Timeline for Reference for matrices with all eigenvalues 1 or -1
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 12, 2023 at 20:27 | comment | added | Geoff Robinson | You work by induction on $n$. Identify the $n$-long integerr column vectors with $\mathbb{Z}^{n}$. We may choose a non zero integral column vector $v$ with $Nv = \pm v$, and we may arrange so that the entries of $v$ have gcd 1, which means that $v$ may be extended to a $\mathbb{Z}$-basis for the integer column vectors. This reduces the problem to rank $n-1$ with a little thought. | |
| Jan 12, 2023 at 11:48 | comment | added | ghc1997 | @GeoffRobinson Thank you, could you tell me where I can find the proof for this result: an invertible integral matrix with all eigenvalue ±1 is conjugate to an upper triangular matrix with all diagonal entries ±1 via an integral matrix | |
| Jan 12, 2023 at 11:28 | comment | added | Geoff Robinson | I meant that the matrix $N$ is conjugate in the above way to an upper triangular matrix with all diagonal entries $\pm1.$ | |
| Jan 11, 2023 at 20:02 | comment | added | ghc1997 | Hi @GeoffRobinson, would it be okay for you to provide a reference for your result? (I found it a bit difficult to prove, as such a matrix is not similar to its Jordan normal form via an invertible integral matrix in general) | |
| Jun 20, 2021 at 16:01 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Feb 20, 2021 at 15:04 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Oct 23, 2020 at 15:04 | comment | added | user44191 | It's not hard to construct two projections preserving the generalized eigenspaces (by adding/subtracting $I_n$ and dividing by $2$/$-2$ if your matrix is diagonalizable), so projections may be a place to start. | |
| Oct 23, 2020 at 14:02 | history | bumped | CommunityBot | This question has answers that may be good or bad; the system has marked it active so that they can be reviewed. | |
| Sep 23, 2020 at 20:38 | comment | added | Geoff Robinson | Your matrix is conjugate (via an invertible integral matrix) to an upper triangular matrix with each main diagonal entry $\pm 1$. I'm not sure that much more can be said, since any matrix with that property has all eigenvalues $\pm 1$. | |
| Sep 23, 2020 at 10:03 | comment | added | Federico Poloni | @CarloBeenakker OP's matrix has integer entries, if I understand correctly, so it cannot be unitary apart from trivial cases (signed permutation matrices). | |
| Sep 23, 2020 at 9:54 | comment | added | Carlo Beenakker | Hermitian unitary matrices have all eigenvalues equal to $\pm 1$; in the physics context, these are studied as scattering matrices of systems with a chiral symmetry (the trace of this matrix then counts the number of topologically protected "zero-modes"). | |
| Sep 23, 2020 at 9:30 | history | edited | Mare | CC BY-SA 4.0 |
added 41 characters in body
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| Sep 23, 2020 at 9:26 | comment | added | Mare | @YCor Thanks, this is already helpful for me. | |
| Sep 23, 2020 at 9:23 | comment | added | YCor | A square matrix with only 1 as eigenvalue (in every field extension) is called unipotent. So $M$ has all eigenvalues $\pm 1$ iff $M^2$ is unipotent. I don't know it this has a name. | |
| Sep 23, 2020 at 9:20 | history | asked | Mare | CC BY-SA 4.0 |