$A^{2n}=I$ but $A^{n}\neq I, -I$

Question

Let $A$ be a $n\times n$ real matrix. such that $A^{2n}=I$ but $A^{n}\neq I, -I$, $n\geq 2?$ I have a example for $A^2=I$ but $A\neq I, -I$ but could not find a similar example for this question. I tried permuting basis vectors but could not an example. Any help is appreciated

Answer 1

Take $A$ to be diagonal, and its entries being $2n$-th roots of unity, and one of them not being a $n$-th of unity.

As far as real entries are concerned: let $r$ be a primitive $2n$-th root of unity. Take $A$ as a block-diagonal matrix, the second block being $I_{n-2}$ and the first block being the $2\times 2$ real matrix associated with the complex number $r$.

Answer 2

Essentially this is just the answer by Mindlack, only pimped to work over $\mathbb{R}$

Pick a $2$-dimensional subspace $V$ (which you can by assumption) and consider $\mathbb{R}^n$ as $V \oplus \mathbb{R}^{n-2}$. then any automorphism $\varphi'$ on $V$ induces an automorphism $\varphi= \varphi\oplus id$ on $\mathbb{R}^n$, and in particular a matrix. Hence we only need to find a $\varphi'$ that does the job. Now observe that by adding up with identities the requirement of $A^n \neq -id$ collapses. hence it suffices to find a $\varphi$ such that $\varphi^{2n}= id$ but now you can just embed $\mathbb{C}\hookrightarrow \mathbb{R}^{2\times 2}\cong \mathrm{Aut}(\mathbb{R}^2)\cong \mathrm{Aut}(V)$ and pick a $2n$-th root of unity in that image (this maps $1$ to $\begin{pmatrix}1 & 0\\0 & 1 \end{pmatrix}$ and $i$ to $\begin{pmatrix}0 & 1\\-1 & 0 \end{pmatrix}$, to prove that this indeed is an embedding of a subring is a beautiful exercise). This does the job.

Morally, it suffices to find a matrix that does the stuff on a subspace, and then lift it to the space you want to have.