3
$\begingroup$

Let $A \in GL(d,\mathbb{R})$, which is the space of $d\times d$ invertible matrices. Assume that the characteristic polynomial of $A$ is irreducible over $Q$. I want to find $A$-invariant subspaces $\operatorname{Gr}(k)$, which is a Grassmannian of $k$-planes of $\mathbb{R}^{d}$.

My attempt: I think one needs to find the eigenspaces $\wedge^{k} A$ that are $A$-invariant subspaces $\operatorname{Gr}(k)$. If the previous sentence is correct, I have the following problems:$A$-invariant subspaces $\operatorname{Gr}(k)$ are $k$-dimensional, but the eigenspaces $\wedge^{k} A$ are $\binom nk$-dimensional.

$\endgroup$
7
  • $\begingroup$ This is determined by $A$'s real canonical form, although how precisely seems tricky. WLOG, the linear operator is multiplication by $T$ on a direct sum of modules of the form $\mathbb{R}[T]/\pi(T)^n$ where $\pi$ is linear or irreducible quadratic. Invariant subspaces are not simply sums of these summands though - for instance isoclinic rotations have infinitely many invariant subspaces. $\endgroup$
    – anon
    Apr 18, 2021 at 20:05
  • $\begingroup$ @runway44 : Thanks for your comment. Assume that the characteristic polynomial is irreducible over $Q$. Then, can we consider the eigenspaces of the exterior products? If not, under what assumptions we can consider it? $\endgroup$
    – Adam
    Apr 18, 2021 at 20:17
  • $\begingroup$ If the char poly is irreducible over $\Bbb Q$ then there are no repeated roots. If the char poly has no repeated roots, then (ignoring dimensions) the $A$-invariant subspaces are precisely the direct sums of the minimal ones, and there are finitely many minimal ones (you get a minimal one, in this case, by summing a conjugate pair of eigenspaces in $\Bbb C^n$ and then intersecting with $\Bbb R^n$). $\endgroup$
    – anon
    Apr 18, 2021 at 21:26
  • $\begingroup$ @runway44 : Thanks for your comment again. Sorry, I don't understand what you meant by "the minimal ones"? what do you mean by summing a conjugate pair?ou meant Invariant subspaces are simply sums of the summands?If not, do you know under what assumptions we have it? Thanks in advance. $\endgroup$
    – Adam
    Apr 18, 2021 at 22:09
  • $\begingroup$ A minimal invariant subspace, as you might expect, is a proper nontrivial invariant subspace that does not contain any smaller nontrivial invariant subspaces. Extend the action of $A$ from $\Bbb R^n$ to $\Bbb C^n$, and let $V_\lambda$ be a nontrivial eigenspace for a complex number $\lambda$. Then $V_{\overline{\lambda}}$ is also a nontrivial eigenspace, where $\overline{\lambda}$ is the conjugate, and $(V_\lambda\oplus V_{\overline{\lambda}})\cap\Bbb R^n$ is a minimal invariant subspace of $\Bbb R^n$, and this is how all of them arise. $\endgroup$
    – anon
    Apr 18, 2021 at 22:39

0

Reset to default

You must log in to answer this question.

Browse other questions tagged .