanged. When the original loop reaches $i$, the new loop will reach $j$ and go out of it to the same node as it went out from $i$ to before the permutation, if that node wasn't $j$, in which case it will go to $i$. When the original loop enters $j$, the new loop enters $i$, and same as before. So basically the result is just that $i$ and $j$ swap names, and the loop is the same as before taking the name swap into account. So this kind of operations do not alter the strong connectedness of the graph.

3. Suppose $A$ is as follows: $$A=\left(\begin{array}{c|c}\Huge{A_{11}} & \Huge{A_{12}} \\\hline \Huge{0} & \Huge{A_{22}}\end{array}\right).$$ Suppose the $\Huge{0}$ is $m\times m$ with $m\geq\frac{n}{2}$. Then we have $m$ nodes that are unconnected to other $m$ nodes, going out. But we don't have $2m$ nodes, or have exactly that many, so those $m$ nodes are cut off from all the other $n-m$, going out. So suppose there is a loop. If it starts at one of the $m$ nodes, it can never reach the other $n-m$, and if it starts at one of those, it can reach the $m$ nodes but never get back, so maybe we have a path through all the nodes, but it can't be a loop, i.e. a closed path. So the graph is not strongly connected.

4. Now the definition doesn't say anything about the size of those blocks, so the problem I still have is that if $m<\frac{n}{2}$, the argument above fails because we have at least one node besides the $m$ nodes and the other $m$ nodes that can't be reached from the first $m$, and that node could be the missing link. Of course, when I said "can't be reached" up till now, I meant "be reached directly", i.e. not passing through other nodes.
5. Of course, if the above is concluded, I have proved that reducibility implies non-strong-connectedness of the graph, so that a strongly connected graph implies irreducibility. But the converse I haven't even tried.

So the questions are: how do I finish the above at points 3-4 and how do I prove the converse? Or maybe I'm missing something in the definition, in which case what is it?

Update 2: I think I am missing something, as a $3\times3$ matrix with a 0 in the bottom-left corner and no other zeroes does have a strongly connected graph, since the only missing arrow is $3\to1$, but we have the loop $1\to3\to2\to1$. So when is a matrix reducible?

Update 3: Browsing the web, I have found some things. I bumped first into this link. Now if that is the definition of reducible matrix, then either I misunderstand the definition of the block triangular form, or the if and only if there doesn't hold, since a matrix with a square block of zeroes bottom-left definitely doesn't satisfy the disjoint set condition but definitely does satisfy the permutation condition, with no permutation at all. Maybe the first condition is equivalent to the non-strong-connection of the graph. Yes, because in that case there are $\mu$ nodes from which you can't reach the remaining $\nu$, so the graph is not strongly connected. So at least that condition implies the non-strong-connectedness of the graph. The converse seems a bit trickier. Looking for that definition, I bumped into this link. Note that no matrix there has a lone corner zero (which would be top-right as the link deals with lower triangular matrixes), and all of them satisfy the disjoint set condition in the link above. So what is the definition of block triangular matrix? If it is that there must be square blocks whose diagonal coincides with part of the original matrix's diagonal and below them there must only be zeroes, then I have finished, since the if and only if in the link above is valid, so reducibility implies non-strong-connectedness of the graph, and whoops, I'm not done yet, I still need the converse, so can someone finally come help me on that? And if it isn't, then what the bleep is it and how do I make this damn proof?

• Has anybody seen this? – MickG Apr 19 '14 at 13:18
• @andreascanceido have you seen this? Any ideas? – MickG Apr 20 '14 at 20:23
• Can someone at least solve my markup problems???? – MickG Apr 23 '14 at 17:42
• Reading this question I asked on meta, and the link in the link there, it seems that multirow is unusable here for now, and trying to define the macros for multirow and multicolumn spanning yields an \fbox-like block of code and turns the rendering of the matrix into the same kind of thing, so definitely no. I'll have to resign to mere \Huge entries without multispanning and for block matrixes I'll have to do them in LaTeX and add them as pictures, hoping NEVER to find myself needing such a matrix in a comment. – MickG Apr 25 '14 at 14:00
• I don't know to prove it, but I think you might find the proof in "Combinatorial Matrix Theory" of Brualdi – dineshdileep Apr 25 '14 at 15:34

First of all, let me recall the definition I've been given in Uni of reducible matrix:

A matrix $A\in\mathbb{C}^{n\times n}$ is reducible if there exists a permutation matrix $\Pi$ such that $\Pi\cdot A\cdot\Pi^T$ is in block upper triangular form. In other words, if it can be placed into such a form by simultaneous row and column permutation, to use Wolfram's phrasing.

However, Wolfram (compare here) has a different opinion. The definition given there is:

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 $a_{ij}=0$.

Or the same thing phrased differently. Wolfram then states the equivalence of these definitions and the equivalence of the Disjoint Index Set Condition (the condition in Wolfram's definition, henceforth DISC) with the opposite of the strong connected graph (henceforth SCG). My definition of reducible matrix, i.e. the Upper Triangular Permutation Condition, will henceforth be referred to as UTPC, and their opposites as n-SCG, n-UTPC, n-DISC, with n for not. Now one problem I had was the definition of block UT, which I defined as:

A matrix is block UT if it has a block of zeroes in the bottom-left corner and non-zero blocks elsewhere.

With such a definition, a full matrix with a lone zero in the bottom-left corner would be block UT, but of course wouldn't satisfy the DISC, so I had to review my definition. I came to think it might be:

A matrix is block UT if it has non-zero blocks whose diagonals coincide with pieces of the matrix's diagonal and cover it all, and below those blocks there are only zeroes.

I'll try to add a picture to visually represent this. Right now I have no time, I hope not to forget about this. Assuming that this is the definition, it is quite evident that block UT form implies the DISC, and vice versa. So if I prove that permutations as in the UTPC do not alter the satisfying of the DISC, then I have proved that DISC and UTPC are equivalent, as Wolfram states. This will be the first step. Subsequently, I will have to prove the equivalence of DISC and SCG. Then, by transitivity of equivalence, I will have answered my question and proved this theorem.

Now, what happens when I apply a simultaneous row and column permutation? Let's restrict ourselves to the simple case of just two rows and the corresponding columns being swapped, as any matrix $\Pi$ can be written as the product of such simple ("elementary") permutation matrixes, and the transpose is the product of the transposes in inverse order, so applying the product matrix is like applying the "elementary" ones one after the other, and if one doesn't alter the DISC, than many subsequent clearly won't. What happens is basically a change of names in the indexes. Let's suppose we swap $i,j$ such that neither is in $I$. Then with the same indexes I satisfy the condition, since $a_{ki}=a_{kj}=0$ for all $k\in I$ implies that after swapping the columns it still works and $a_{ik}$ and $a_{jk}$ we don't care about since $i,j\not\in I$. In any case $a_{ik}=a_{jk}=0$ for all $k\in J$ stays true after the row and column swapping, so if $i,j\in I$ the DISC stays unchanged. If $i\in I$ and $j\in J$, then for $\Pi\cdot A\cdot\Pi$ we will have $i\in J$ and $j\in I$. That is because before swapping, we had $a_{ik}=0$ for all $k\in J$, $j$ being included, and $a_{kj}=0$ for all $k\in I$, $i$ therein included; swapping the columns makes $a_{jk}=0$ for all $k\in J$. Then swapping the columns means the zero in the place $a_{ij}$, brought by the row swapping into place $a_{jj}$, get to place $a_{ji}$. So if we move $j$ to $I$, the DISC is satisfied if and only if we can prove $i$ can be placed in $J$. But then the column swapping ensures that, by the same line of thought as this of the row swapping for $j$ moving to $I$. So the proof is concluded.

Proving the equivalence of DISC and n-SCG seems easier than proving UTPC$\iff$n-SCG, which is why I chose this path. Suppose we have DISC. Then we have $\mu$ nodes associated with the indexes in $I$ from which neither of the remaining $\nu$ nodes associated with the indexes in $J$ can be reached. Then it is definitely impossible to trace a loop through all the nodes in the graph, since at most one can hope to connect all the $\mu$ and $\nu$ nodes separately, and one of the $\nu$ nodes to one of the $\mu$, but not to close the loop by going "from $\mu$ to $\nu$".

The converse is a bit trickier 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. Such a proof is very long, boring and unelegant, and sounds also very nerdy, because of all the cases it considers, especially for $n$ growing. I wonder how the number of cases grows with $n$. Perhaps exponentially or factorially. From 2 to 3 we have seen an enormous change. In any case, I sure hope a more elegant proof exists and someone knows it and posts it here as a second answer. If that doesn't happen, I will isolate this bit and ask another question, and link that question in a second answer here. I will wait at most 12 days before doing so. Of course, marking that as a duplicate when I've had to answer this would sound totally absurd :). Anyway it seems reasonable to assume that the given proofs are error-free and that this last proof, with all its bad qualities, can be extended, with lots of patience, to greater dimensions. I wonder if it is possible to prove it by induction on $n$. I will not think about that, as I have already thought about this problem even too much, since it was a theorem given in Numerical Calculus without demonstration, I won't be asked for its proof in an exam, and this definitely suffices to get en extra point if I'm good enough to get asked the "30L" question, and besides I have many other proofs and exercises to worry about. So my writing here is concluded, save maybe for the other question linked to this one in a possible second answer, if I am forced to ask such a question.

As I had expected, no-one has answered here. So here is the second answer, and the link which - hopefully - will finally conclude all this mess: Proving that irreducibility of a matrix implies strong connectedness of the graph.