Peter, I would prefer something specialised to quadratics.

Dear Federico, thanks a lot for your suggestion. I will try to provide a proof for this intuitive result.

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.

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 $\Omega$.

Thanks Louis for your reply. This was my first strategy to attack the problem. For sure, it is very simple and fast, but it is based on fixing the eigenvalues that are potentially equal to a value that I don't know. If from the statistical test I get the result that $k$ eigenvalues are equal, what shoud I do, fixing them to the average of these $k$ eigenvalues? Maybe, an iterative procedure could help in soving the problem, but I was wondering whether there was some analytical result in matrix algebra. Or maybe, some specific algorithm already treated in the literature.

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.