# Prove that H(x,y) is a conjugacy between two systems

Prove that $H(x,y)=(x,-y)$ provides a conjugacy between $$X'=\begin{bmatrix}1&1\\-1&1\end{bmatrix}X$$ and $$Y'=\begin{bmatrix}1&-1\\1&1\end{bmatrix}Y$$

I know that for H to be a conjugacy, it must hold that $\phi^Y(t,H(x_0))=H(\phi^X(t,x_0))$ where $\phi^X$ and $\phi^Y$ are the flows of the first and second system respectively, but I don't understand how to find the flows of the systems. My textbook brushes over the process saying it is trivial or that it easily follows, but I can't figure out how to find the flows, or apply them. This is the work I have so far:

the eigenvalues of both systems are $\lambda=1\pm i$ so both systems are hyperbolic because they have the same number of eigenvalues with a positive real part, therefore they are conjugate to each other. The only thing I don't know what to do is find the actual conjugacy... or prove that the given H is the conjugacy.

• So, do you know how to solve systems of linear equations? Or about matrix exponent? – Evgeny Sep 16 '16 at 11:20
• @Evgeny, yes I can solve systems, but I don't know about matrix exponents – bowen.jane Sep 16 '16 at 13:07
• Basically, if you can solve this system the expression that you get defines a flow for the vector field defined by linear system of ODEs. The only thing that you must take into account is that your constants $C_1, C_2, \dots, C_n$ somehow should be related to $x_0$. In that case you have a formula that describes the flow. – Evgeny Sep 16 '16 at 18:36
• @Evgeny so if the general solution of both systems is $e^t(c_1\begin{bmatrix}cos(t)\\-sin(t)\end{bmatrix}+c_2\begin{bmatrix}sin(t)\\cos(t)\end{bmatrix}$ how do I find the right constants that relate to $x_0$? – bowen.jane Sep 16 '16 at 19:04
• Well, that supposed to be simpler part of problem. But okay, if your formula describes a flow, plugging $t = 0$ in flow always gives you $\phi(0, x_0) = x_0$. Can you take it from here? – Evgeny Sep 17 '16 at 5:21