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Show that for $k = 1, \ldots, d$, the set of hyperbolic linear vector fields with $\dim E^{s} = k$ is open in $\mathcal{L}(\mathbb{R}^{d},\mathbb{R}^{d})$.

Definition

  1. $A\colon \mathbb{R}^{d} \longrightarrow \mathbb{R}^{d}$ is linear hyperbolic if all its eigenvalues ​​have non-zero real part.

  2. If $A\colon \mathbb{R}^{d} \longrightarrow \mathbb{R}^{d}$ is a linear hyperbolic field, then there is a direct sum decomposition $ \mathbb{R}^{d} = E^{s} \oplus E^{u}$ satisfying:

    • the eigenvalues ​​of $A \vert E^{s}$ are the eigenvalues ​​of $A$ whose real part is negative.
    • the eigenvalues ​​of $A \vert E^{u}$ are the eigenvalues ​​of $A$ whose real part is positive.

Attempt: I tried to use the fact that the set of hyperbolic linear vectors is open and dense in $\mathcal{L}(\mathbb{R}^{d},\mathbb{R}^{d})$, but I was unsuccessful.

Any help would be appreciated!

John B
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christopher silva
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    All you need to show is that the eigenvalues depend continuously on the coefficients. You can find references for that fact here https://math.stackexchange.com/questions/556137/eigenvalues-are-continuous – Severin Schraven Aug 05 '20 at 18:58
  • why is that enough? – Kauane Araujo Aug 05 '20 at 19:06
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    @KauaneAraujo Because for a hyperbolic linear map we have that $\operatorname{dim}(E^s)$ equals the number of eigenvalues with negative real part. Hence, if you want to keep the dimension fix, then you just need to ensure that number of eigenvalues with negative real part does not change. If you know continuity, then this follows immediately. – Severin Schraven Aug 05 '20 at 19:15
  • @Severin Schraven Got it. thanks! – Kauane Araujo Aug 05 '20 at 19:18
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    @KauaneAraujo Great :) glad I could help. – Severin Schraven Aug 05 '20 at 21:41

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