The next problem states that given a $m\times m$ non singular matrix $T$, such that $T^2=I$, is called a reflection for $\dot{x}=f(x) \Leftrightarrow f(Tx)=-Tf(x)$ for all $x\ \epsilon\ R^{m}$. Need to prove that $\phi(t,T\xi)\equiv T\phi(-t,\xi)$ where $\phi(t,\xi)$ is the general solution of $x=f(x)$, and also to show that a reflection of a solution in the x1 axis is also a solution.
I know that $\dot{x}=f(x)$ and has a solution of $\phi(t,\xi)$, i also know that $\phi(0,\xi)=\xi$ because is the identity of the flow, so I tought that if we multiply by $-T$ the function and the solution we have $-Tf(x)$ and $-T\phi(t,\xi)$, and this new transformation given by $-T$ should preserve the new function with its new solution because is just a reflection.
So if we have $-T\phi(t,\xi)$, and if we put $t=0$ for the solution, we will have $-T\phi(0,\xi)=-T\xi$, that it would have been the same to express as $-\phi(0,T\xi)=-T\xi$ or $\phi(0,-T\xi)=-T\xi$, in other words $\phi(t,-T\xi)$ or $-\phi(t,T\xi)$ should be valid for $-Tf(x)$, but i need to arrive to $\phi(t,T\xi)\equiv T\phi(-t,\xi)$, not the previous one's. Maybe im approaching wrong and I should use another technique, or i could be missing something.
Any feedback is appreciated. Thanks.