# Find an $n\times n$ integer matrix with determinant 1 and $n$ distinct eigenvalues

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.

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

• 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). – derpy Apr 14 '14 at 16:31
• Oh cool I wasn't able to see a nice way to take care of the even case. This works well. – rVitale Apr 14 '14 at 16:36
• Yours is the credit for finding an elegant solution. :) – derpy Apr 14 '14 at 16:38
• Thanks! let me add your idea to the answer, so we have a full answer. – rVitale Apr 14 '14 at 16:40
• @derpy: Note that your $A$ has determinant $1$ because its eigenvalues occur in complex conjugate pairs ( and none are real). – 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.