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

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.

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