8

Is it possible to characterize the set of real matrices which have real eigenvalues geometrically? That is, is it possible to say that a linear map $T$ has real eigenvalues if and only if it satisfies some property ${\cal P}$ which has a geometric flavor?

I understand, of course, that different people will disagree on what ``geometric'' means. Still, I'm quite interested in seeing any possible answers. It seems plausible that properties of matrices which are independent of choice of basis have geometric interpretations.

Examples of the sort of thing I'm looking for:

  • A linear map $T$ has eigenvalues within the unit circle if and only if $\lim_k T^k x = 0$ for any $x$.
  • A linear map on $\mathbb{R}^n$ has all of its eigenvalues equal to zero if $T^n x = 0$ for any $x \in \mathbb{R}^n$.
  • A linear map $T$ has $1$ in its set of eigenvalues if and only if it has a fixed point.
  • A linear map $T$ will have a root of unity in its set of eigenvalues if and only if it has a periodic orbit.
aro
  • 81

2 Answers2

4

How about this? A pair of complex conjugate eigenvalues corresponds to a $2$-dimensional invariant subspace that is stretched and rotated nontrivially. So all real eigenvalues means no such subspaces. So for any nonzero vector $v$, there is no nonzero $w$ orthogonal to it so that, up to scaling, the linear map rotates the plane $\text{Span}(v,w)$.

Ted Shifrin
  • 115,160
  • You may easily have a 2-dimensonal eigenspace corresponding to a real eigenvalue. This eigenspace will be a 2-dimensional invariant subspace. – lhf May 28 '13 at 03:49
  • 3
    @lhf Of course. But it's not being rotated — it's being stretched only! – Ted Shifrin May 28 '13 at 03:58
1

$T^2$ embeds in a continuous 1-parameter real semigroup of transformations $(T^2)^\alpha $. This is a dynamical interpretation, not purely geometric but close.

zyx
  • 35,436