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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.

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well, then $A$ would have a 26 degree minimal polynomial, which is wierd. –  Bombyx mori Jul 31 '12 at 11:56
    
Linear doesn't mean at most degree 1 polynomial ? –  Xoff Jul 31 '12 at 12:00
    
It means a linear transformation, you can think about it as a matrix. –  Bombyx mori Jul 31 '12 at 12:02
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A degree 12 polynomial cannot be irreducible: the only irreducible real polynomials are linear polynomials and quadratic polynomials with non-real roots. –  Hurkyl Jul 31 '12 at 14:07
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Every real polynomial factors into linear polynomials and quadratic polynomials with non-real roots. In this case, the quadratic polynomials are easy to find. The polynomial $(x-\zeta_{26})(x-\zeta_{26}^{-1})$ has real coefficients, and so is a factor of $x^{26} - 1$ over the reals. $\zeta_{26}$ is any primitive 26-th root of unity, I don't care which one. e.g. $\zeta_{26} = \exp(\pi i / 13)$ will do. –  Hurkyl Jul 31 '12 at 14:31

3 Answers 3

up vote 3 down vote accepted

The decomposition of $x^{26}-1$ into irreducible factors over $\mathbb R$ is $$ (x-1)\cdot(x+1)\cdot\prod\limits_{k=1}^{12}p_k(x),\qquad p_k(x)=x^2-2\cos(k\pi/13)x+1. $$ This almost determines the minimal polynomial $\mu_A(x)$ of $A$, since $\mu_A(x)$ divides $x^{26}-1$. Hence $\mu_A(x)$ is the product of at most three factors $p_k(x)$ and possibly a factor $x-1$ and possibly a factor $x+1$. In any case, $\mu_A(x)$ has no repeated irreducible factor hence $A$ is equivalent over $\mathbb R$ to a matrix diagonal by blocks, with (1.) possibly an even number of blocks of size $1\times1$ equal to $+1$ or to $-1$ and (2.) possibly some blocks of size $2\times2$ equal to rotation matrices $$ \begin{pmatrix}\cos(k\pi/13)&\sin(k\pi/13)\\ -\sin(k\pi/13)&\cos(k\pi/13)\end{pmatrix}. $$ This yields the desired decomposition as follows: choose each plane left invariant by one of the rotation matrices, if there are some, and complete by any planes made of eigenvectors with eigenvalues $\pm1$, if necessary. (Note that the result is such that $AV_i=V_i$ for every $i$.)

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Suppose $A^2 = I$. Let $x$ be any nonzero vector. Then the space spanned by $x$ and $Ax$ is invariant. Repeat.

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This makes sense. Pick up one vector, and then using a linearly independent one. Since we have 6 dimensions this process must terminate at the 3rd one. Thanks! –  Bombyx mori Jul 31 '12 at 12:14
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Sorry but I fail to see how this is a solution. What happens if $A^2\ne I$? –  Did Jul 31 '12 at 12:24
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@user32240: Wow!! To sum up, you think that $A^{26}=I$ implies $A^2=I$ (wrong) and this probably explains why you accepted this answer, but you also think that $A^2=I$ implies $A=I$ or $A=-I$ (wrong as well). –  Did Jul 31 '12 at 12:36
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@user32240: Let me give you some food for thought: rotation... –  Did Jul 31 '12 at 13:09
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@user32240: Hmmm... Trying to think would be more profitable than putting words in my mouth. Anyway, let me continue with my hint: consider $B=\begin{pmatrix}\cos(\pi/13)& \sin(\pi/13)\\ -\sin(\pi/13) &\cos(\pi/13)\end{pmatrix}$. Questions: (1.) $B^{26}=I$? (2.) $B^2=I$? (3.) Which minimal polynomial? If you find the answers, you are up for some surprises... –  Did Jul 31 '12 at 14:20

Consider $$B=\sum_{i=0}^{12}A^{2i}$$

You will find easy invariant subspace by considering $B(x)$, because $A^2(B(x))=B(x)$. So take $B(x)$ and $A(B(X))$ to build a 2 dimensional invariant subspace. There are some degenerate case (for example $B(x)=0$), but they should not be too hard...

EDIT :

A has 6 eigen values $\lambda_i$ in $\mathbb C$. As $A^{26}=I$, $\lambda_i^{26}=1$, they are non zero values. Either

  1. $\lambda_i\in{\mathbb R}$
  2. $\lambda_i\notin{\mathbb R}$ and $\exists j \lambda_j=\bar{\lambda_i}$

So you can build 3 2-dimensional spaces by grouping the complex values with their respective conjugate, or any real value with any other real value...

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