We denote by $A$ a real square matrix of size 2 and we set, for all $x$ in $\mathbb{R}^2$, $f(x) = Ax$. For $a$ in $\mathbb{R}^2$, we denote by $u_a(t)$ the solution on $\mathbb{R}$ of the Cauchy problem $$X' = AX, \quad X(0) = a$$ In other words, $u_a$ is the unique function of class $C^1$ from $\mathbb{R}$ to $\mathbb{R}^2$ such that $u_a(0) = a$ and, for every real $t$, $u_a'(t) = A u_a(t)$. We assume $A$ is diagonal of the form $$A = \operatorname{diag}(\lambda_1, \lambda_2) = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$ Let $a$ and $b$ be two elements of $\mathbb{R}^2$ and let $t$ be a real number. Show that $$\operatorname{det}\left(u_a(t), u_b(t)\right) = \exp\left(t \operatorname{div}_f(a)\right) \operatorname{det}\left(u_a(0), u_b(0)\right)$$
We denote by $A$ a real square matrix of size 2 and we set, for all $x$ in $\mathbb{R}^2$, $f(x) = Ax$. For $a$ in $\mathbb{R}^2$, we denote by $u_a(t)$ the solution on $\mathbb{R}$ of the Cauchy problem
$$X' = AX, \quad X(0) = a$$
In other words, $u_a$ is the unique function of class $C^1$ from $\mathbb{R}$ to $\mathbb{R}^2$ such that $u_a(0) = a$ and, for every real $t$, $u_a'(t) = A u_a(t)$.
We assume $A$ is diagonal of the form
$$A = \operatorname{diag}(\lambda_1, \lambda_2) = \begin{pmatrix} \lambda_1 & 0 \\ 0 & \lambda_2 \end{pmatrix}$$
Let $a$ and $b$ be two elements of $\mathbb{R}^2$ and let $t$ be a real number. Show that
$$\operatorname{det}\left(u_a(t), u_b(t)\right) = \exp\left(t \operatorname{div}_f(a)\right) \operatorname{det}\left(u_a(0), u_b(0)\right)$$