A rational "fifth root" of the scalar matrix $2I$

Question

I am working on the following problem from a past exam.

Find a necessary and sufficient condition for there to exist a square matrix $A$ of order $n$ whose entries are all rational, such that $A^5 = 2I_n$, where $I_n$ stands for the identity matrix of order $n$.

By an argument using determinants, one easily sees that $n\in 5\mathbb Z$ is a necessary condition. Moreover, if we assume such an $A =: A_5$ exists for $n=5$, then for general $n = 5k$, $A = \bigoplus^kA_5$ is what we want.

Thus the problem boils down to whether there exists a $5\times5$ rational matrix $A$ such that $A^5 = 2I_5$. What I would like to ask is whether such an $A$ exists, and how one can find such $A$ by hand most efficiently. I would be grateful for your help.

Answer 1

You want a matrix $A$ that satisfies the polynomial equation $A^5-2A^0=0$. An easy way to obtain this is to take the companion matrix of that polynomial $P=X^5-2$; this gives the example of user7530 (and the example by anon also gives this matrix on the most obvious basis). The polynomial$~P$ will be the minimal one satisfied by$~A$ (which may not be surprising here because $X^5-2$ is irreducible over$~\Bbb Q$, but it holds independently of that fact), and also the characteristic polynomial of the matrix$~A$.

Answer 2

Hint: multiplication by $2^{1/5}$ is a linear map on the dimension-$5$ vector space ${\Bbb Q}[2^{1/5}]$.

Answer 3

I would approach the problem by thinking about how $A$ needs to act on basis vectors $e_i$. Since we are applying $A$ five times and there are five basis vectors, we would like $A$ to be a cyclic permutation of the basis vectors, plus an extra scaling by two at some point: $$e_1 \mapsto e_2 \mapsto e_3 \mapsto e_4 \mapsto e_5 \mapsto 2e_1,$$ for instance.

This gives us the matrix

$$\left[\begin{array}{ccccc}0 & 0& 0& 0 & 2\\1&0&0&0&0\\0&1&0&0&0\\0&0&1&0&0\\0&0&0&1&0\end{array}\right]$$ which has the property you want.