User eslam - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T09:05:27Z http://mathoverflow.net/feeds/user/12032 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/51205/the-eigenvalues-of-the-sum-of-two-nilpotent-matrices The eigenvalues of the sum of two nilpotent matrices Eslam 2011-01-05T15:06:33Z 2011-01-10T16:26:13Z <p>I have a matrix that is given by $A e^{i q} + A^* e^{-i q}$ with $A$ a nilpotent $n\times n$ matrix. The eigenvalues I get turn out always to be independent on q but I cannot prove it. I want to know how can I prove this (or if this alone is not sufficient for the eigenvalues to be independent on q)</p> http://mathoverflow.net/questions/51205/the-eigenvalues-of-the-sum-of-two-nilpotent-matrices/51655#51655 Answer by Eslam for The eigenvalues of the sum of two nilpotent matrices Eslam 2011-01-10T14:42:21Z 2011-01-10T16:26:13Z <p>I finally found out the property that resulted in the fact that eigenvalues in my case are independent of $q$. In my case, I found that $Tr((A e^{i q} + A^* e^{-i q})^m)$ is independent on $q$ for any $m$. In which case, it is not difficult to prove that the eigenvalues should be independent on $q$ having $\sum_i \lambda_i^m \frac{\partial \lambda_i }{\partial q}= 0$ for all $m$ we can form a linear combination of these equations such that $\sum_i P(\lambda_i) \frac{\partial \lambda_i }{\partial q}= 0$ where P is some polynomial that we can choose such that $P(\lambda_i) = \frac{\partial \lambda_i }{\partial q}$ given any $q$ and this way we can see that $\frac{\partial \lambda_i }{\partial q}$ vanishes. However, I am still interested to see how such a constraint can be implemented in David's solution (what does it correspond to regarding F).</p> http://mathoverflow.net/questions/51219/degeneracies-and-symmetry-group-representations Degeneracies and Symmetry group representations Eslam 2011-01-05T17:02:19Z 2011-01-05T17:02:19Z <p>If I have a matrix that has only one non-trivial symmetry and thus the irreducible representations of the symmetry group are two one-dimensional representations (the one that assigns 1 to both operations and the one that assigns one to the identity and -1 to the other symmetry operation) and if my symmetry group reduces in this case to $c_1 R_1 \oplus c_2 R_2$, where $R_1$ and $R_2$ are the two representations. What can I deduce about the number of eigenvalues and their degeneracies.</p> http://mathoverflow.net/questions/51219/degeneracies-and-symmetry-group-representations Comment by Eslam Eslam 2011-01-05T17:35:47Z 2011-01-05T17:35:47Z Thanks you for the reply. http://mathoverflow.net/questions/51205/the-eigenvalues-of-the-sum-of-two-nilpotent-matrices/51214#51214 Comment by Eslam Eslam 2011-01-05T17:31:47Z 2011-01-05T17:31:47Z the condition I added is not only that it is even but that only component which are multiples of $\frac{n}{2}$ are non-vanishing. http://mathoverflow.net/questions/51205/the-eigenvalues-of-the-sum-of-two-nilpotent-matrices/51214#51214 Comment by Eslam Eslam 2011-01-05T16:51:58Z 2011-01-05T16:51:58Z Thanks for the reply. Seemingly all the details of my particular problem has to do with the fact that my eigenvalues come out independent on q for all even dimensions I tried so far. For example one more thing I deduced is that in my problem (whether or not the matrices are nilpotent) the eigenvalues can be written as $\epsilon_k(q) = \sum_{l=0}^{\infty} a_{kl} cos(\frac{n}{2}lq)$ where n is the matrix dimension (which is always even) http://mathoverflow.net/questions/51205/the-eigenvalues-of-the-sum-of-two-nilpotent-matrices/51209#51209 Comment by Eslam Eslam 2011-01-05T15:58:03Z 2011-01-05T15:58:03Z thanks for your reply. Seemingly I forgot to add very important piece of information which is that $A e^{i q} + B e^{-i q}$ is hermitian (and thus $A$ and $B$ are hermitian conjugates of each other). Also in the case I am considering $n$ is always even (although I am not sure if this is relevant)