4
$\begingroup$

In a homological algebra problem I am in the situation that I have an invertible (over $\mathbb{Z}$) integer matrix $X$ and a permutation matrix $Y$ such that $N:=XY$ is a matrix with all eigenvalues equal to 1 or -1.

Question 1 : Does such a situation appear already in other situations/fields in algebra/combinatorics? Is there an interpretation for this or does it have a combinatorial meaning?

Question 2: Do (invertible integer) matrices with all eigenvalues equal to 1 or -1 have a name, or do matrices $X$ as above have a name? Are they studied in the literature?

Improve this question
$\endgroup$
10
  • 6
    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Sep 23, 2020 at 9:23
  • 2
    $\begingroup$ 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"). $\endgroup$
    – Carlo Beenakker
    Commented Sep 23, 2020 at 9:54
  • 3
    $\begingroup$ @CarloBeenakker OP's matrix has integer entries, if I understand correctly, so it cannot be unitary apart from trivial cases (signed permutation matrices). $\endgroup$
    – Federico Poloni
    Commented Sep 23, 2020 at 10:03
  • 2
    $\begingroup$ 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 $\pm 1$. $\endgroup$
    – Geoff Robinson
    Commented Sep 23, 2020 at 20:38
  • 1
    $\begingroup$ 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. $\endgroup$
    – Geoff Robinson
    Commented Jan 12, 2023 at 20:27

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .