# Linear ODE When the Jacobian is Defective

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.

• 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 $x'=-x, \,y'=-y$. Jan 21 '20 at 21:08
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 $$B$$.