Prove that the rotation of sums is equal to the rotation of products

Question

So the question starts off:

Prove $$\ e^{t_1+t_2} = e^{t_1}e^{t_2}$$ E(t) is a unique solution to $\dot{E} = E, E(0) = 1$. Let $E_1(t) = E(t_1 + t)$, and E_2(t) = E(t_1)E(t)

$$\dot E_1 (t) = \dot E_1 (t_1 + t) = E(t_1+t) = E_1 (t); E_1(0) = E(t_1)$$ $$\dot E_2 (t) = E(t_1)\dot E (t) = E(t_1)E(t) = E_2(t); E_2(0) = E(t_1)$$

Therefore(by uniqueness and existence theorem I think): $$E_1(t) = E_2(t)$$ Which implies: $$E(t_1+t) = E_1(t) = E_2(t) = E(t_1)E(t)$$ Then by setting $t = t_2$ we get our desired result.

Then the question asks to prove: $$R(t_1+t_2) = R(t_1)R(t_2)$$ where $$R(t) = \begin{bmatrix}\cos(t)&-\sin(t)\\\sin(t)&\cos(t)\end{bmatrix} $$ Given: $R_1(t) = R(t+t_2)$ and $R_2(t) = R(t)R(t_2)$ Prove by a similar argument to the one above.

Any suggestions on how to approach this?

Answer 1

Taking the approach that $R'' = -R$ we see that

$$\ddot R_1(t) = \ddot R(t+t_2) = -R(t+t_2) = -R_1(t)$$ $$R_1(0) = R_1(t_2)$$ $$\ddot R_2(t) = R(t_2)\ddot R(t) = R(t_2)(-R(t)) = -R_2(t)$$ $$R_2(0) = R(t_2)$$

By the existence and uniqueness theorem, $$R_1(t) = R_2(t)$$ $$R(t+t_2) = R_1(t) = R_2(t) = R(t_2)(R(t))$$ $$R(t+t_2) = R(t_2)(R(t))$$

Setting t = $t_1$: $$R(t_1+t_2) = R(t_2)(R(t_1))$$

All you have you to do now is prove that $R'' = -R$

Answer 2

If you want to repeat the argument, you can look at the equation $R''=-R$.

Another way is to notice that $$ R(t)=e^{it}\begin{bmatrix}1/2&i/2\\ -i/2&1/2\end{bmatrix} + e^{-it}\begin{bmatrix} 1/2&-i/2\\ i/2&1/2\end{bmatrix}. $$ Then \begin{align} R(s)R(t)&=e^{it}e^{is}\begin{bmatrix}1/2&i/2\\ -i/2&1/2\end{bmatrix}^2+e^{-it}e^{-is}\begin{bmatrix}1/2&-i/2\\ i/2&1/2\end{bmatrix}^2+2\operatorname{Re}e^{it}e^{-is}\begin{bmatrix}1/2&i/2\\ -i/2&1/2\end{bmatrix}\begin{bmatrix}1/2&-i/2\\ i/2&1/2\end{bmatrix}\\ \ \\ &=e^{i(t+s)}\begin{bmatrix}1/2&i/2\\ -i/2&1/2\end{bmatrix}+e^{-i(t+s)}\begin{bmatrix}1/2&-i/2\\ i/2&1/2\end{bmatrix}+2\operatorname{Re}e^{i(t-s)}\begin{bmatrix}0&0\\ 0&0\end{bmatrix}\\ \ \\ &=\begin{bmatrix} \operatorname{Re} e^{i(t+s)}&-\operatorname{Im} e^{i(t+s)}\\ \operatorname{Im} e^{i(t+s)}&\operatorname{Re} e^{i(t+s)} \end{bmatrix}\\ \ \\ &=\begin{bmatrix} \cos(t+s) &-\sin(t+s)\\ \sin(t+s)&\cos(t+s)\end{bmatrix}\\ \ \\ &=R(t+s) \end{align}