(a) Let $R(t, \omega)$ denote the rotation matrix $$ R(t, \omega)=\left[\begin{array}{cc} \cos (\omega t) & -\sin (\omega t) \\ \sin (\omega t) & \cos (\omega t) \end{array}\right] . $$ Find a $2 \times 2$ matrix $G(\omega)$ such that $R(t, \omega)=e^{t G(\omega)}$.
(b) Set $A(t)=R(t, \omega) B[R(t, \omega)]^{-1}$ for $B \in \mathcal{B}$. Let $\dot{x}=A(t) x$ and define $y(t)=$ $[R(t, \omega)]^{-1} x(t)$. Show that $$ \dot{y}=[B-G(\omega)] y $$ and find the corresponding expressions for $y(t), x(t)$, for any given $x(0)$.
(c) For $$A(t)= \begin{pmatrix} -1-9(\cos(6t))^2+12\sin(6t)\cos(6t) & 12(\cos(6t))^2+9\sin(6t)\cos(6t)\\ -12(\sin(6t))^2+9\sin(6t)\cos(6t) & -1-9(\sin(6t))^2+12\sin(6t)\cos(6t)\\ \end{pmatrix}$$
, find $B, \omega$, and the eigenvalues of $B-G(\omega)$.
(d) Based on the solution obtained in (b), to generate a $2 \times 2$ matrix $A(t)$ with negative real eigenvalues such that (1) holds, it suffices to find a $2 \times 2$ matrices $B$ and $G(\omega)$ such that there exists a nonzero vector $v$ and a $\lambda>0$ for which $[G-B(\omega)] v=\lambda v$. Prove that this requirement can be met if and only if $B \in \mathcal{B}$, where $\mathcal{B}$ is the collection of matrices having negative eigenvalues s.t $xBx>0$.
My try:
a) Proof. (a) Since $R(t, \omega)$ has two complex eigenvalues, the most efficient way I can thought of is to regard it as an element of Lie group $S O(2)$, and it suffices to find the generator $X$ for the corresponding lie algebra so(2), i.e. $$ \frac{d}{d \alpha} R(\alpha)=X R(\alpha) $$ Then we get $$ \left[\begin{array}{cc} -\sin \alpha & -\cos \alpha \\ \cos \alpha & -\sin \alpha \end{array}\right]=\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]\left[\begin{array}{cc} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{array}\right] $$ $X$ is a generator for the 1-dimensional lie algebra, so $$ G(\omega)=\omega \cdot\left[\begin{array}{cc} 0 & -1 \\ 1 & 0 \end{array}\right]=\left[\begin{array}{cc} 0 & -\omega \\ \omega & 0 \end{array}\right] $$ (b) We take the derivative for $x(t)=R(t, \omega) y(t)$, then $$ \begin{aligned} R^{\prime}(t, \omega) y(t)+R(t, \omega) \dot{y}(t) &=R(t, \omega) B y(t) \\ R(t, \omega) \dot{y}(t) &=\left[R(t, \omega) B-R^{\prime}(t, \omega)\right] y(t) \\ \dot{y}(t) &=\left[B-\left[R(t, \omega)^{-1}\right] R^{\prime}(t, \omega)\right] y(t) . \end{aligned} $$ It suffices to show $G(\omega)=\left[R(t, \omega)^{-1}\right] R^{\prime}(t, \omega)$ and notice that $$ \left[R(t, \omega)^{-1}\right] R^{\prime}(t, \omega)=e^{-t G(\omega)} \cdot\left(e^{t G(\omega)} \cdot G(\omega)\right)=G(\omega) . $$ Now we have
$$ \begin{aligned} &y(t)=\exp [B-G(\omega)] \cdot[R(t, \omega)]^{-1} x(0) \\ &x(t)=R(t, \omega) \exp [B-G(\omega)] \cdot[R(t, \omega)]^{-1} x(0) \end{aligned} $$
(c) For example (2), we know $\omega=6$ and $$ \begin{aligned} B=[R(t, 6)]^{-1} A(t) R(t, 6) &=\left[\begin{array}{cc} \cos (6 t) & \sin (6 t) \\ -\sin (6 t) & \cos (6 t) \end{array}\right] A(t)\left[\begin{array}{cc} \cos (6 t) & -\sin (6 t) \\ \sin (6 t) & \cos (6 t) \end{array}\right] \\ &=\left[\begin{array}{cc} \cos (6 t) & -\sin (6 t) \\ \sin (6 t) & \cos (6 t) \end{array}\right] \end{aligned} $$
Now I am stuck with mainly d) and partially c). Please help me if you can.