If two linear differential equations have matrices who have the same eigenvalues, are they topologically equivalent?

Question

I have two differential equations

$\displaystyle \frac{dX}{dt}=AX$ and

$\displaystyle \frac{dY}{dt}=CY$

If $A$ and $C$ have the same eigenvalues, are they topologically equivalent?

@ChrisEagle: It refers to the (flows of) the ODEs rather than to the matrices. http://en.wikipedia.org/wiki/Topological_conjugacy#Topological_equivalence — Hans Lundmark, Feb 27 '13 at 00:07

score 2 · Accepted Answer · answered Feb 26 '13 at 21:29

2

If $A$ and $C$ have the same Jordan canonical form (in particular if they have distinct eigenvalues and those are the same for both) then they are similar: there is an invertible matrix $S$ such that $A = S C S^{-1}$. Then $X$ satisfies $\dfrac{dX}{dt} = A X$ if and only if $Y = S^{-1} X$ satisfies $\dfrac{dY}{dt} = C Y$.

answered Feb 26 '13 at 21:29

Robert Israel

448,999

And if they don't have the same Jordan form, but merely the same eigenvalues? – Chris Eagle Feb 27 '13 at 02:04
Then they are not similar. – Robert Israel Feb 27 '13 at 02:26

If two linear differential equations have matrices who have the same eigenvalues, are they topologically equivalent?

1 Answers1