# Explicit topological conjugation between two ODEs

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...

• 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 '14 at 20:44
• Got something from the answer below? – Did Mar 17 '14 at 8:17

$$\dot y=\begin{bmatrix} -2 & 1\\ 1 & -2 \end{bmatrix}y$$ 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.