Timeline for Can the eigenvalues of a real symmetric tensor be complex?
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
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| Apr 24, 2021 at 20:09 | comment | added | Carlo Beenakker | understood, I will delete my last comment. | |
| Apr 24, 2021 at 19:59 | comment | added | Zach Teitler | A lot of this discussion was my repeated mistakes, and you and others correcting me. I'm sorry if I acted rashly, but I didn't feel like my thoughtless errors were a positive contribution, and frankly I'm a little embarrassed. | |
| Apr 24, 2021 at 19:52 | comment | added | Zach Teitler | I deleted my comments. | |
| Feb 18, 2021 at 5:09 | comment | added | Matt | I would be interested in a definition of eigenvalues and eigenvectors that entails only real eigenvalues for a symmetric tensor. If there is any references regarding eigenvalues of tensors you find interesting I would be grateful to hear more about it. | |
| Feb 17, 2021 at 1:08 | comment | added | Michael Engelhardt | @ZachTeitler - OP is explicitly interested in the paper he cites, and there the normalization happens to be $x^2 =1$. We may deplore it, but that's what the OP is looking for. | |
| Feb 17, 2021 at 0:43 | vote | accept | Matt | ||
| Feb 16, 2021 at 16:16 | comment | added | LSpice | Ah, good, I missed the $\sum x_i^2 = 1$ condition. But still, doesn't this counterexample only show that the argument is wrong, not explicitly (yet) that a real symmetric tensor can have complex eigenvalues? | |
| Feb 16, 2021 at 15:30 | comment | added | Carlo Beenakker | ah, that is a well known issue, there is no unique definition of an eigenvalue/eigenvector for a tensor; I am not sure one definition is more natural than another. | |
| Feb 16, 2021 at 15:28 | comment | added | Michael Engelhardt | @ZachTeitler - the paper cited by the OP explicitly stresses that $x^2 =1$ is intended, not $\bar{x} x=1$. | |
| Feb 16, 2021 at 15:13 | comment | added | Carlo Beenakker | @LSpice --- in that "cheap" example, the condition $\sum_i x_i^2=1$ forces $\lambda=\pm 1$; and it would also force $t=\pm 1$ in Zach Teitler's example. | |
| Feb 16, 2021 at 15:11 | comment | added | LSpice | @ZachTeitler, ah, good point! So, for example, a much cheaper example: even just $T_{111} = 1$ and all other entries equal $0$ has $\sum T_{i j k}x_k x_j = \lambda x_i$, where $x_1 = \lambda$ and all other entries equal $0$. | |
| Feb 16, 2021 at 14:57 | comment | added | Carlo Beenakker | well, the argument provided by the OP implies both real eigenvalues and real eigenvectors for any real symmetric $T$; this counterexample invalidates that, doesn't it? | |
| Feb 16, 2021 at 14:53 | comment | added | LSpice | Isn't the original question about a complex eigenvalue, not just a complex eigenvector with real eigenvalue? | |
| Feb 16, 2021 at 14:52 | history | edited | LSpice | CC BY-SA 4.0 |
Link to @lambda's comment
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| Feb 16, 2021 at 12:15 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
added 162 characters in body
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| Feb 16, 2021 at 12:09 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |