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Let $A \in M_{n}(\mathbb{Z}) $ be a matrix with $\det A \neq 0$ and with the following property: for any positive integer $k$ the equation $X^{k}=A$ has at least one solution in $M_{n}(\mathbb{Z})$. Prove that $A$ is the unit matrix.

A.M.M. Dec. 2008, Problem 11401 , author: Marius Cavachi

Claudiu Mindrila
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1 Answers1

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The same property descents to any $M_n(\mathbb Z/p\mathbb Z)^*$, whenever $p$ is co-prime to $\det A$.

These are all finite groups, and in a finite group $G$, this property can only hold for the identity element, because $A = X^{|G|} = 1$.

Thus $A$ is the identity matrix modulo all but finitely many primes, then it is clearly also the identity matrix in $\mathbb Z$.

MooS
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    SImple and elegant argument! – Crostul Sep 21 '17 at 13:31
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    Note that you have to throw away those primes, that divide $\det(A)$. But it still works, because we still have infinitely many primes left. – MooS Sep 21 '17 at 13:32
  • Well, such a matrix must have determinant $1$ because $\det A = ( \det X )^k$ for all $k$. – Crostul Sep 21 '17 at 13:33
  • Yes, so you either argue that way or you throw away all bad primes (While it turns out that there are no bad primes anyway). – MooS Sep 21 '17 at 13:35
  • @Crostul Actually I like it more to throw away bad primes, because when you argue $\det A=1$, you basically use the (obvious) base case $n=1$. My proof however does it all in one go without using the base case.

    Nevertheless, this is of course just a matter of taste.

     – MooS Sep 21 '17 at 13:43