Consider $y'(t)=A(t)y(t), \ A\in\mathbb{R}^{n\times n}$ and $A_{ij}(t)$ are continuous. If $A(t)$ has the form
$A(t)=\begin{bmatrix}A_{11}(t) & A_{12}(t) \\ 0 & A_{22}(t)\end{bmatrix}$
where $A_{11}\in\mathbb{R}^{n_1\times n_1},A_{11}\in\mathbb{R}^{n_2\times n_2} $ with $n_1+n_2=n$. Show that the state-transition matrix $G(t,t_0)$ is of the form
$G(t,t_0)=\begin{bmatrix}G_{11}(t,t_0) & G_{12}(t,t_0) \\ 0 & G_{22}(t,t_0)\end{bmatrix}$
I tried taking $\frac{d}{dt}G(t,t_0)=A(t)G(t,t_0)$ and expanding but I didn't reach any result, any tips would be great.
From $\partial_t G(t,t_0) = A(t)G(t,t_0)$, we have $$ \begin{bmatrix}\partial_tG_{11} & \partial_tG_{12} \\ \partial_tG_{21} & \partial_tG_{22}\end{bmatrix} = \begin{bmatrix}A_{11}G_{11}+A_{12}G_{21} & A_{11}G_{12}+A_{12}G_{22} \\ A_{22}G_{21} & A_{22}G_{22}\end{bmatrix}. $$ Separating into homogeneous and inhomogeneous parts gives $$ \begin{bmatrix}\partial_tG_{11}-A_{11}G_{11} & \partial_tG_{12}-A_{11}G_{12} \\ \partial_tG_{21}-A_{22}G_{21} & \partial_tG_{22}-A_{22}G_{22}\end{bmatrix} = \begin{bmatrix}A_{12}G_{21} & A_{12}G_{22} \\ 0 & 0\end{bmatrix}. $$ Turning to our initial condition, $G(t_0,t_0) = I_n$, we have $$ G_{11}(t_0,t_0) = I_{n_1}\,,\,G_{12}(t_0,t_0) = 0\,,\, G_{21}(t_0,t_0) = 0\,,\, G_{22}(t_0,t_0) = I_{n_2}. $$ $G_{11}$ and $G_{22}$ have nonzero initial conditions, so clearly they won't be identically zero. $G_{12}$ has zero initial conditions, but is an inhomogeneous equation, so it also will not be identically zero. However, $G_{21}$ solves a homogeneous equation with zero initial conditions, so $G_{21} = 0$.