Let $V$ be an $n$-dimensional real vector space, and let $1<k<n-1$ be fixed. Let $T: V\to V$ be a linear map, and suppose that there exists a $k$-dimensional $T$ invariant subspace of $V$.
Does there exist an $(n-k)$ $T$-invariant subspace of $V$?
The smallest possible dimensions where a counter-example might be is when $k=2,n=5$.
Two comments:
By duality, $T$ has a $k$-dimensional invariant subspace if and only if the dual map $T^*:V^* \to V^*$ has an $n-k$-dimensional invariant subspace. (If $U$ is $T$-invariant, then the subspace of $V^*$ whose restriction to $U$ is zero is $T^*$-invariant).
I excluded the cases $k=1$ and $k=n-1$, since I know that the answer is positive for those. Indeed, since the characteristic polynomials of $T$ and $T^{*}$ are identical, we have
$$T \, \text{ has an eigenvector if and only if } \, T^{*} \, \text{ has an eigenvector } \tag{1}.$$
By the previous comment, we also have
$$T \, \text{ has an eigenvector if and only if } \, T^{*} \, \text{ has a co-dimension one invariant subspace } \tag{2}.$$
Combining $(1)$ and $(2)$, we conclude that $T^*$ has an eigenvector if and only if $T^*$ has a co-dimension one invariant subspace. Since any map is the dual of its dual, this holds for any endomorphism $T$.