Pretty much what the title suggests: for any positive integer $n$, I'm looking for an $n$-by-$n$ matrix with integer entries, determinant $1$ and $n$ eigenvalues.

In case it is absolutely useless to come up with such a matrix, I'm looking for a proof that such a matrix exists.

  • 1
    $\begingroup$ Have you tried looking at diagonal matrices? $\endgroup$ – LASV Apr 14 '14 at 15:57
  • $\begingroup$ A simple example would be $\begin{pmatrix}0 & 1\\ -1 & 0 \end{pmatrix}$ $\endgroup$ – ah11950 Apr 14 '14 at 15:58
  • 1
    $\begingroup$ @LASV The entries have to be integers, so that won't work. $\endgroup$ – user142299 Apr 14 '14 at 16:05
  • $\begingroup$ Ah, I should have added that it has to be for arbitrary $n$. I'll do that. $\endgroup$ – Bryder Apr 14 '14 at 16:08
  • $\begingroup$ The matrix entries have to be integers, but not the eigenvalues, right? $\endgroup$ – Robert Lewis Apr 14 '14 at 16:13

When $n$ is odd use the $n \times n$ matrix for the cyclic permutation $(a_1,a_2,...,a_n) \rightarrow (a_n,a_1,...,a_{n-1})$.. Then $A^n=I$ and since $n$ is odd, $det(A)=1$ and the entries of $A$ are all $1$ or $0$. The eigenvalues are the distinct $n$'th roots of unity (it has characteristic polynomial $x^n-1$). If you allow determinant $-1$ this will work in the even case as well.

By derpy's suggestion below we can do the odd and even case at once using the matrix for the map:

$(a_1,a_2,...,a_n) \rightarrow ((-1)^{n+1}a_n,a_1,...,a_{n-1})$.

Then when $n$ is odd we get the cyclic permutation, and the quasi-cyclic one for $n$ even. In each case $det(A)=1$ and the eigenvalues are distinct roots of $\pm 1$.

  • 2
    $\begingroup$ I think in the even case you could simply replace one of the 1's with a -1. This corresponds to the "quasi"-permutation $ A : (a_1,a_2,\dotsc,a_{n-1},a_n) \mapsto (-a_n,a_1,\dotsc,a_{n-2},a_{n-1}) $, so that $ A^n = -I $, and since $ n $ is even we have $ \det\left(-I\right) = 1 $. The eigenvalues are still distinct (the $n$ roots of -1). $\endgroup$ – derpy Apr 14 '14 at 16:31
  • $\begingroup$ Oh cool I wasn't able to see a nice way to take care of the even case. This works well. $\endgroup$ – rVitale Apr 14 '14 at 16:36
  • $\begingroup$ Yours is the credit for finding an elegant solution. :) $\endgroup$ – derpy Apr 14 '14 at 16:38
  • $\begingroup$ Thanks! let me add your idea to the answer, so we have a full answer. $\endgroup$ – rVitale Apr 14 '14 at 16:40
  • $\begingroup$ @derpy: Note that your $A$ has determinant $1$ because its eigenvalues occur in complex conjugate pairs ( and none are real). $\endgroup$ – Geoff Robinson Apr 14 '14 at 16:46

Take any representation of degree $n$ of any symmetric group $S_m$. All the matrix entries will be integers but determinant could be $\pm 1$. As suggested by others, we can change all the signs in the first row, if needed, and get integer matrices of determinant $+1$. As all matrices are of finite order, they will be diagonalizable and hence have $n$ eigenvalues.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.