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Mar 7, 2021 at 11:05 history bumped CommunityBot This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Feb 5, 2021 at 11:31 comment added Dima Pasechnik Algebraically, the repeated eigenvalue is present iff the discriminant of the characteristic polynomial $f(t)$ of $\Omega$ vanishes identically. It means that $f$ and $f'$ have a common root. More generally, eigenvalue repeats $k$ times iff $f$ and its $k$th derivative $f^{(k)}$ have a common root. One doesn't need roots, of course, one can use resultants instead.
Feb 5, 2021 at 10:15 answer added Federico Poloni timeline score: 1
Feb 5, 2021 at 10:04 comment added Federico Poloni What kind of iterative algorithm do you plan to start once you have set those close eigenvalues to their mean? At that point diagonalizing $\tilde{\Omega}$ will give you back the same eigenvectors and eigenvalues, so I don't think there are further improvements that you can make.
Feb 5, 2021 at 9:44 comment added meie73 Yes, the test suggests $k$, and those eigenvalues that are (statistically) equal. Specifically, I know these eigenvalues are equal (each other), but I don't know to which value. My first idea was to fix these eigenvalues to their mean and start an iterative algorithm, substantially based on repeating the suggestion made by Louis here below. However, I don't know whether it is stable and whether it returns the closest matrix to $\Omega$ according to some metric.
Feb 5, 2021 at 9:26 comment added Federico Poloni But then your test tells you $k$, at least? So the problem is, given $k$ and $\Omega$, find the closest matrix to $\Omega$ (in some metric) with $k$ repeated eigenvalues?
Feb 5, 2021 at 9:22 comment added meie73 I don't know anything more that the matrix $\Omega$. Essentially, it is an estimated covariance matrix of a multivariate model. But if some eigenvalues are equal, I might have problems in terms of identification of certain parameters of the model bla bla bla. Let a statistical test says $k$ eigenvalues, that look like very similar, are effectively equivalent (from a statistical point of view). I would like to see the consequences on my identification problem when the eigenvalues are exactly the same. For this reason, I would like to work with the "closest" matrix to the original $\Omega$.
Feb 5, 2021 at 9:08 comment added Federico Poloni This sounds more like a modelling problem. What do you know, apart from the matrix entries? Do you have a 'typical' level of noise, for instance? What would you do if the eigenvalues are logarithmically distributed, e.g., $diag(1, 1.1, 1.11, 1.111, 1.11111, ...)$?
Feb 5, 2021 at 9:02 comment added meie73 Hi Federico, it is exactly the point. Actually, a priori I don't know how many eigenvalues are equal. I am interested in a very general strategy for obtaining such $\tilde{\Omega}$, any metric could be justifiable.
Feb 4, 2021 at 21:47 comment added Federico Poloni So you are looking for the closest matrix to a given one with repeated eigenvalues? How many? In which metric?
Feb 4, 2021 at 17:57 answer added Louis Deaett timeline score: 0
Feb 4, 2021 at 17:41 review First posts
Feb 4, 2021 at 18:35
Feb 4, 2021 at 17:37 history asked meie73 CC BY-SA 4.0