Timeline for Can the derivative of eigenvectors with respect to its components be taken as zero if all eigenvalues are equal?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Nov 18 at 21:20 | vote | accept | user544899 | ||
| Nov 12 at 22:53 | answer | added | Fred Hucht | timeline score: 2 | |
| Nov 12 at 22:32 | history | edited | user544899 | CC BY-SA 4.0 |
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| Nov 12 at 5:31 | history | edited | Daniele Tampieri | CC BY-SA 4.0 |
Minor typo fixes and grammar improvements
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| Nov 12 at 1:26 | review | Close votes | |||
| Nov 16 at 3:08 | |||||
| Nov 11 at 19:52 | comment | added | user544899 | @LiviuNicolaescu I am not sure I understand your comment, so I have edited the question and added my particular use case. I hope that makes my question clearer. Thanks | |
| Nov 11 at 19:50 | history | edited | user544899 | CC BY-SA 4.0 |
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| Nov 11 at 19:33 | comment | added | Liviu Nicolaescu | Suppose that the matrix is the identity matrix. All eigenvalues are equal to $1$ and any vector is an eigenvector. S you are asking what is the derivative of a vector with respect to its components. | |
| S Nov 11 at 19:24 | review | First questions | |||
| Nov 12 at 1:04 | |||||
| S Nov 11 at 19:24 | history | asked | user544899 | CC BY-SA 4.0 |