All Questions
7 questions
1
vote
1
answer
198
views
Directed graph whose adjacency matrix admits only 0 as eigenvalue
Let $G$ be a directed graph and let $P_i$
be its vertices. Let $A$
be the corresponding adjacency matrix of $G$, i.e. $a_{i,j}=1$
if and only if there is a directed edge from $P_i$
to $P_j$, ($a_{i,...
0
votes
0
answers
327
views
Eigenvector to zero eigenvalue of general Laplacian
I was wondering what we can say about the eigenvectors of a matrix $A$ fullfilling $Ax =0$ where $A$ is symmetric with a diagonal equal to one and every row sums up to 0. Obviously this is a ...
0
votes
1
answer
97
views
If there are eigenvectors with largest components $i$ resp. $j$, then is there an eigenvector with two largest components $i$ and $j$?
Let $G=(V,E)$ be a connected (finite simple) graph with vertex set $V=\{1,...,n\}$ and let $\theta_2\in\Bbb R$ be the second-largest eigenvalue of its adjacency matrix. I wonder about the following ...
1
vote
2
answers
153
views
From one eigenvector to many, in a very local graph?
Let $\Gamma$ be an undirected graph of bounded degree $d$ with $V = \{1,2,\dotsc,N\}$ as its set of vertices, and edges only between vertices that are at a distance $\leq M$ apart (where $M$ is much ...
1
vote
1
answer
2k
views
Eigenvalues of adjacency matrix of a k-regular graph
If $A_G$ is the adjacency matrix of a k-regular graph, let $B = J+xA_G$, where J is the matrix whose elements are all 1s and $x\in R$ is a scalar. If $\lambda_1\geq\lambda_2\geq \dots \geq \lambda_n$ ...
1
vote
0
answers
158
views
Interpreting (Fiedler) spectral bisectioning
I would appreciate help on how to interpret the results of spectral bisectioning of a graph.
Given a $G=(V,E)$ with size $N$ represented by $Q$ its Laplacian matrix where the eigenvalues are ordered ...
1
vote
0
answers
138
views
Relationship betwen eigenvectors
Suppose that we have two matrices A and B. Matrix B is taken from A with one row and one column deleted. On the other hand A is n*n matrix and B is (n-1)*(n-1) matrix and is created by deleting last ...