Eigen problem with constrained (equal) eigenvalues Let $\Omega$ be a symmetric and positive definite matrix. From a test of hypothesis I know that some eigenvalues are likely to be equal (the test also suggests which eigenvalues). Do you have any suggestions for obtaining the matrix, say $\tilde{\Omega}$, that can be obtained by fixing those eigenvalues to be equal? In other words, I would expect a matrix, quite "close" to $\Omega$, but not exactly the same.
Any suggestions is really appreciated.
 A: The average (centroid) $\lambda = \frac{\lambda_1 + \dots + \lambda_k}{k}$ minimizes the sum of squared differences $f(\lambda) = \sum_{i=1}^k (\lambda - \lambda_i)^2$.
This suggests an algorithm, which is essentially what Louis Deaett suggests, if I understand correctly:

*

*compute the diagonalization $\Omega = V\Lambda V^{-1}$.

*permute eigenvalues and eigenvectors so that the sum of squared distances from the centroid $f(\lambda)$ is minimized by the first $k$ eigenvalues. It is not clear how to do this step in optimal time, even if the solution may be evident in the 'eyeball norm' in many practical cases.

*replace those $k$ eigenvalues with their mean, to obtain $\tilde{\Omega} = V\tilde{\Lambda}V^{-1}$.

I conjecture that this algorithm gives you the optimal answer in the Frobenius norm $\|\Omega - \tilde{\Omega}\|_F^2 =  \sum_{i,j} (\Omega_{ij} - \tilde{\Omega}_{ij})^2$; based on experience on similar problems, it looks like Weyl's inequalities for eigenvalues could be used to give a proof.
A: I'm not sure if this is what you're after, but your positive definite (symmetric) matrix is going to be diagonalized by some unitary matrix $U$.  So $U^*\Omega U=\Lambda$ for some diagonal matrix $\Lambda$ of its eigenvalues.  This gives $U\Lambda U^*= \Omega$.
It sounds like perhaps what you are after is $\tilde\Omega=U\tilde\Lambda U^*$, where the difference between $\Lambda$ and $\tilde\Lambda$ is that in $\tilde\Lambda$, the two eigenvalues that were ``close'' have been replaced by two values that are equal.
