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Let $A: \mathbb{R}^6 \rightarrow \mathbb{R}^6$ be linear transformation, $A^{26} = I$. Find linear spaces $V_1, V_2, V_3$, such that: $\mathbb{R}^6 = V_1 \oplus V_2 \oplus V_3$, dim $V_1$ = dim $V_2$ = dim $V_3$ and $AV_1 \subset V_1, AV_2 \subset V_2, AV_3 \subset V_3$. Can you help me?

I see that $\mu_1\mu_2\ldots\mu_6 = \textrm{det}A$ $\in \{-1,1\}$ and $\mu_i \in \{e^{\frac{2k\pi i}{26}}\mid k = 0,1,\ldots,25\}$. I need to know more about $\sigma (A)$. Is it possible to show $1$ or $-1$ has to be in the spectrum?

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If $A$ has a complex eigenvalue $\lambda=a+bi (b\neq 0)$, then the corresponding (complex) eigenvector is of the form $X+iY$ where $X,Y$ have real coordinates, and it is easy to see that $X$ and $Y$ must be linearly independent. Then the subspace $V_{\lambda}={\textsf{span}}(X,Y)$ is two-dimensional and invariant by $A$. The other eigenvalues are $\pm 1$, so it should be easy for you to finish from here.

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  • $\begingroup$ Why the other eigenvalues are necessarily $1$ and $-1$ ? $\endgroup$ – user371231 Sep 19 '18 at 18:53
  • $\begingroup$ The other eigenvalues are the ones whose imaginary part is zero, in other words, the real ones. But the only real roots of unity are $\pm 1$. $\endgroup$ – Ewan Delanoy Sep 20 '18 at 7:58
  • $\begingroup$ Actually what I want to mean it may happen that the ch polynomial is product of three irreducibles of degree $2.$ If they are distinct then it is ok for me. But if it happens that all the irreducible factors are same then it is not clear that Eigen vectors will work only. $\endgroup$ – user371231 Sep 20 '18 at 15:12
  • $\begingroup$ It does not matter if all eigenvalues are the same, i.e. if there is a single complex eigenvalue. My argument still holds. When you pick an eigenvector, you can forget about the rest of the world. $\endgroup$ – Ewan Delanoy Sep 21 '18 at 8:11

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