# Discrete Symmetry from an ODE

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.

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Your goal $\phi(t,T\xi)\equiv T\phi(-t,\xi)$ will be achieved if you can show two things:
1. $\phi(0,T\xi) = T \phi(0,\xi)$ -- that is, the functions agree at time $t=0$
2. $\frac{d}{dt}\phi(t,T\xi)\equiv \frac{d}{dt} \left(T\phi(-t,\xi)\right)$ -- that is, the functions have the same $t$-derivative.
Item 1 is really a tautology: both sides are equal to $T\xi$ by the definition of $\phi$.
In 2, the left-hand side is $f(T\xi)$. To calculate the right-hand side, notice that derivative commutes with linear transformation $T$, and then use the chain rule: $$\frac{d}{dt} \left(T\phi(-t,\xi)\right) = T \frac{d}{dt} \phi(-t,\xi) = -T f(\xi)$$ Thus, both sides agree.