An example for an invertible matrix with no $3$-dimensional invariant subspaces

Question

Is there an example for an invertible linear transformation $\mathbb R^n \to \mathbb R^n$ with no $3$-dimensional invariant subspaces (where $n \ge 4$)?

Every real linear transformation admits $2$-dimensional invariant subspaces.

Answer 1

Let $V$ be a finite dimensional vector space and let $T \colon V \rightarrow V$ be a linear map. If $T$ has a one-dimensional invariant subspace (i.e, an eigenvalue), then since the characteristic polynomials of $T$ and $T^{*}$ are identical, we see that $T^{*}$ has some eigenvector $0 \neq \varphi \in V^{*}$. But then $\ker \varphi$ is a co-dimension one $T$-invariant subspace. This shows that if $T$ has a one-dimensional $T$-invariant subspace, it also has a co-dimension one $T$-invariant subspace. By duality, if $T$ has a co-dimension one $T$-invariant subspace, it follows that $T$ has a one-dimensional $T$-invariant subspace.

Now, let $T \colon \mathbb{R}^4 \rightarrow \mathbb{R}^4$ be any linear map which doesn't have an eigenvalue. Such a map won't have any three-dimensional $T$-invariant subspaces. For example, you can take the map represented in the standard basis by

$$ \begin{pmatrix} 0 & -1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & -1 \\ 0 & 0 & 1 & 0\end{pmatrix}. $$