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A complete basis is formed by augmenting the eigenvectors with generalized eigenvectors, which are necessary for solving defective systems of ordinary differential equations and other problems. An n × n {\displaystyle n\times n} defective matrix always has fewer than n {\displaystyle n} distinct eigenvalues , since distinct eigenvalues always ...
It then follows that the eigenvectors of A form a basis if and only if A is diagonalizable. A matrix that is not diagonalizable is said to be defective. For defective matrices, the notion of eigenvectors generalizes to generalized eigenvectors and the diagonal matrix of eigenvalues generalizes to the Jordan normal form.
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 ...
Notation: The index j represents the jth eigenvalue or eigenvector. The index i represents the ith component of an eigenvector. Both i and j go from 1 to n, where the matrix is size n x n. Eigenvectors are normalized. The eigenvalues are ordered in descending order.
For example, the fourth-order Hilbert matrix has a condition of 15514, while for order 8 it is 2.7 × 10 8. Rank A matrix A {\displaystyle A} has rank r {\displaystyle r} if it has r {\displaystyle r} columns that are linearly independent while the remaining columns are linearly dependent on these.
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.
In linear algebra, a generalized eigenvector of an matrix is a vector which satisfies certain criteria which are more relaxed than those for an (ordinary) eigenvector. [1]Let be an -dimensional vector space and let be the matrix representation of a linear map from to with respect to some ordered basis.
For example, the eigenstate of ^ corresponding to the eigenvalue can be labelled as | . Such an observable is itself a self-sufficient CSCO. Such an observable is itself a self-sufficient CSCO. However, if some of the eigenvalues of a n {\displaystyle a_{n}} are degenerate (such as having degenerate energy levels ), then the above result no ...