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I am trying to construct an example of a linear operator $T : \mathbb{Q}^3 \rightarrow \mathbb{Q}^3$ for which the only $T$-invariant subspaces are the whole space and the zero subspace.

If we first look at an example from the 2x2 case let $T$ be the linear operator on $\mathbb{R}^2$ represented in the standard ordered basis by $$ A = \left( \begin{array}{ccc} 0 & -1 \\ 1 & 0 \end{array} \right) $$

Then if $W$ is any other invariant subspace not equal to $\{0\}$ or the whole space then $W$ must have dimension $1$ and so is spanned by some nonzero vector $\alpha$. But $W$ invariant under $T$ implies that $\alpha$ is a eigenvector, but $A$ has no real real eigenvalues.

If we try to apply the above logic to a 3x3 matrix then I am stuck on what to do if I assume the dimension of the invariant subspace is 2.

Question: In any case is it still clear that if $A$ represents some linear operator $T : \mathbb{Q}^3 \rightarrow \mathbb{Q}^3$ then for $T$ to have no nontrivial invariant subspaces should A not have any real eigenvalues?

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In general, not having real eigenvalues is insufficient to conclude not having invariant subspaces. For example, in 4 dimensions, a rotation in the $e_1$-$e_2$ plane compose with a rotation in the $e_3$-$e_4$ plane has no real eigenvalues, but has two invariant two dimensional subspaces. –  Willie Wong Jul 22 '11 at 13:53
    
thank you this is very instructive –  user7980 Jul 22 '11 at 13:57

1 Answer 1

up vote 3 down vote accepted

Over the reals, you won't find any examples in dimension 3 or any odd dimension because every operator in such a space has an eigenvector (since every real polynomial of odd degree has a real root).

Over the rationals, you only need to find a polynomial of degree 3 with rational coefficients having no rational root and take its companion matrix. The simplest one I can think of is $x^3-x-1$.

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but the root is not necessarily in $\mathbb{Q}$. –  Willie Wong Jul 22 '11 at 13:48
3  
I think you should also mention that an operator on a (non-zero) real vector space $V$ always has an invariant subspace of dimension $1$ or $2$. Therefore, if the dimension of a real vector space $V$ exceeds $2$, an operator $T$ on $V$ always has a non-trivial, proper invariant subspace. –  Amitesh Datta Jul 22 '11 at 13:52
    
thank you for the answer and all the comments this is extremely helpful –  user7980 Jul 22 '11 at 13:57

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