Let $M \in M_{n}(\mathbf{R})$. We denote by (S) the differential system: $$(\mathrm{S}) \quad X' = MX$$ where $X$ is a function from the variable $t$ in $\mathbf{R}$ to $\mathbf{R}^{n}$, differentiable on $\mathbf{R}$. Let $T \in M_{n}(\mathbf{R})$. We assume that $M$ is similar to $T$ in $M_{n}(\mathbf{R})$ and we denote by $(\mathrm{S}^{*})$ the differential system $$(\mathrm{S}^{*}) \quad Y' = TY$$ Prove that the coordinates of a solution $X$ of (S) are linear combinations of the coordinates of a solution $Y$ of $(\mathrm{S}^{*})$.
Let $M \in M_{n}(\mathbf{R})$. We denote by (S) the differential system:
$$(\mathrm{S}) \quad X' = MX$$
where $X$ is a function from the variable $t$ in $\mathbf{R}$ to $\mathbf{R}^{n}$, differentiable on $\mathbf{R}$.
Let $T \in M_{n}(\mathbf{R})$. We assume that $M$ is similar to $T$ in $M_{n}(\mathbf{R})$ and we denote by $(\mathrm{S}^{*})$ the differential system
$$(\mathrm{S}^{*}) \quad Y' = TY$$
Prove that the coordinates of a solution $X$ of (S) are linear combinations of the coordinates of a solution $Y$ of $(\mathrm{S}^{*})$.