Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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
up vote 4 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 $x^3-x-1$.

share|cite|improve this answer
but the root is not necessarily in $\mathbb{Q}$. – Willie Wong Jul 22 '11 at 13:48
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.