Category Archives: Graphs on Napkins

… and other handy surfaces. This is a collection of casual exposition about algebraic graph theory and other maths on my mind and scrap paper.

Non-existence of strongly regular graphs with parameters $(49,16,3,6)$

Some people make a habit of occasionally visiting Wikipedia’s front page and following hyperlinked phrases down a rabbit hole of articles. I do a similarly thing with Andries Brouwer’s tables of strongly regular graphs, which leads to a cascade of papers, checking BCN and computing stuff in Sage.  How does that happen exactly? As usual in math, we now need to start with some definitions.

Definitions

A strongly regular graph is a $k$-regular graph where every pair of adjacent vertices has $a$ common neighbours and every pair of non-adjacent vertices has $c$ common neighbours. The tuple $(n,k,a,c)$ is called the parameter set of the strongly regular graph (or SRG). For example, the Petersen graph is a SRG with parameter set $(10,3,0,1)$. We immediately, see that these number $n,k,a$ and $c$ have to satisfy some conditions, some trivial ($n>k$ and $a,c \leq k$) and others less so.

Brouwer’s tables consists of all possible parameter sets for $n$ up to $1300$ which have passed some elementary checks. The parameters which are in green are classes where it has been confirmed that there does in fact exist at least one SRG with those parameters. The ones in red have been shown by other means (failing to satisfy the absolute or relative bound, for example) to not be parameter sets of any SRGs. The entries in yellow are parameter sets where the existence of a SRG is still open. This leads us to …

Non-existence of SRGs

Finding whether or not a putative SRG exists is a pretty interesting problem; the existence of the $57$-regular Moore graph which is a SRG with parameters $(3250, 57,0,1)$ is a long-standing open problem in algebraic graph theory. Today, we’ll look at the parameter set $(49,16,3,6)$, which holds the distinction of being the first red entry in Brouwer’s tables which is not a conference graph and also not ruled out by the absolute bound. This means that it was ruled out by a somewhat ad-hoc method; in this case, it was ruled out by Bussemaker, Haemers, Mathon and Wilbrink in this 1989 paper. However, computers have come along way since 1989, so instead of following their proof, I will give another proof, which uses computation (which was done in Sage).

That’s way too many 2’s, this SRG can’t exist

Suppose $X$ is a SRG with parameter set$(49,16,3,6)$. This graph has eigenvalues $$16, 2^{(32)}, \text{ and } -5^{(16)}$$ where the multiplicities are given as superscripts. We’ll proceed by trying to build $X$. First, we “hang a graph by a vertex” as follows:

The main thing to note right away is that neighhourhood of a vertex $x$ in $X$, denoted $\Gamma_1(x)$ in the diagram, is a $3$-regular graph on $16$ vertices. This is excellent news because there are only 4060 such graphs and one can generate them using geng, or download them from Gordon Royle’s website.

Next, we apply interlacing. We denote by $\lambda_i(G)$ the $i$th largest eigenvalue of $G$. Since $\Gamma_1(x)$ is an induced subgraph of $X$, interlacing theorem tell us $$ \lambda_i(X) \geq \lambda_i(\Gamma_1(x)) \geq \lambda_{33 + i}(X)$$ for $i=1,\ldots,16$. In particular, we see that the second largest eigenvalue of $\Gamma_1(x)$ is at most  $2$. In general, this gives us all kinds of useful information (for example, that $\Gamma_1(x)$ has to be connected), but since we only care about cubic graphs on $16$ vertices, we can just go ahead and check which of them have this property. In turns out that only 35 cubic graphs on 16 vertices have this property.

If I have an induced subgraph $Y$ of $X$ on 18 vertices, the  interlacing theorem would give the following:  $$ \lambda_2(X) \geq \lambda_2(\Gamma_1(x)) \geq \lambda_{33}(X).$$ Recalling that the second and 33rd largest eigenvalues of $X$ are both $2$, we see that any subgraph of $X$ on 18 vertices must have its second eigenvalue equal to $2$. Since $\lambda_i(G)$ already has 16 vertices, we’re not far off. Let $y$ be a vertex in the second neighbourhood of $x$ and we will consider the subgraph $Y$ induced by $x$, the neighbours of $x$ and $y$.

In terms of the computation, we try to add $x$ and $y$ to each of the remaining 35 cubic graphs. We can always choose $y$ to be adjacent to a fixed vertex in $\Gamma_1(x)$ and so there are ${16 \choose 5}$ ways of adding the other $5$ edges incident to $y$. Thus we have 35(3003) candidates for $Y$. A quick computation later, we see that none of those graphs have $2$ as the second largest eigenvalue, which contradicts the existence of $X$.

So what? And also, now what?

It’s fun proving the non-existence of a SRG that is already known not to exist, but clearly what we actually want to do is convert some yellow entries of Brouwer’s table to either red or green. The method here (as well as that of the original paper of  Bussemaker, Haemers, Mathon and Wilbrink) works because the multiplicity of $2$ as an eigenvalue is really large. The degree of the graph is equal to the multiplicity of its smallest eigenvalue, which smells of some kind of duality. This happens, for example, in the complement of Clebsch graph (which, by the way, is formally self-dual). The next open case where this occurs is $(100,33,8,12)$.  In this case, the first neighbourhood of a vertex would be a $8$-regular graph on 33 vertices, with a fairly large spectral gap. We probably don’t want to be generating all such graphs, so some new idea is needed.

Why are Hermitian matrices diagonalizable?

“Since we are working with a Hermitian matrix, we may take an eigenbasis of the space …”

“Wait, sorry, why are Hermitian matrices diagonalizable, again?”

“Umm … it’s not quick to explain.”

This exchange happens often when I give talks about spectra of graphs and digraphs in Bojan’s graph theory meeting. I know several ways to prove that Hermitian matrices are diagonalizable, but I couldn’t think of a simple, succinct statement which gives the right intuitive thing. So, in this post, I’d like to arrive at such a one-liner.

Definitions

We consider $n\times n$ matrices with entries in $\mathbb{C}$. A matrix $H$ is Hermitian if $H^* = H$, where $H^* = \bar{H}^T$ is the conjugate transpose of $H$. A matrix $H$ is diagonalizable if $H$ is similar to a diagonal matrix: i.e. there exist invertible matrix $P$ such that $P^{-1} H P = D$ where $D$ is a diagonal matrix.

An eigenvalue of $H$ is $\lambda$ such that there exist $v \in \mathbb{C}^n$ such that $$Hv = \lambda v$$ and $v$ is said to be an eigenvector of $H$. The characteristic polynomial of $H$ is $$\phi(H,t) = \det(tI – H).$$ If $\lambda$ is a number such that $\phi(H,\lambda) = 0$, then $\lambda I – H$ has a non-trivial kernel and so there exists a vector $v$ such that $(\lambda I – H)v = 0$. Thus, every root of $\phi(H,t)$ is an eigenvalue of $H$. However, the roots of the characteristic polynomial are not the same as the multiset of eigenvalues because there is a question of multiplicity.

Geometric vs algebraic multiplicities

The geometric multiplicity, $m_g(\lambda)$, of an eigenvalue $\lambda$ of $H$ is the dimension of the subspace of $\mathbb{C}^n$ generated by all eigenvectors $H$ with eigenvalue $\lambda$ (this space is called the eigenspace of $\lambda$). The algebraic multiplicity, $m_a(\lambda)$, is the multiplicity of $\lambda$ as a root of $\phi(H,t)$.

Proposition. A Hermitian matrix $H$ is diagonalizable if and only if $m_a(\lambda) = m_g(\lambda)$ for each eigenvalue $\lambda$ of $H$.

Proof. Suppose $H$ is a $n\times n$ Hermitian matrix.

“$\Leftarrow$” It is easy to see that the characteristic polynomial has degree $n$ and hence $n$ roots. Since the algebraic and geometric multiplicities agree, we see that $H$ has $n$ orthonormal eigenvectors (details left as an exercise) which can be used to form the columns of a matrix $P$. Then $P^*P = I $ and $P^*HP$ is a diagonal matrix with the eigenvalues of $H$ on the diagonal.

“$\Rightarrow$” Let $P$ be a matrix such that $P^{-1}HP = D$, where $D$ is a diagonal matrix. Then $$HP =PD$$ and we may consider the $j$th column of both sides. On the left side, we get $Hv$ where $v$ is the $j$th column of $P$. On the right side, we get $D_{jj}v$, and so $v$ is an eigenvector of $H$ with eigenvalue $D_{jj}$. Since $H$ and $D$ are similar, $$\phi(H,t) = \phi(D,t) = \prod_{j = 1}^n (t-D_{jj}). $$ Thus the multiset $\{D_{jj}\}_{j=1}^n$ is the set of eigenvalues of $H$ with both geometric and with algebraic  multiplicities. $\Box$

Diagonalizability

We will now show that Hermitian matrices are diagonalizable by showing that every eigenvalue has the same algebraic and geometric multiplicities.

Theorem. If $H$ is a Hermitian matrix with eigenvalue $\lambda$, then $m_g(\lambda) = m_a(\lambda)$.

Proof. We take $H$ to be a $n\times n$ Hermitian matrix and $\lambda$ an eigenvalue of $H$. We proceed by induction on $n$.

If $m_a(\lambda) = 1$, we are done, since there must be exactly one eigenvector of $\lambda$. We may assume $a = m_a(\lambda) >1$. Let $x$ be an eigenvector of $H$ such that $Hx = \lambda x$. We may assume that $x$ is normalized, i.e. $x^*x = 1$. We may extend $\{x\}$ to an orthonormal basis of $\mathbb{C}^n$, say $\{x, v_2, \ldots, v_n\}$.

Let $V= \langle v_2, \ldots, v_n \rangle = \langle x \rangle ^{\perp}$. We may consider $\mathbb{C}^n$ as the direct sum $\langle x \rangle \oplus V$. Let $v \in V$. Observe that $$x^*(Hv) = x^*H^*v=  (Hx)^*v = (\lambda x)^*v = \bar{\lambda} x^*v = 0.$$ Thus $Hv \in V$ and $V$ is a $H$-invariant subspace of $\mathbb{C}^n$.

Let $P$ be the unitary matrix with $\{x, v_2, \ldots, v_n\}$ as its columns. The above gives that $$ P^* HP = \left( \begin{array}{cccc} \lambda & 0 & \cdots & 0 \\ 0 & & & \\ \vdots & & B &  \\0 &&& \end{array} \right)$$ and $B$ is Hermitian since the left side is. We see that $$\phi(H,t) = \phi(P^* HP, t) = (t-\lambda)\phi(B,t).$$ Thus, $\lambda$ is an eigenvalue of $B$ with algebraic and geometric multiplicity  $a-1$ (by induction) and $B$ has pair-wise orthogonal eigenvectors $x’_2, \ldots, x’_a$ of $\lambda$. For $j = 2, \ldots, a$, let $x_j = P\left( \begin{array}{c} 0 \\ x’_j \end{array} \right)  P^*$. It is easy to see that $x, x_2, \ldots, x_a$ are pair-wise orthogonal eigenvectors of $H$ with eigenvalue $\lambda$, which proves the theorem. $\Box$

The Take-away

There are many (mostly equivalent ways) to show this; we could have used induction to prove $H$ is diagonalizable, without consider geometric vs algebraic multiplicities, we could have proved the decomposition into Jordan blocks, or we could have proven the spectral decomposition theorem.

The crux of the proof is that, when $H$ is Hermitian, the vector space $W^{\perp}$ is $H$-invariant when $W$ is. In our proof, this allowed us to, colloquially speaking, keep pulling eigenvectors out of $\mathbb{C}^n$. In general, given an $H$-invariant subspace $W$ of $\mathbb{C}^n$, we can consider the action of $H$ (by multiplication on the left) on $W$ and find the minimal polynomial of $H$ over $W$. If $\psi_1$ and $\psi_2$ are the minimal polynomials of $H$ over $W$ and $W^{\perp}$, respectively, then $\phi(H,t) = \psi_1 \psi_2$.


Intuitively, a Hermitian matrix $H$ is diagonalizable because we can break $\mathbb{C}^n$ into $H$-invariant, pairwise orthogonal, subspaces and diagonalize $H$ over each subspace.


To see that this a property that is not true of all matrices, consider the following matrix: $$N = \left(\begin{array}{cccc} 0 & 1 & 1 & 1 \\ 0& 0& 1 & 1 \\ 0& 0 & 0 & 1 \\ 0&0&0&0 \end{array} \right).$$ For graph theorists, this is the adjacency matrix of a transitive tournament on $4$ vertices. It is also a nilpotent matrix; that is, $N^4 = 0$. The characteristic polynomial of $N$ is $t^4$ and $N$ has $0$ as an eigenvalue with algebraic multiplicity $4$. However, $N$ has only one linearly independent eigenvector $e_1$ (the elementary basis vector, $(1\,\, 0\,\, 0\,\, 0)^T$).

Here, $\langle e_1 \rangle$ is an eigenspace of $N$ and hence $N$-invariant. Observe that $e_2 \in \langle e_1 \rangle^{\perp}$ but $Ne_2 = e_1$ and so $ \langle e_1 \rangle^{\perp}$ is certainly not $N$-invariant.

References

Roman, Steven. Advanced Linear Algebra. (Springer Graduate Texts in Mathematics, Vol. 135)

Prasolov, V. V. Problems and Theorems in Linear Algebra. (AMS Translations of Mathematical Monographs, Vol. 134)

Graph theory: helping tourists and map-makers since 1735

On pi day this year, I gave a talk with the same title as this post at A Taste of Pi, an event organized by SFU to introduce high school students to math. I’ve decided to tell a story about the history of graph theory as a gentle introduction to the area.

Slides to my pi day talk: here

To make graph theory accessible to high school students, I made a lot of pdf images. I’ve decided to convert some sequences of my pdf animations into gifs. Here are my two favourites. It might be necessary to click on them, to get them to animate.

Cardioids:

One way to draw an eulerian planar graph is to simply draw closed curves on a piece of paper and add vertices where the curves intersect. The result is a plane drawing of an eulerian planar graphs … whose faces can be coloured with 2 colours.

But we need not draw any random doodle (although that is also fun), we can draw an approximation to a cardioid as follows:

  1. We start with a circle C with a defined point P.
  2. Draw a circle whose centre is at a point Q  on C and  pass through point P.
  3. Repeat step 2 until satisfied. (Repeat for all choices of Q to be a cardioid.)
cardioid colouring
Drawing a cardioid to obtain an eulerian planar graph, embedded in the plane, and its 2-face-colouring.

2-face-colourability of eulerian planar graphs:

It is well-known that the faces of planar eulerian graphs are 2-colourable. If we take the planar dual Y of an eulerian plane graph X, we see each face of Y has even degree, since each vertex of X has even degree. Since the facial walks of a planar graph generates its cycle space, this implies that Y is bipartite.

Here is an illustration of the simpler, inductive proof:

A pictorial proof the planar eulerian graphs can be 2-coloured.
A pictorial proof the planar eulerian graphs can be 2-coloured.

Graphs on napkins

Some people believe in the statement: “if it’s not recreate-able, it’s not true.” … I don’t.
But instead of having dozens of photos all labelled “incomprehensible (but surely important) scribblings with math symbols”, I’m converting them into casual expositions. The posts will be generally aimed at an upper-year undergraduate level and will be about various topics in algebraic graph theory, like cores of cubelike graphs, generalized quadrangles and association schemes.