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Questions related to the spectrum of graphs, defined using one of the possible variants of the discrete Laplace operator or Laplacian matrix. See https://en.wikipedia.org/wiki/Discrete_Laplace_operator
17
votes
Accepted
Are these three different notions of a graph Laplacian?
These are usually known as the Laplacian, the normalized Laplacian and the unsigned Laplaian. All three are positive semidefinite. If the graph is regular, they all provide the same information.
If t …
14
votes
Accepted
When the Lovász theta-function saturates its upper bound
Suppose $G$ is a $k$-regular graph on $n$ vertices, with least eigenvalue $\tau$.
Lovasz proved that
$$
\theta(G) \le \frac{n}{1-\frac{k}{\tau}}.
$$
Further if the automorphism group of $G$ acts …
9
votes
Accepted
Complex Eigenvalues of Directed Graphs
Let $D$ be the Paley tournament on seven vertices. Its vertices are the integers mod seven
and there is an arc from $i$ to $j$ is $j-i$ is a non-zero square mod seven. The characteristic polynomial of …
9
votes
Accepted
Spectrum of an adjacency matrix
Since the eigenvalues are real, and since their sum is the trace of $A$, which is zero, we see that either all eigenvalues are zero, or there are both positive and negative eigenvalues. So no non-empt …
8
votes
Matching polynomials and Ramanujan graphs
One approach that goes some way to explaining this is through the path-tree of a graph. This is defined as follows. Choose a vertex $u$ in the graph $G$, The vertices of the path-tree $T(G,u)$ are the …
8
votes
Connection between eigenvalues of matrix and its Laplacian.
Essentially your question is equivalent to asking for the relation between the spectrum of $A+D$ and $A$, where are $A$ is symmetric, $D$ is diagonal and both matrices are real.
And, by change of basi …
7
votes
Connectivity of weighted graph and zero Laplacian eigenvalues
Start with the unweighted case. We have $L=BB^T$ where $B$ is (what I call) the incidence matrix of an orientation of $G$. So $B$ has one 1 and one $-1$ in each column with all other entries zero. The …
7
votes
The first eigenvalue of a graph - what does it reflect?
(This is just an overlong comment.)
A basic problem is that the complete bipartite graphs $K_{1,ab}$ and $K_{a,b}$ have the same spectral radius, and these graphs would not usually be viewed as simil …
7
votes
Accepted
When does graph Laplacian have eigenvalue -1?
We have $\det(tI-L) =\det(D^{-1}(tD-A))$. The matrix $D-A$ is positive semidefinite; it is the usual Laplacian in graph theory. The matrix $A+D$ is also positive semidefinite, and if the underlying gr …
6
votes
Integral roots of circulant matrix
Let $C$ be circulant of order $n\times n$. The first row of $C$ defines a complex function from the cyclic group $\mathbb{Z}_n$; denote its value on $i$ by $\rho(i)$.
Define two elements $a$ and $b$ o …
6
votes
What is the state of the art on triangle-free strongly regular graphs?
The complete bipartite graphs $K_{n,n}$ are strongly regular and triangle-free.
This nitpicking aside, your summary is accurate.
6
votes
Accepted
spectrum and degree sequence
No. One simple class of examples are Latin square graphs. If $L$ is an $n\times n$ Latin
square with entries from $\{1,\ldots,n\}$, the vertices of Latin square graph are the $n^2$
triples; two triple …
5
votes
About the second largest adjacency eigenvalue of Abelian Cayley graphs
Let $M$ be a $d\times m$ matrix over $GF(2)$ and let $X(M)$ be the graph on the binary vectors of length $d$, where two vectors are adjacent in their difference is a column of $M$. (This is a Cayley g …
5
votes
Repeated Second Eigenvalue of the Adjacency Matrix of a Graph
It you allow weighted adjacency matrices and if you insist (among there things) that the eigenspace associated to $\lambda_2$ satisfies the "strong Arnold condition", then you are dealing the Colin d …
5
votes
Accepted
Characteristic polynomial of hypercube graph
View the vertices as elements of $\mathbb{Z}^n$. If $a\in\mathbb{Z}^n$, define a function $f_a$ on the vertices by
$$
f_a(x) = (-1)^{a^Tx}.
$$
This function is an eigenvectors and if $a$ has weight …