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Let $V$ be an $n$-dimensional real vector space, and let $1<k<n-1$ be fixed. Let $T: Vto V$ 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

$$T , text has an eigenvector if and only if , T^* , text has an eigenvector tag1.$$

By the previous comment, we also have

$$T , text has an eigenvector if and only if , T^* , text has a co-dimension one invariant subspace tag2.$$

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 $T$.

The result is true over any field. Let $V$ be an $n$-dimensional vector space over an arbitrary field $mathbbF$, let $T colon V rightarrow V$ be an operator and let $U$ be a $k$-dimensional $T$-invariant subspace of $V$. Denote by
$$ U^0 := , varphi(u) = 0 ,,,forall u in U $$
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.

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

$$beginpmatrixU_mathbbR & 0 \
0 &U_mathbbC
endpmatrix$$

where $U_mathbbR$ is upper triangular, and $U_mathbbC$ block-upper triangular with blocks of size $2$. Say $U_mathbbR$ is of size $r_1$ and $U_mathbbC$ 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, $U_mathbbC$ has invariant subspaces of all even dimensions. And because the sum of an invariant subspace of $U_mathbbR$ and an invariant subspace of $U_mathbbR$ 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 $kle n$ 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.