Determinants and the Lindström-Gessel-Vienot Lemma

Solomon is currently a student at the University of Pennsylvania where he is pursuing a Ph.D in Computer Science. He has a keen interest in all things Computer Science and Maths, so he decided to specialize in Programming Languages, an area which applies ideas from diverse areas of Computer Science, Mathematics and Philosophy. One of his favourite things is when connections are made between seemingly disparate areas of inquiry, so he has enjoyed reading the Graphical Linear Algebra blog for the connections between string diagrams and linear algebra. As the Lindström-Gessel-Vienot Lemma shows us, there is also a surprising combinatorial connection. Such occurrences make Solomon very happy!

Consider the familiar diagram for the $2 \times 2$ matrix $\begin{pmatrix} a & b \\ c & d \end{pmatrix}$ :

matrix

Suppose that we wanted to push some flow from the input nodes $X_1$ and $X_2$ to the output nodes $Y_1$ and $Y_2$ . Moreover, we want to do this in such a way that every output node receives some flow without any congestion in the network. More specifically, want to push flow from the input nodes to the output nodes in such a way that:

no two different paths that are part of the flow intersect at
a vertex (not even endpoints), and
for each output node $Y_j$ , there is a path from some input
node $X_i$ to $Y_j$ .

For example, in the diagram above, we can push $a$ units of flow from $X_1$ to $Y_1$ and $d$ units of flow from $X_2$ to $Y_2$ .

flowex1

We can also push $c$ units of flow from $X_1$ to $Y_2$ and $b$ units of flow from $X_2$ to $Y_1$ .

flowex2

We are not allowed to push $a$ units of flow from $X_1$ to $Y_1$ and $b$ units of flow from $X_2$ to $Y_1$ , since then we would have two different paths intersecting at $Y_1$ .

flowcex1

Also, we are not allowed to push only $a$ units of flow from $X_1$ to $Y_1$ , since then there is no path from some $X_i$ to $Y_2$ .

We call a flow that satisfies conditions (1) and (2) a viable flow. Now we will measure the signed weight a viable flow. The weight of a viable flow is simply the product of all weights on the edges that are part of the flow. To get the signed weight, we multiply the result by -1 if the flow links $X_1$ to $Y_2$ . Otherwise, we multiply the result by +1.
For example, in our $2 \times 2$ matrix example, the first viable flow has weight $+ad$ . The unsigned weight of this flow is $ad$ since we push $a$ units of flow from $X_1$ to $Y_1$ and $d$ units of flow from $X_2$ to $Y_2$ . To get the signed weight of this flow, we multiply the unsigned weight $ad$ by +1 since $X_1$ is linked to $Y_1$ . A similar computation shows that the signed weight of the second viable flow is $-bc$ .
Interestingly, when we add signed weights of all the viable flows in the diagram, we get the determinant of the matrix associated with the diagram i.e. $ad-bc$ . Coincidence?

The Lindström-Gessel-Vienot Lemma (LGVL) tells us that this in fact is not a
coincidence. Specifically, LGVL associates to each finite directed acyclic
graph (dag) $G$ a matrix whose determinant is exactly equal to the
sum of the signed weights of the viable flows in $G$ . To make this relationship precise we shall give the statement of LGVL, but first we need to introduce some definitions and notation that we will need in order to give the statement of the theorem.

First, we introduce permutations. For any sets $A=\{a_1, \ldots, a_n\}$ and $B=\{b_1, \ldots, b_n\}$ we call any bijection $\sigma : A \longrightarrow B$ a permutation from $A$ to $B$ . Now assume that $\sigma$ sends $a_1, \ldots, a_n$ to $\sigma(a_1), \ldots, \sigma(a_n)$ respectively. For example, let $A = B = \{1,2,3,4,5\}$ , and let $\sigma$ send $1,2,3,4,5$ to $3,5,2,1,4$ respectively. Write $a_1, \ldots, a_n$ in a row in that order, skip a line, then write $b_i$ below $a_i$ for each $b_i$ . Draw a line from $a_i$ to $\sigma(a_i)$ , and count the number $m$ of crossings in between the ${a_i}'$ s and the ${b_i}'$ s. Define the sign of $\sigma$ be $(-1)^m$ . For the permutation $\sigma$ we have just described, there are six crossings, so the sign of the permutation is $(-1)^6 = 1$ .

The sign of a permutation $\sigma$ can also be computed by raising -1 to the
number $inv(\sigma)$ of inversions of $\sigma$ , where $inv(\sigma)$ is the number of ordered pairs $(i, j)$ such that $i < j$ but $\sigma(i) > \sigma(j)$ . In the example above, the number of inversions of $\sigma$ is 6 = $|\{(1,3), (1,4), (2,3), (2,4), (2,5), (3,4)\}|$ , which gives the same result as before.

For the next few pieces of notation as well as the statement of LGVL, assume that $G$ is a finite directed acyclic graph. Assume further that $G$ has input nodes $X_1, \ldots, X_n$ and output nodes $Y_1, \ldots Y_n$ .

setup

As before, we are interested in viable flows, i.e. flows that have the property that:

no two different paths that are part of the flow intersect at
a vertex (not even endpoints), and
for each output node $Y _j$ , there is a path from some input
node $X_i$ to $Y_j$ .

Each viable flow can described by an $n$ -tuple $P = (P_1, \ldots, P_n)$ of paths , where $P_i$ is a path from the input node $X_i$ to the output node $Y_{\sigma(i)}$ for some permutation $\sigma$ . Let $\sigma(P)$ be this permutation.

For each directed path $P$ between any two vertices in $G$ , let $\omega (P)$ be the product of the weights of the edges of the path. For any two vertices $X_i$ and $Y_j$ , write $e(X_i, Y_j)$ for the sum

$e(X_i,Y_j)=\sum\limits _{P:X_i\to Y_j} \omega (P),$

where $P$ ranges over all paths from $X_i$ to $Y_j$ .
We are finally in a position to give the statement of LGVL, which is that

$\left \vert {\begin{pmatrix}e(X_{1},Y_{1})&e(X_{1},Y_{2})&\cdots &e(X_{1},Y_{n})\\e(X_{2},Y_{1})&e(X_{2},Y_{2})&\cdots &e(X_{2},Y_{n})\\\vdots &\vdots &\ddots &\vdots \\e(X_{n},Y_{1})&e(X_{n},Y_{2})&\cdots &e(X_{n},Y_{n})\end{pmatrix}} \right \vert = \sum\limits _{{P = (P_{1},\ldots ,P_{n})}}{\mathrm {sign}}(\sigma (P))\prod\limits _{{i=1}}^{n}\omega (P_{i}),$

where $P$ ranges over all viable flows in $G$ !

LGVL holds for any finite directed acyclic graph, not just graphs that look like the $2 \times 2$ example. For example consider the following graph:

secondexgraph

or equivalently, the following string diagram:

secondex
The matrix for this is

$\begin{pmatrix} ag+bh & bi & 0\\ ch & ci + dj & dk\\ 0 & ej & ek + fl \end{pmatrix}$ .

It has determinant $agciek + agcifl + agdjfe + bhdjfe$ , and it is easy to check that each of these terms indeed gives a viable flow from the ${s_i}'$ s to the ${t_i}'$ s. Note also that the sign of each term is +1 since $s_i$ is sent to $t_i$ in each of the viable flows.

Now on the one hand, LGVL gives us a way of computing the sum of the signed weights of viable flows in a dag as the determinant of a certain matrix. On the other hand, given a matrix $M$ , we may compute the determinant of $M$ by doing the following. First, we find a special dag $G$ . Next, we find a set of input vertices $\{X_i\}$ and output vertices $\{Y_j\}$ in $G$ . We check to see that $M_{ij} = e(X_i, Y_j)$ . Then, we (easily!) compute the sum of the signed weights of viable flows in $G$ from $\{X_i\}$ to $\{Y_j\}$ . Finally we conclude that the result is equal to the determinant of $M$ by LGVL. This compositional approach to computing the determinant often leads to useful and surprising results, as we shall see below.

Consider the determinant of the following matrix:

$A = \begin{pmatrix} (1,1) & (1,2) & ... & (1,n) \\ (2,1) & (2,2) & ... & (2,n) \\ \vdots & \vdots & \ddots & \vdots \\(n-1,1) & (n-1,2) & ... & (n-1,n)\\(n,1) & (n,2) & ... & (n,n)\end{pmatrix},$

where $(k, \ell)$ denotes the greatest common divisor of $k$ and $\ell$ .

It is difficult to come up with a closed formula for this determinant by using say row-reductions or the co-factor expansion. However, using the strategy involving LGVL gives a simple and surprising result.

But first, a few more definitions! Given any positive integer $k$ , let $\phi(k)$ denote the number of positive integers less than or equal to $k$ which are relatively prime to $k$ . For example, $\phi(6) = 2$ since the only numbers less than 6 that are relatively prime to 6 are 1 and 5. On the other hand $\phi(7)=6$ since all the positive integers less than 7 are relatively prime to 1. In fact, it is easy to see that $\phi(p)=p-1$ for any prime $p$ .
Also, for the purposes of the proof, we shall need to use the fact that

$\sum\limits_{d|k} \phi(d) = k,$

which is a property first established by Gauss. For example

$6 = \phi(1) +\phi(2) + \phi(3) + \phi(6) = 1 + 1 + 2 + 2$ .

Now we proceed to the actual computation. Consider the graph $G$ which has $3n$ vertices: $\{s_1, \ldots, s_n\}$ are the input vertices, $\{t_1, \ldots, t_n\}$ are the output vertices and $\{v_1, \ldots, v_n\}$ are intermediate vertices. If $d$ divides $k$ , put the edge $s_k \to v_d$ with weight $\phi(d)$ and the edge $v_d \to t_k$ with weight 1.

smith

What is the weight $e(s_k, t_{\ell})$ of all the paths from $s_k$ to $t_{\ell}$ ? Well each path from $s_k$ to $t_{\ell}$ can have only two edges: an edge $s_k \rightarrow v_d$ of weight $\phi(d)$ and an edge $v_d \rightarrow t_{\ell}$ of weight 1, for some $v_d$ , where $d$ divides $k$ and $d$ divides $\ell$ . Of course the weight of this path is $\phi(d)$ multiplied by 1, which is just $\phi(d)$ . Therefore,

$e(s_k, t_{\ell}) = \sum\limits_{\substack{d|k \\ d|\ell}}\phi(d) = \sum\limits_{d|(k, \ell)}\phi(d) = (k, \ell)$

The last equality is justified by Gauss’ result.

By the LGVL, it follows that

$\det(A) = \sum\limits_{P = (P_1,\ldots,P_n)} \mathrm{sign}(\sigma(P)) \prod\limits_{i=1}^n \omega(P_i),$

where $P = (P_1, \ldots P_n)$ ranges over all viable flows from the input vertices $S = \{s_1, \ldots, s_n\}$ to the output vertices $T = \{t_1, \ldots, t_n\}$ .
Now the only path from $s_1$ to $T$ is $s_1 \to v_1 \to t_1$ since 1 has only one divisor, namely 1. As a result, for each prime $1 \leq p \leq n$ , any path from $s_p$ to $T$ in a viable flow must pass through $v_p$ . This is because $p$ has only two divisors, namely 1 and $p$ , but the path through 1 has already been blocked by 1.

Consequently, for any product $pq$ of two primes, any path from $s_{pq}$ to $T$ must pass through $v_{pq}$ . Repeating this argument for all the other numbers less than or equal to $n$ , we conclude that all the viable flows from $S$ to $T$ link $s_k$ to $v_k$ .

Now if there is a viable flow which links $v_k$ to $t_{\ell}$ for some $k \neq \ell$ , then there must be some $k', \ell'$ such that $v_{k'}$ is linked to $t_{\ell}'$ with $k' > \ell'$ (why?). But this means that $k'$ divides $\ell'$ , a contradiction since $k' > \ell$ ! So we must have $v_k = t_k$ , for all $k$ .

Thus, there is a unique viable flow from $S$ to $T$ i.e. the flow for which $s_k$ is sent to $t_k$ via $v_k$ . Thus, we conclude that the determinant of $A$ is $\phi(1) \cdot \ldots \cdot \phi(n)$ , since the weight of each path $P_k$ in this flow is $\phi(k)$ as we showed earlier:

$\left \vert \begin{pmatrix}(1,1) & (1,2) & ... & (1,n) \\ (2,1) & (2,2) & ... & (2,n) \\ \vdots & \vdots & \ddots & \vdots \\(n-1,1) & (n-1,2) & ... & (n-1,n)\\(n,1) & (n,2) & ... & (n,n)\end{pmatrix} \right \vert = \phi(1) \cdot \ldots \cdot \phi(n)$

5 thoughts on “Determinants and the Lindström-Gessel-Vienot Lemma”

dez says:

August 1, 2017 at 6:22 am

Cute!

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markusq says:

August 2, 2017 at 5:24 pm

Sweet. Thank you for sharing this!
I noticed two minor typos (omitted words) and had a few questions.
“signed weight a viable flow” –> “signed weight of a viable flow”
“Each viable flow can described by” –> “Each viable flow be can described by”
I think the variable P is overloaded between this and the following paragraphs (as both an arbitrary viable flow and an arbitrary path between two given endpoints). Am I missing something, or are these actually distinct objects?
You pass lightly over a potentially interesting point: “…given a matrix M, we may compute the determinant of M by doing the following. First, we find a special dag G…” Is this inverse operation always possible?

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- markusq says:
  
  August 2, 2017 at 7:10 pm
  
  Oops. I messed up. “be can” –> “can be”. 🙂
  
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. says:

July 21, 2018 at 2:01 am

”It has determinant agciek + agcifl + agdjfe + bhdjfe, and it is easy to check that each of these terms indeed gives a viable flow from the {s_i}’s to the {t_i}’s.“
should be
”It has determinant agciek + agcifl + agdjfl + bhdjfl, and it is easy to check that each of these terms indeed gives a viable flow from the {s_i}’s to the {t_i}’s.“

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Armando Figueroa says:

July 9, 2021 at 11:39 pm

Hi Pawel. Thanks for this beautiful blog!! I have learned and enjoyed a lot.
I am a newbie on this matter. I hope someday you can explain us, in your diagrammatic picture, the characteristic polynomial and trace invariants for the Hermitian case. Are your diagrams connected with the so-called trace diagrams?: https://mathoverflow.net/questions/6139/how-can-i-learn-about-doing-linear-algebra-with-trace-diagrams

Thank you in advance !!

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