If a linear map $T$ has a $k$-dimensional invariant subspace, does it admit an $n-k$ invariant subspace?

Question

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:

1. 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).

2. 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 $T$.

Answer 1

The result is true over any field. Let $V$ be an $n$-dimensional vector space over an arbitrary field $\mathbb{F}$, let $T \colon V \rightarrow V$ be an operator and let $U$ be a $k$-dimensional $T$-invariant subspace of $V$. Denote by $$ U^0 := \{ \varphi \in V^{*} \, | \, \varphi(u) = 0 \,\,\,\forall u \in U \} $$ the annihilator of $U$. Note that if $\varphi \in U^{0}$ then $$ (T^{*}(\varphi))(u) = \varphi(Tu) = 0 $$ because $Tu \in U$ and $\varphi \in U^{0}$ so $T^{*}(\varphi) \in U^{0}$ and hence $U^{0}$ is an $n-k$-dimensional $T^{*}$-invariant subspace.

Now we can use the relatively well-known but non-trivial fact that any matrix is similar to its transpose (see this answer). An invariant way of stating this is that there exists an isomorphism $S \colon V \rightarrow V^{*}$ such that $S^{-1} \circ T^{*} \circ S = T$. Set $W = S^{-1}(U^{0})$. Then $W$ is an $n-k$-dimensional subspace of $V$ and $$ T(W) = T(S^{-1}(U^{0})) = S^{-1}(T^{*}(U^{0})) \subseteq S^{-1}(U^{0}) = W $$ so $W$ is $T$-invariant.

Answer 2

I hope I'm not mistaken but I believe this result is true for real numbers. In summary, we prove the following (which yelds a positive answer to our question as I'll explain) : If $T$ has a real eigenvalue, then $T$ has invariant subspaces of all dimensions. And if $T$ has no real eigenvalue, then $T$ has invariant subspaces of all even dimensions.

Using the Jordan normal form for real matrices (or a somewhat weaker version), we may find a basis of $V$ in which the matrix of $T$ is of the form

$$\begin{pmatrix}U_{\mathbb{R}} & 0 \\ 0 &U_{\mathbb{C}} \end{pmatrix}$$

where $U_{\mathbb{R}}$ is upper triangular, and $U_{\mathbb{C}}$ block-upper triangular with blocks of size $2$. Say $U_{\mathbb{R}}$ is of size $r_1$ and $U_{\mathbb{C}}$ is of size $2 r_2$. Now, upper triangular matrices have invariant subspaces of all possible dimensions (the subspace generated by the $k$ first coordinates is an invariant subspace of dimension $k$). And similarly, $U_{\mathbb{C}}$ has invariant subspaces of all even dimensions. And because the sum of an invariant subspace of $U_{\mathbb{R}}$ and an invariant subspace of $U_{\mathbb{R}}$ is an invariant subspace of $T$, then we get subspaces of all possible dimensions than can be written as a sum of a number $\le r_1$ and an even number $\le 2r_2$).

If $r_1 > 0$ (which implies $T$ has real eigenvalues), then $T$ has invariant subspaces of all possible dimensions (just like if you have $n$ euros in coins of $2$ and coins of $1$, you can make any amount $k\le n$ provided you have at least of coin of $1$). Now if $r_1 = 0$, you can only get subspaces of even dimensions. But then again, $n=2 r_2$ is even, and since there is a subspace of dimension $k$, it means that $k$ is also even. Indeed, if $V$ were a subspace of odd dimension, then $T_{|V}$ would have an eigenvalue (because its characteristic polynomial has odd degree), contradicting $r_1 =0$. So $n-k$ is even, and there is a subspace of that dimension.