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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 ...
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.
Suppose the eigenvectors of A form a basis, or equivalently A has n linearly independent eigenvectors v 1, v 2, ..., v n with associated eigenvalues λ 1, λ 2, ..., λ n. The eigenvalues need not be distinct. Define a square matrix Q whose columns are the n linearly independent eigenvectors of A,
Once the eigenvalues are computed, the eigenvectors could be calculated by solving the equation (), = using Gaussian elimination or any other method for solving matrix equations. However, in practical large-scale eigenvalue methods, the eigenvectors are usually computed in other ways, as a byproduct of the eigenvalue computation.
To solve this particular ordinary differential equation system, at some point in the solution process, we shall need a set of two initial values (corresponding to the two state variables at the starting point). In this case, let us pick x(0) = y(0) = 1.
Here it is assumed that floating point operations are optimally rounded to the nearest floating point number. 2. The upper triangle of the matrix S is destroyed while the lower triangle and the diagonal are unchanged. Thus it is possible to restore S if necessary according to for k := 1 to n−1 do !
The following discussion uses the simplest case, where the system has two lumped springs and two lumped masses, and only two mode shapes are assumed. Hence M = [m 1, m 2] and K = [k 1, k 2]. A mode shape is assumed for the system, with two terms, one of which is weighted by a factor B, e.g. Y = [1, 1] + B[1, −1].
The Lanczos algorithm is most often brought up in the context of finding the eigenvalues and eigenvectors of a matrix, but whereas an ordinary diagonalization of a matrix would make eigenvectors and eigenvalues apparent from inspection, the same is not true for the tridiagonalization performed by the Lanczos algorithm; nontrivial additional steps are needed to compute even a single eigenvalue ...