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Consider the linear ODE \begin{align} \dot x = J x, \: x\in \mathbb{R}^n \end{align} with the Jacobian $J$. Page 35 of Ordinary Differential Equations with Applications by Chicone asserts that the space $\mathbb{R}^n$ can always be decomposed as a direct sum of linear subspaces: the stable eigenspace (stable manifold) corresponding to the eigenvalues of $J$ with negative real parts, the unstable eigenspace (unstable manifold) corresponding similarly to the eigenvalues of $J$ with positive real parts, and the center eigenspace (center manifold) corresponding to the eigenvalues with zero real parts.

My question is: what are the implications of $J$ being defective? For example, suppose \begin{align} J = \begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix} \end{align} which has the eigenvalue of $-1$ with algebraic multiplicity of $2$, but the geometric multiplicity of $1$. The eigenspace is spanned by $[1,0]$. I was wondering what happens to the Hartman-Grobman Theorem in this case and how the space is partitioned.

Thank you, in advance, for your comments.

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  • $\begingroup$ There are various confusions here: the sentence "The Hartman-Grobman theorem implies that..." is false. In fact it is false even if you erase the "center" part. Up to the necessary corrections, there is not improvement on the conjugacy when the Jordan canonical form is nontrivial (see the following exercise). Exercise: find a topological conjugacy between the flows of $x'=-x+y, \,y'=-y$ and $x'=-x, \,y'=-y$. $\endgroup$
    – John B
    Jan 21 '20 at 21:08
  • $\begingroup$ Thanks for the comment. Maybe this fact is not directly implied by The Hartman-Grobman theorem. My question is about this fact. I can delete the sentence "The Hartman-Grobman theorem implies that..." $\endgroup$
    – Arthur
    Jan 21 '20 at 23:10
  • $\begingroup$ Sorry, I don't understand what you mean. The statement is false, so it is not a "fact". Please clarify your question otherwise (which as it stands is unrelated to the Grobman-Hartman theorem). Anyways, my last comment means that the space is not subpartitioned when the Jordan form is nontrivial (recall that the conjugacy is never unique). $\endgroup$
    – John B
    Jan 22 '20 at 0:07
  • $\begingroup$ Thanks for your clarification. I revised the question and added a reference. Maybe the non-linearity of the ODE and the conjugacy were confusing in the previous version of the question, so I made it linear to focus on the main point of my question. Thanks. $\endgroup$
    – Arthur
    Jan 22 '20 at 3:26
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What happens is that the existence of a topological conjugacy between linear equations whose matrices have only eigenvalues with nonzero real part depends solely on the number of eigenvalues with positive (or negative) real part. More precisely:

There exists a topological conjugacy between the flows of two linear equations $x'=Ax$ and $x'=Bx$ for some $n\times n$ matrices without eigenvalues on the vertical axis if and only if $A$ and $B$ have the same number of eigenvalues with positive (or negative) real part.

The proof becomes somewhat simple only after introducing specific norms related to $A$ and $B$.

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