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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 $T$.

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  • $\begingroup$ It is not necessarily the case that we have an invariant subspace $B$ complementary to $A$; it could be the case that $B$ necessarily contains $A$. $\endgroup$ – Ben Grossmann Mar 29 '17 at 4:43
  • $\begingroup$ Are we allowed to use eigenvalues here? What about Jordan canonical form? $\endgroup$ – Ben Grossmann Mar 29 '17 at 4:43
  • $\begingroup$ Is $V$ necessarily a vector space over $\Bbb C$? $\endgroup$ – Ben Grossmann Mar 29 '17 at 4:45
  • $\begingroup$ @Omnomnomnom There is no condition $V$ is over $\mathbb{C}$. $\endgroup$ – noname1014 Mar 29 '17 at 5:00
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    $\begingroup$ I have edited the question considerably, but kept exactly the intention of the original poster: I only changed some of the phrasing, elaborated on some partial results, and mentioned where we should start to look for possible counter-examples. $\endgroup$ – Asaf Shachar Mar 13 '19 at 9:52
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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.

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    $\begingroup$ Note that the usual proof of the fact that a matrix is similar to its transpose uses the rational canonical form. Over $\mathbb{R}$, you can replace it with the real Jordan form so in this sense, my solution uses the same technology Joel uses. However, instead of using a canonical form to analyze completely the possible dimensions of the invariant subspaces (an analysis which depends highly on the field we are working over) to deduce the result, it "uses" the canonical form to deduce the result directly without providing insight on the possible dimensions of invariant subspaces. $\endgroup$ – levap Mar 14 '19 at 14:53
  • $\begingroup$ It's a very elegant proof ! I've been thinking about how to generalize my proof to an arbitrary field : using the Jordan form, we get a block-upper triangular matrix (whose sizes are degrees of irreducible factors of the minimal polynomial), and the possible size of invariant subspaces are exactly those that can be written as a sum of the size of the blocks. Now if k can be written in such a way, then n-k can also by a complement argument. $\endgroup$ – Joel Cohen Mar 15 '19 at 2:09
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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.

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