Timeline for How to check if a vector is an eigenvector of a matrix using mathematica?
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2020 at 0:06 | comment | added | J. M.'s missing motivation♦ | Yes, that's good for matrices, but don't use that for vectors. | |
| Apr 23, 2020 at 23:45 | comment | added | Neil_P | I'm using ctr+Enter and ctr+, to make a matrix within Mathematica, and passing the values that way. I haven't seen the link you sent, thank you for the info. | |
| Apr 23, 2020 at 18:41 | comment | added | J. M.'s missing motivation♦ |
With[{m = {{4, 5}, {-2, 6}}, v = {1 - 3 I, 2}}, MatrixRank[{m.v, v}] == 1] gives True. Also, why are you inputting eigenvectors in a format like {{1 - 3 I}, {2}}? Have you already seen this?
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| Apr 23, 2020 at 18:27 | comment | added | Neil_P | @J.M., Actually, with some more experimenting (and more Diffeq homework), Carl Woll's answer doesn't work for the matrix {{4, 5},{-2, 6}} with the eigenvector{{1 - 3:ii:},{2}}. It doesn't seem to work with any vector with complex eigenvectors | |
| Apr 22, 2020 at 16:42 | comment | added | Neil_P | @J.M. For whatever reason, I had trouble with it a week ago, but not today. Even with the same matrices. I'll delete this post in a bit. | |
| Apr 20, 2020 at 21:17 | vote | accept | Conor Cosnett | ||
| Apr 20, 2020 at 21:17 | |||||
| Apr 17, 2020 at 3:58 | comment | added | J. M.'s missing motivation♦ | If you think about it, the rank of a matrix ought to be the same whether transposed or not. Can you give an example of a matrix-vector pair where Carl's version fails, but yours works? | |
| Apr 17, 2020 at 3:57 | history | edited | J. M.'s missing motivation♦ | CC BY-SA 4.0 |
added 10 characters in body
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| Apr 16, 2020 at 21:37 | review | Low quality posts | |||
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| Apr 16, 2020 at 20:00 | review | Late answers | |||
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| Apr 16, 2020 at 19:48 | review | First posts | |||
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| Apr 16, 2020 at 19:43 | history | answered | Neil_P | CC BY-SA 4.0 |