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Thus the elements of the spectrum are precisely the eigenvalues of T, and the multiplicity of an eigenvalue λ in the spectrum equals the dimension of the generalized eigenspace of T for λ (also called the algebraic multiplicity of λ). Now, fix a basis B of V over K and suppose M ∈ Mat K (V) is a matrix.
Given an n × n square matrix A of real or complex numbers, an eigenvalue λ and its associated generalized eigenvector v are a pair obeying the relation [1] =,where v is a nonzero n × 1 column vector, I is the n × n identity matrix, k is a positive integer, and both λ and v are allowed to be complex even when A is real.l When k = 1, the vector is called simply an eigenvector, and the pair ...
Similarly, the geometric multiplicity of the eigenvalue 3 is 1 because its eigenspace is spanned by just one vector []. The total geometric multiplicity γ A is 2, which is the smallest it could be for a matrix with two distinct eigenvalues. Geometric multiplicities are defined in a later section.
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.
Note that there are 2n + 1 of these values, but only the first n + 1 are unique. The (n + 1)th value gives us the zero vector as an eigenvector with eigenvalue 0, which is trivial. This can be seen by returning to the original recurrence. So we consider only the first n of these values to be the n eigenvalues of the Dirichlet - Neumann problem.
As stated in the introduction, for any vector x, one has (,) [,], where , are respectively the smallest and largest eigenvalues of .This is immediate after observing that the Rayleigh quotient is a weighted average of eigenvalues of M: (,) = = = = where (,) is the -th eigenpair after orthonormalization and = is the th coordinate of x in the eigenbasis.
The equation above formulates an eigenvalue problem. Any eigenvector for T spans a 1-dimensional invariant subspace, and vice-versa. In particular, a nonzero invariant vector (i.e. a fixed point of T ) spans an invariant subspace of dimension 1.
Naively, if at each iteration one solves a linear system, the complexity will be k O(n 3), where k is number of iterations; similarly, calculating the inverse matrix and applying it at each iteration is of complexity k O(n 3). Note, however, that if the eigenvalue estimate remains constant, then we may reduce the complexity to O(n 3) + k O(n 2 ...