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In the special case of a finite simple graph, the adjacency matrix is a (0,1)-matrix with zeros on its diagonal. If the graph is undirected (i.e. all of its edges are bidirectional), the adjacency matrix is symmetric. The relationship between a graph and the eigenvalues and eigenvectors of its adjacency matrix is studied in spectral graph theory.
In spectral graph theory, an eigenvalue of a graph is defined as an eigenvalue of the graph's adjacency matrix, or (increasingly) of the graph's Laplacian matrix due to its discrete Laplace operator, which is either (sometimes called the combinatorial Laplacian) or / / (sometimes called the normalized Laplacian), where is a diagonal matrix with ...
In mathematics, in graph theory, the Seidel adjacency matrix of a simple undirected graph G is a symmetric matrix with a row and column for each vertex, having 0 on the diagonal, −1 for positions whose rows and columns correspond to adjacent vertices, and +1 for positions corresponding to non-adjacent vertices.
Let = be an positive matrix: > for ,.Then the following statements hold. There is a positive real number r, called the Perron root or the Perron–Frobenius eigenvalue (also called the leading eigenvalue, principal eigenvalue or dominant eigenvalue), such that r is an eigenvalue of A and any other eigenvalue λ (possibly complex) in absolute value is strictly smaller than r, |λ| < r.
The famous Cheeger's inequality from Riemannian geometry has a discrete analogue involving the Laplacian matrix; this is perhaps the most important theorem in spectral graph theory and one of the most useful facts in algorithmic applications. It approximates the sparsest cut of a graph through the second eigenvalue of its Laplacian.
Let A be a square n × n matrix with n linearly independent eigenvectors q i (where i = 1, ..., n).Then A can be factored as = where Q is the square n × n matrix whose i th column is the eigenvector q i of A, and Λ is the diagonal matrix whose diagonal elements are the corresponding eigenvalues, Λ ii = λ i.
The fundamental fact about diagonalizable maps and matrices is expressed by the following: An matrix over a field is diagonalizable if and only if the sum of the dimensions of its eigenspaces is equal to , which is the case if and only if there exists a basis of consisting of eigenvectors of .
Proof: Let D be the diagonal matrix with entries equal to the diagonal entries of A and let B ( t ) = ( 1 − t ) D + t A . {\displaystyle B(t)=(1-t)D+tA.} We will use the fact that the eigenvalues are continuous in t {\displaystyle t} , and show that if any eigenvalue moves from one of the unions to the other, then it must be outside all the ...