Timeline for Can we add two matrices by performing an operation on their eigenvalues & eigenvectors?
Current License: CC BY-SA 2.5
12 events
| when toggle format | what | by | license | comment | |
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| S Sep 11, 2010 at 18:05 | vote | accept | Edward | ||
| Sep 11, 2010 at 18:05 | vote | accept | Edward | ||
| S Sep 11, 2010 at 18:05 | |||||
| Sep 11, 2010 at 17:56 | answer | added | Bill Thurston | timeline score: 36 | |
| Sep 11, 2010 at 17:34 | answer | added | Dick Palais | timeline score: 17 | |
| Sep 11, 2010 at 16:27 | comment | added | darij grinberg | There are some deeply nontrivial bounds on the eigenvalues of $A+B$ for hermitian $A$ and $B$: see terrytao.wordpress.com/2010/01/12/… | |
| Sep 11, 2010 at 16:24 | history | edited | Edward |
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| Sep 11, 2010 at 13:17 | comment | added | Donu Arapura | If the matrices commute, then they can be into simultaneous upper triangular form, so the set eigenvalues of the sum is contained in set of sums of the eigenvalues of the original matrices. However, in general there isn't much you can say. You can try to experiment with a sum of a $2\times 2$ diagonal matrix and a generic matrix to convince yourself of that. | |
| Sep 11, 2010 at 12:59 | comment | added | Helge | And one should also specify what "find" means. | |
| Sep 11, 2010 at 12:59 | comment | added | Helge | The simple answer is "not simpler than diagonalizing the new matrix". The long answer is: "yes, but it depends on the problem at hand, and you do not provide enough information to even guess a good solution for your problem. Let me just say: There is eigenvalue perturbation theory. So if you know an eigenvalue of $A$ with eigenvector $\psi$. Then you approximately know an eigenvalue of $A + t B$ if $t$ is small enough (depending on various things)". | |
| Sep 11, 2010 at 10:20 | comment | added | user5810 | Well, you would also need to know what eigenvalues correspond to what eigenvectors ... | |
| Sep 11, 2010 at 10:19 | comment | added | Robin Chapman | From eigenvalues alone no. If you have eigenvalues and a full set of eigenvectors, you know the matrix. If there are not enough eigenevctors, then the answer will again be no. | |
| Sep 11, 2010 at 9:58 | history | asked | Edward | CC BY-SA 2.5 |