Eigenvectors of graph Laplacians: a landscape

We have the following result

To see this consider the linear system

$$(L-\lambda I)X=0.$$

It can be solved using Gauss’s elimination. This involves algebraic manipulations so that the result $X$ is rational. If $X$ is rational, then multiplying by the product of the denominators of the entries, we obtain an eigenvector with integer entries.

We now show that the eigenvalues of a graph Laplacian are either integers or irrationals. We have the following rational root lemma on the roots of polynomials with integer coefficients, see for example [13]

A consequence of this is

The fact that some graphs have integer spectrum was discussed by Grone and Merris [14]. Many of their results are inequalities for $\lambda_{2}$ and $\lambda_{n-1}$. Our results complement their approach.

The clique $K_{n}$ has eigenvalue $n$ with multiplicity $n-1$ and eigenvalue $0$. The eigenvectors for eigenvalue $n$ can be chosen as $v^{k}=e^{1}-e^{k},~{}~{}k=2,\dots,n$. To see this note that

$$L=nI_{n}-\mathbf{1},$$

where $I_{n}$ is the identity matrix of order $n$ and $\mathbf{1}$ is the $(n,n)$ matrix where all elements are $1$.

A star of $n$ vertices $S_{n}$ is a tree such that one vertex , say vertex 1, is connected to all the others. For a star $S_{n}$, the eigenvalues and eigenvectors are

• •

  $\lambda=1$ multiplicity $n-2$ , eigenvector $e^{2}-e^{k},~{}~{}k=3,\dots,n$

• •

  $\lambda=n$ multiplicity $1$ , eigenvector $(n+1)e^{1}-\sum_{k=2}^{n}e^{k}$

• •

  $\lambda=0$ multiplicity $1$ , eigenvector ${\hat{1}}$

Consider a bipartite graph $K_{n_{1},n_{2}}$. The Laplacian is

$$L=\begin{pmatrix}n_{2}&0&\dots&0&-1&\dots&&-1\\ 0&n_{2}&0&\dots&-1&\dots&&-1\\ \dots&\dots&\dots&\dots&\dots&\dots&&\\ 0&\dots&0&n_{2}&-1&\dots&&-1\\ -1&\dots&&-1&n_{1}&0&\dots&0\\ -1&\dots&&-1&0&n_{1}&\dots&0\\ \dots&&&&&\dots&\dots&\dots\\ -1&\dots&&-1&0&0&\dots&n_{1}\end{pmatrix},$$

(3)

where the top left bloc has size $n_{1}\times n_{1}$, and the bottom right bloc $n_{2}\times n_{2}$. The eigenvalues with their multiplicities denoted as exponents are

$$0^{1},~{}~{}n_{1}^{n_{2}-1},~{}~{}n_{2}^{n_{1}-1},~{}~{}(n1+n2)^{1}.$$

Eigenvectors for $n_{1}$ can be chosen as $e^{n_{1}+1}-e^{i}~{}~{}(i=n_{1}+2,\dots,n_{1}+n_{2})$. The eigenvector for $n=n_{1}+n_{2}$ is $(1/n_{1},\dots,1/n_{1},-1/n_{2},\dots,-1/n_{2})^{T}$.

Similarly, the spectrum of a multipartite graph $K_{n_{1},n_{2},\dots n_{p}}$ is

$$0^{1},~{}~{}(n-n_{1})^{n_{1}-1},~{}~{}(n-n_{2})^{n_{2}-1},\dots,~{}~{}(n-n_{p})^{n_{p}-1},~{}~{}n^{p}.$$

The eigenvectors associated to $n-n_{1}$ are composed of $1$ and $-1$ in two vertices of part 1 padded with zeros for the rest.

For a cycle, the Laplacian is a circulant matrix, therefore its spectrum is well-known. The eigenvalues are

$$\mu_{k}=4\sin^{2}\left[{(k-1)\pi\over n}\right],~{}~{}k=1,\dots,n.$$

(4)

They are associated to the complex eigenvectors $v^{k}$ whose components are

$$v_{j}^{k}=\exp{\left[i(j-1)(k-1)2\pi\over n\right]}~{}~{},j=1,\dots n.$$

(5)

The real eigenvectors $w^{k},~{}x^{k}$ are,

	$\displaystyle w^{k}=(0,~{}\sin(a_{k}),~{}\sin(2a_{k}),~{}\dots,~{}\sin((n-1)a_{k}))^{T},$		(6)
	$\displaystyle x^{k}=(1,~{}\cos(a_{k}),~{}\cos(2a_{k}),~{}\dots,~{}\cos((n-1)a_{k}))^{T},$		(7)
	$\displaystyle a_{k}={2(k-1)\pi\over n}$		(8)

Ordering the eigenvalues, we have

	$\displaystyle\lambda_{1}=\mu_{1}=0,$		(9)
	$\displaystyle\lambda_{2}=\lambda_{3}=\mu_{2},$		(10)
	$\displaystyle\lambda_{2k}=\lambda_{2k+1}=\mu_{k+1},$		(11)
	$\displaystyle\dots$		(12)

For $n=2p+1$

$$\lambda_{2p}=\lambda_{2p+1}=\mu_{p+1}$$

For $n=2p$

$$\lambda_{2p}=\mu_{p}=4$$

is an eigenvalue of multiplicity 1; an eigenvector is $(1,-1,\dots,1,-1)^{T}$. In all other cases, the eigenvalues have multiplicity two so that all vertices are soft nodes.

Remark that the maximum number of 0s is $n/2$. To see this, note that if two adjacent vertices have value 0 then their neighbors in the cycle must have 0 as well and we only have 0s , but the null vector is not an eigenvector. This means that we have at most $n/2$ 0s. This bound is reached for $n$ even.

For chains $C_{n}$, there are only single eigenvalues, they are [15]

$$\lambda_{k}=4\sin^{2}({\pi(k-1)\over 2n})~{}~{},k=1,\dots,n.$$

(13)

The eigenvector $v^{k}$ has components

$$v_{j}^{k}=\cos{\left({\pi(k-1)\over n}(j-{1\over 2})\right)}~{}~{},j=1,\dots n.$$

(14)

Obviously the cosine is zero if and only if:

$$(k-1)(2j-1)=n(1+2m),$$

(15)

where $m$ is an integer. There is no solution for $n=2^{\alpha}$, for $\alpha$ a positive integer. Apart from this case, there is always at least one soft node. If $n$ is a prime number, the middle vertex $j=(n+1)/2$ is the only soft node. For $k$ odd, all vertices $j$ such that $2j-1$ divides $n$ have a zero value, including the middle vertex.

For $n$ odd, chains and cycles share $(n-1)/2$ eigenvalues and eigenvectors. To see this consider a chain with $n=2p+1$. All $k=2q+1$ give a chain eigenvalue $\lambda_{k}=4\sin^{2}({\pi q\over 2p+1})$ that is also a cycle eigenvalue. The eigenvector components $v_{j}^{q}$ are such that $v_{1}^{q}=v_{2p+1}^{q}$.

An important theorem due to Merris [10] connects equal component vertices.

This transformation preserves the eigenvalue and eigenvector. It applies to multiple graphs. Fig. 1 shows examples of the transformation.

We have the following corollary of the theorem.

This allows to generate many more graphs that have an eigenvalue $\lambda$.

An elementary transformation inspired by Merris’s principle of reduction and extension [10] is to add a soft node to an existing soft node. This does not change the eigenvalue. We have the following result.

The general case presented by Merris [10] amounts to applying several times this elementary transformation.
The transformation is valid for graphs with arbitrary weights and the extended edges can have arbitrary weights.

Fig. 2 illustrates this property on the two graphs labeled $5.6$ and $5.23$ in the classification given in [1]. An immediate consequence of this elementary transform is that any soft node can be extended into an arbitrarily large graph of soft nodes while preserving the eigenvalue and extending the eigenvector in a trivial way. Fig. 3 shows two graphs that have the same eigenvalue $\lambda=1$ and that are connected by the articulation transform.

A consequence of the contraction principle of Merris [10] is that coalescing two soft nodes of a graph leaves invariant the eigenvalue. This is especially important because we can ”solder” two graphs at a soft node.

This transformation is valid for graphs with arbitrary weights.

We have the following theorem.

Let us detail the eigenvector relation for the Laplacian for $G^{\prime}$. Consider any new vertex $j$ linked to the $p$ soft nodes and to $d$ new nodes. The corresponding line of the eigenvector relation for the Laplacian for $G^{\prime}$ reads

$$(d+p)x^{\prime}_{j}+\sum_{i\sim j,i\geq n}(-1)x^{\prime}_{i}=\lambda^{\prime}x^{\prime}_{j}.$$

This implies

$$(d+p-\lambda^{\prime})x^{\prime}_{j}=\sum_{i\sim j,i\geq n}x^{\prime}_{i}.$$

An obvious solution is

$$\lambda^{\prime}=\lambda=p,~{}~{}~{}x^{\prime}_{i}=x^{\prime}_{n}~{}~{}\forall i\geq n+1.$$

The value $x^{\prime}_{n}$ is obtained by examining line $n-1$. We have

$$\alpha+\sum_{i=n}^{n-k-1}(-1)x_{i}^{\prime}=0$$

so that

$$x^{\prime}_{n}={x_{n}\over k}.$$

In fact, we can get all solutions by satisfying the two conditions

$$\forall j\geq n~{}~{}dx^{\prime}_{j}=\sum_{i\sim j}x^{\prime}_{i},~{}~{}x_{n}=\sum_{i\geq n}x^{\prime}_{i}.$$

(16)

$\Box$

Fig. 5 shows examples of expansion from a single soft node for different values of $d$. Here the eigenvalue is $1$. Fig. 6 shows examples of expansion from two soft nodes. The eigenvalue is $2$. For $d=2$, the values at the edges at the bold edges are such that their sum is equal to $1$. For $d=2$, the values at the triangle are all equal to $t$, the same holds for the square with a value $s$. These values verify $3t+4s=1$.

We have the following transformation that leaves the eigenvalue unchanged [12].

This result was proved in [12] for a graph with weights 1. Here we generalize it to a graph with arbitrary weights.

The eigenvalue relation at vertex $i$ reads

$$(d_{i}-\lambda)x_{i}=\sum_{m\sim i,m\neq j}w_{i,m}x_{m}+w_{i,j}x_{j}$$

Since $x_{i}=-x_{j}$, this implies

$$(d_{i}+w_{i,j}-\lambda)x_{i}=\sum_{m\sim i,m\neq j}w_{i,m}x_{m}.$$

Introducing the two new vertices $k,l$ such that ${x}^{\prime}_{k}={x}^{\prime}_{l}=0$ connected to $i$ by edges of weights $w_{i,k}=\alpha w_{i,j},~{}~{}w_{i,l}=(1-\alpha)w_{i,j}$, the relation above leads to

$$(d_{i}+w_{i,k}+w_{i,l}-\lambda)x^{\prime}_{i}=\sum_{m\sim i}w_{i,m}x^{\prime}_{m}+w_{i,k}x^{\prime}_{k}+w_{i,l}x^{\prime}_{l},$$

which shows that $x^{\prime}$ is eigenvector of the new graph.

$\Box$

See Fig. 7 for an illustration of the theorem.

Fig. 8 shows an example of the action of inserting a soft node.

When the graph is weighed, the result is still valid. Consider that we add only one soft vertex connected to $i$ by a weight $w_{i,k}$. The eigenvalue of the new graph is $\lambda+w_{i,k}$.
This can transform a graph with an integer eigenvalue to a graph with an irrational eigenvalue.

Connecting a soft node to all the vertices of a graph augments all the non zero eigenvalues by 1. This result was found by Das [11]. We recover it here and present it for completeness.

See Fig. 9 for examples.

Important examples are the ones formed with the special graphs considered above. There, adding a vertex to an $n-1$ graph, one knows explicitly $n-1$ eigenvectors and eigenvalues.

The theorem 3.2 by Das [11] can be seen as a direct consequence of adding a soft node and an articulation to a graph.

First we define perfect and alternate perfect matchings.

We have the following result [12] inspired by the alternating principle of Merris [10].

This is a second operator which shifts eigenvalues by $\pm 2$. Examples are given in Fig. 10.

The cartesian product $G\square H$ of two graphs $G=(V,E)$ and $H=(W,F)$ has set of vertices $V\times W=\{(v,w),v\in V,~{}w\in W\}$. It’s set of edges is $\{\{(v_{1},w_{1}),(v_{2},w_{2})\}\}$ such that $v_{1}~{}v_{2}\in V$ and $w_{1}w_{2}\in W$. We have the following result, see Merris [10].

Fig. 11 illustrates the theorem.

Important examples are the ones formed with the special graphs considered above. There, one knows explicitly the eigenvectors and eigenvalues. For example, the cartesian product $C_{n}\times C_{m}$ of two chains $C_{n}$ and $C_{m}$ with $n$ and $m$ nodes respectively has eigenvalues

$$\lambda_{i,j}=\lambda_{i}+\lambda_{j},$$

where $\lambda_{i}$ (resp. $\lambda_{j}$) is an eigenvalue for $C_{n}$ (resp. $C_{m}$). The eigenvectors are

$$v^{i,j}=\cos[{\pi(i-1)\over n}(p-{1\over 2})]\cos[{\pi(j-1)\over m}(q-{1\over 2})],$$

where $i,p\in\{1,\dots,n\},~{}~{}~{}j,q\in\{1,\dots,m\}$.

We recall the definition of the complement of a graph $G$.

We have the following property, see for example [1].

An example is shown in Fig. 12. The eigenvalues and eigenvectors are given in table 1.

Many times, $G_{c}$ is not connected. An example where $G_{c}$ is connected is the cycle 6….

We introduce the notions of $\lambda$, $\lambda$ soft and $\lambda$ soft minimal graphs. The transformations of the previous section will enable us to prove the relation between these two types of graphs.

Clearly, for a given $\lambda$, there is at least one $\lambda$ minimal graph. As an example the 1 soft minimal graph is shown below.

In the following section, we study the properties of a $\lambda$ subgraph $G$ included in a $\lambda$ graph $G"(V",E")$. Consider two graphs $G(V,E)$ with $n$ vertices and $G^{\prime}(V"-V,E^{\prime})$ such that $E$ only connects elements of $V$ and $E^{\prime}$ only connects elements of $V^{\prime}$. Assume two graphs $G(n)$ and $G^{\prime}(n^{\prime})$ are included in a large graph $G"$ and are such that $G(V,E)$ Assume $p$ vertices of $G$ are linked to $p^{\prime}$ vertices of $G^{\prime}$. We label the $p$ vertices of $G$, $n-p+1,\dots,n$ and the $p^{\prime}$ vertices of $G^{\prime}$, $1,\dots,p^{\prime}$. We have

	$\displaystyle LX=\lambda X,$		(17)
	$\displaystyle L"X"=\lambda X",$		(18)

where $L"$ is the graph Laplacian for the large graph $G"$; $L"$ can be written as

$$L"=\begin{pmatrix}L&0\\ 0&L^{\prime}\end{pmatrix}+\begin{pmatrix}0&0&0&0\\ 0&a&-b&0\\ 0&-b^{T}&c&0\\ 0&0&0&0\\ \end{pmatrix}.$$

A first result is

At this point, we did not assume any relation between the eigenvectors $X$ for $G$ and $X"$ for $G"$. We have the following

We can then assume that the eigenvectors of $L"$ have the form

$$L"\begin{pmatrix}X\\ X^{\prime}\end{pmatrix}=\lambda\begin{pmatrix}X\\ X^{\prime}\end{pmatrix},$$

where $LX=\lambda X$. Substituting $L"$, we get

$$\lambda\begin{pmatrix}X\\ X^{\prime}\end{pmatrix}=\begin{pmatrix}L&0\\ 0&L^{\prime}\end{pmatrix}\begin{pmatrix}X\\ X^{\prime}\end{pmatrix}+\begin{pmatrix}0&0&0&0\\ 0&a&-b&0\\ 0&-b^{T}&c&0\\ 0&0&0&0\\ \end{pmatrix}\begin{pmatrix}X\\ X^{\prime}\end{pmatrix}.$$

Using the relation (17) we obtain

	$\displaystyle\begin{pmatrix}0&0\\ 0&a\end{pmatrix}X+\begin{pmatrix}0&0\\ -b&0\end{pmatrix}X^{\prime}=0,$		(20)
	$\displaystyle L^{\prime}X^{\prime}-\begin{pmatrix}0&b^{T}\\ 0&0\end{pmatrix}X+\begin{pmatrix}c&0\\ 0&0\end{pmatrix}X^{\prime}=\lambda X^{\prime}.$		(21)

There are $p$ non trivial equations in the first matrix equation and $p^{\prime}$ in the second one. Using an array notation (like in Fortran), the system above can be written as

	$\displaystyle aX(n-p+1:n)-bX^{\prime}(1:p^{\prime})=0,$		(22)
	$\displaystyle-b^{T}X(n-p+1:n)+cX^{\prime}(1:p^{\prime})+(L^{\prime}X^{\prime})(1:p^{\prime})=\lambda X^{\prime}(1:p^{\prime}),$		(23)
	$\displaystyle(L^{\prime}X^{\prime})(p^{\prime}+1:n^{\prime})=\lambda X^{\prime}(p^{\prime}+1:n^{\prime}),$		(24)

Extracting $X$ from the first equation, we obtain

$$X(n-p+1:n)=a^{-1}bX^{\prime}(1:p^{\prime}),$$

(25)

and substituting in the second equation yields the closed system in $X^{\prime}$

	$\displaystyle(-b^{T}a^{-1}b+c)X^{\prime}(1:p^{\prime})+(L^{\prime}X^{\prime})(1:p^{\prime})=\lambda X^{\prime}(1:p^{\prime}),$		(26)
	$\displaystyle(L^{\prime}X^{\prime})(p^{\prime}+1:n^{\prime})=\lambda X^{\prime}(p^{\prime}+1:n^{\prime}),$		(27)

where we used the fact that the matrix $a$ of the degrees of the connections is invertible by construction.

We then write equations (26,27) as

$$({\bar{\Delta}}+L^{\prime})X^{\prime}=\lambda X^{\prime},$$

(28)

where

$${\bar{\Delta}}=\begin{pmatrix}\Delta&0\\ 0&0\end{pmatrix}$$

This is an eigenvalue relation for the graph Laplacian $({\bar{\Delta}}+L^{\prime})$. Four cases occur.

• (i)

  $\lambda=0$ then $X^{\prime}$ is a vector of equal components and $X$ also.

• (ii)

  $\lambda\neq 0$ is an eigenvalue of $L^{\prime}$. Then one has the following

  Theorem 6.7

  Assume a graph $G"$ is $\lambda$ for an eigenvector $X"=(X,X^{\prime})^{T}$ and contains a $\lambda$ graph $G$ for the eigenvector $X$. Consider the graph $G^{\prime}$ with vertices $V(G")-V(G)$ and the corresponding edges in $G"$.
  If $G^{\prime}$ is $\lambda$ then $G"$ is obtained from $G$ using the articulation or link transformations.

  Proof. Since $\lambda\neq 0$ is an eigenvalue of $L^{\prime}$, we can choose $X^{\prime}$ an eigenvector for $\lambda$ so that $L^{\prime}X^{\prime}=\lambda X^{\prime}$, then $\Delta X^{\prime}=0$.
  A first possibility is $X^{\prime}=0$, this corresponds to an articulation between $G$ and $G^{\prime}$.
  If $X^{\prime}\neq 0$, $L^{\prime}X^{\prime}=\lambda X^{\prime}$, implies that $X^{\prime}$ is not a vector of equal components so that $X^{\prime}\notin{\rm Null}(\Delta)$. The only possibility for $\Delta X^{\prime}=0$ is $\Delta=0$ so that

  $$c=b^{T}a^{-1}b.$$

  The term $(b^{T}a^{-1}b)_{ij}$ is

  $$(b^{T}a^{-1}b)_{ij}=\sum_{k=1}^{p}{b_{ki}b_{kj}\over a_{kk}}.$$

  Since the matrix $c$ is diagonal, we have

  $$\sum_{k=1}^{p}{b_{ki}b_{kj}\over a_{kk}}=0,\forall i\neq j$$

  Then $b_{ki}b_{kj}=0$ so that a vertex $k$ from $G$ is only connected to one other vertex $i$ or $j$ from $G^{\prime}$. Then $p=p^{\prime}$. This implies $a_{ii}=c_{ii}=1,\forall i\in\{1,,\dots,p\}$. The graphs $G$ and $G^{\prime}$ are then connected by a number of edges between vertices of same value. $\Box$

• (iii)

  $\lambda\neq 0$ is not an eigenvalue of $L^{\prime}$ and $L^{\prime}$ and ${\bar{\Delta}}$ share a common eigenvector $X^{\prime}$ for eigenvalues $\lambda^{\prime}$ and $\lambda-\lambda^{\prime}>0$.
  For $\lambda-\lambda^{\prime}=1$, a possibility is to connect a soft node of $G$ to $G^{\prime}$. For $\lambda-\lambda^{\prime}=p$ integer, a possibility is to connect $p$ soft nodes of $G$ to $G^{\prime}$.
  We conjecture that there are no other possibilities.

• (iv)

  $\lambda\neq 0$ is not an eigenvalue of $L^{\prime}$ and $L^{\prime}$ and ${\bar{\Delta}}$ have different eigenvectors. Then there is no solution to the eigenvalue problem (28).
  To see this, assume the eigenvalues and eigenvectors of $L^{\prime}$ and ${\bar{\Delta}}$ are respectively $\nu_{i},V^{i}$, $\mu_{i},W^{i}$ so that

  $$L^{\prime}V^{i}=\nu_{i}V^{i},~{}~{}{\bar{\Delta}}W^{i}=\mu_{i}W^{i},~{}~{}i=1,2,\dots n$$

  The eigenvectors can be chosen orthonormal and we have

  $$QV=WQ$$

  where $Q=(q_{k}^{j})$ is an orthogonal matrix, $V$ and $W$ are the matrices whose columns are respectively $V^{i}$ and $W^{i}$. We write

  $$W^{j}=\sum_{k}q_{k}^{j}V^{k}.$$

  Assuming $X^{\prime}$ exists, we can expand it as $X^{\prime}=\sum_{i}\alpha_{i}V^{i}$ Plugging this expansion intro the relation $({\bar{\Delta}}+L^{\prime})X^{\prime}=\lambda X^{\prime}$ yields

  $$\sum_{i}\left(\alpha_{i}\nu_{i}V^{i}+\alpha_{i}\sum_{j}q_{j}^{i}\mu_{j}\sum_{k}q_{k}^{j}V^{k}\right)=\sum_{i}\lambda\alpha_{i}\nu_{i}V^{i}$$

  Projecting on a vector $V^{m}$ we get

  $$\alpha_{m}\nu_{m}+\alpha_{m}\sum_{j}q_{j}^{m}\mu_{j}q_{m}^{j}=\lambda\alpha_{m}\nu_{m}$$

  A first solution is $\alpha_{m}=0,\forall m$ so that $X^{\prime}=0$, an articulation. If $\alpha_{m}\neq 0$ then we get the set of linear equations linking the $\nu_{i}$ to the $\mu_{i}$.

  $$\sum_{j}q_{j}^{m}\mu_{j}q_{m}^{j}=(\lambda-1)\nu_{m},~{}~{}m=1,\dots n$$

  Since $Q$ is a general orthogonal matrix, the terms $q_{j}^{m}$ are irrational in general. Therefore we conjecture that there are no solutions.