Talk:Eigenvalue perturbation

message: I have added two examples to explain that the perturbation of eigenvectors is complex in the case of repeated eigenvalues. I have added that the generalized eigenvalue problem arises in  mechanical vibrations.
BRousselet (talk) 09:23, 3 August 2022 (UTC)
thread_title:  Some improvements 

message: I don't have time to edit this article, but one can derive a formula for the perturbations of eigenvalue/eigenspaces to all orders using Theorem V.3.3 of Bhatia's matrix analysis book.  Note that if the distinct eigenvalues are {lamda_i} then the projection onto the eigenspace is f_i(A), where f_i is a function which is 1 in a neighborhood of lamda_i and 0 in neighborhoods of lamda_j, for j not equal to i. Then you can apply the formula in that book and get the derivative of the projection onto the eigenspace and consequently the derivative of the eigenvalue using the formula lamda_i = Tr(A x the projection onto the lamda_i eigenspace.)  You can repeat the proceedure and get the derivatives to all orders if one likes.  It would be better to simply find a reference that does this and cite it.  — Preceding unsigned comment added by 99.235.250.152 (talk) 03:29, 3 May 2013 (UTC)
thread_title: Reference

message: Theorem:  λi and xi, the generalized eigenvalues and eigenvectors of K over M, are real-analytic functions of K and M over the domain where
#K and M are symmetric matrices
#M is positive definite
#The eigenvalues λi are distinct.
Some of these lemmas may be described in our article holomorphic functional calculus.
Lemma 1:  If F is a real-analytic function defined on an open subset U of the reals, then F(X) (as defined in matrix function) is a real-analytic function of X, over the domain where all the eigenvalues of X are in U.
Lemma 2: The function taking the coefficients of an nth degree polynomial to its n roots, in order, is real-analytic over the domain where (1) the xn coefficient is non-zero, and the n roots are distinct.
Lemma 3:  Let S be the unit n-sphere, considered as column vectors.  Then the function taking x in S to x xT is analytic, and it has a locally analytic inverse function.
Proof of theorem:  We apply Lemma 1 to construct an analytic function M−1/2.  (We already knew that a positive semi-definite matrix has a unique principal square root.  The lemma is only needed to demonstrate the map is analytic.)  The generalized eigenvalues of K over M are the eigenvalues of  and the generalized eigenvectors xi of K over M are related to the eigenvectors of A by   (This reduction is only necessary because I don't know how to describe generalized Frobenius covariants.)
The eigenvalues λi are the roots of
:\left| \mathbf K - \lambda \mathbf M \right| = 0, 
or
:\left| \mathbf A - \lambda I \right| = 0,
so, by Lemma 2, the eigenvalues are analytic.
Noting the representation of the Frobenius covariants of a matrix A as a polynomial in A with coefficients rational functions of the eigenvalues, the Frobenius covariants of A are analytic in A.
Finally, we use Lemma 3 to demonstrate that the eigenvectors are locally analytic.
QED.
Of course, this would need to be in a reliable source to be included.  It seems to be the case that, even where the eigenvalues are equal, the generalized eigenvalues can be reordered to have directional derivatives in all directions (in 2n2-space), but that doesn't help with eigenvectors.  — Arthur Rubin  (talk) 11:58, 18 April 2015 (UTC)
replies: text: Thanks for the effort.  My understanding is that we need to show that the eigenvalues \lambda_i(\mathbf{A}, \mathbf{M}) are Lipschitz continuous, right? (which I believe they are, at least based on similarities with what Terry Tao has pointed out.)  Are you trying to show this fact? If we show this in your proof? Also, can you take a look at Matrix Perturbation Theory, page 15--10, Fact 1? It seems that is what we want. Taha (talk) 04:36, 19 April 2015 (UTC)
replies: text: I'm not sure Lipschitz continuity of the eigenvectors is adequate to allow the "existing" "proof" to be supported.  However, the rest of that "fact 1", seems to be the desired statement about generalized eigenvalues. — Arthur Rubin  (talk) 06:26, 19 April 2015 (UTC)
thread_title: Theorem and outline of proof

message: The page currently deals with the generalised eigenvalue problem. Usually, I'm not interested in the generalised version, but just the usual eigenvalue problem. Of course the generalised version is a generalised version and thus the usual eigenvalue problem is merely a special case - but I can't help thinking that both for my own reference and for new users, the article would be much more useful if it presented the results for the usual eigenvalue problem first, and then had a section on the generalised problem below it. 131.112.112.6 (talk) 08:51, 14 March 2016 (UTC)
thread_title:  Why not start with the non-generalised version? 

message: The proof applies to the obvious case where the perturbed Hilbert space and the original Hilbert space are the same. This is kind of obvious, but there is a physics example which kind of demonstrates this and it follows below.
My argument is based on physics reasoning. Consider the free motion Hamiltonian,
H = 1/(2*m) P^2 .
and make the quadratic potential as perturbation,
dH = 1/2 * m * omega * X^2.
Any non-zero omega would change the spectrum from continuous to discrete. 
Is it relevant that these are operators on a continuum (improper) basis?
The spectrum of the quantum harmonic oscillator is (semi-)infinite and discrete.  — Preceding unsigned comment added by 70.27.100.205 (talk) 01:07, 25 March 2016 (UTC)
thread_title:  Hilbert spaces? 

message: ... does not make sense to me.  There are 2 pairs of equations.  The first pair contains 2 almost identical equations, one having a coefficient of λ in the right hand term and the other not.  Either this is not true or I haven't understood something.  — Preceding unsigned comment added by 195.213.180.112 (talk) 14:26, 20 May 2022 (UTC)
thread_title:  Results section....