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I am figuring out (trying) to find an explicit topological conjugation between two ODEs.

$$\frac{dx}{dt} = -x \quad (\text{flow: } e^{-t})$$

and

$$\frac{dx}{dt} = Ax, \qquad A = \begin{pmatrix} -2 & 1 \\ 1 & -2 \end{pmatrix}$$ (a $2 \times 2$ matrix with hyperbolic eigenvalues)

My question at hand is it is easy to find a homeomorphism between two matrices, but a matrix and a non-matrix has me a bit confused at the moment. I got stuck after finding the matrix exponential of the $2 \times 2$ matrix because I do not know what to do afterwards...

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I know very little about this subject, but I am assuming $x$ is a vector of length 2 here? In which case, $-x = -Ix$ where I is the identity matrix, so you really do have two matrices. –  Steven Gubkin Feb 5 at 20:44
    
Got something from the answer below? –  Did Mar 17 at 8:17
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1 Answer 1

Assuming that you agree with the comment of @Steven Gubkin, which has to be ... OK, and maybe this is known to the OP ... but here it goes.

First note that by a similarity matrix you can transform

$$ \dot y=\begin{bmatrix} -2 & 1\\ 1 & -2 \end{bmatrix}y $$ to $$ \dot z=Bz=\begin{bmatrix} -3 & 0\\ 0 & -1 \end{bmatrix}z. $$

The previous transformation is done by defining $z=Cy$ with $C=\begin{bmatrix} -1 & 1\\ 1 & 1\end{bmatrix}$, and noting that $CAC^{-1}=B$.

Next you already have the second equation as you want it, so $x_2=z_2$. For the first coordinate just define $x_1=z_1^{1/3}$ and perform the required computations.

Side note: It is important to note that you cannot require a smooth conjugacy, as in that case you'd need that the spectrum of both matrices are the same.

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