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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.

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

2 Answers 2

up vote 3 down vote accepted

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 $\pm 1$.

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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 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 at 16:36
    
Yours is the credit for finding an elegant solution. :) –  derpy Apr 14 at 16:38
    
Thanks! let me add your idea to the answer, so we have a full answer. –  rVitale Apr 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 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.

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