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Eigenvalue problems occur naturally in the vibration analysis of mechanical structures with many degrees of freedom. The eigenvalues are the natural frequencies (or eigenfrequencies) of vibration, and the eigenvectors are the shapes of these vibrational modes.
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.
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 ...
In numerical linear algebra, the Jacobi eigenvalue algorithm is an iterative method for the calculation of the eigenvalues and eigenvectors of a real symmetric matrix (a process known as diagonalization).
There is an eigenvalue bound for independent sets in regular graphs, originally due to Alan J. Hoffman and Philippe Delsarte. [ 13 ] Suppose that G {\displaystyle G} is a k {\displaystyle k} -regular graph on n {\displaystyle n} vertices with least eigenvalue λ m i n {\displaystyle \lambda _{\mathrm {min} }} .
The eigenvalues of A must also lie within the Gershgorin discs C j corresponding to the columns of A. Proof. Apply the Theorem to A T while recognizing that the eigenvalues of the transpose are the same as those of the original matrix. Example. For a diagonal matrix, the Gershgorin discs coincide with the spectrum. Conversely, if the Gershgorin ...
For a free particle, the Hamiltonian = is invariant under translations. Translation commutes with the Hamiltonian: [ H , T ^ ] = 0 {\displaystyle [H,\mathbf {\hat {T}} ]=0} . However, if we express the Hamiltonian in the basis of the translation operator, we will find that H {\displaystyle H} has doubly degenerate eigenvalues.
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.