I got stuck in this problem:
Let $A:\mathbb{R}^{6}\rightarrow \mathbb{R}^{6}$ be a linear transformation. Assume $A^{26}=I$, prove that $R^{6}=\oplus_{i=1}^{3} V_{i}$, with $AV_{i}\subset V_{i}$(the explicit condition is $V_{i}$ are 2-dimensional invariant subspaces of $\mathbb{R}^{6}$ under $A$).
My thought is $A$ must have a minimal polynomial of degree less or equal to 6. Thus since it divides $x^{26}-1$, the only choices are: $$x-1,x+1,x^{2}-1$$ since the rest term $$(x^{13}-1)/(x-1)*(x^{13}+1)/(x+1)$$ has factors irreducible and degree higher than 6. And the claim is trivial in the case $A=\pm I$. But I do not know how to deal with the case $A^{2}-I=0$ - $A$ can only have eigenvalues $1$ and $-1$, but how this helps to solve the problem?
Edit:
In the light of did's comments $\sum^{12}_{i=0}x^{i}$ and $\sum^{12}_{i=0}(-1)^{i}x^{i}$ can be reducible over the reals in pairs of 6 quadratics, and the corresponding $A$'s are rotations. But I still feel rather confused as if problem is solved at this stage by suggesting $A$'s minimal polynomial must be a product of $$x-1,x+1,x^{2}-1, x^{2}-\cos[\theta]x+1$$ which are dealt with respectively by $I,-I$, selecting linearly independent vectors and run with $A$, and selecting the rotational invariant subspace. Since obviously cases like $$(x\pm 1)(x^{2}-\cos[\theta]x+1)$$ or even $$(x+1)(x-1)^{2}$$ could happen.