I have this system of differential equations: $$ X'(t)= A(t) X(t), $$
where $A$ is an $n\times n$ matrix of continuous functions.
We know that if $\{X_1,\ldots, X_n\}$ is a system of solutions, then the Wronskian is $$W(t)=\det( X_1(t),\ldots,X_n(t)).$$ How do we prove that $$W'(t)=\text{tr}(A)\, W(t)?$$
Thank you.
$\det$ is n-linear so, in a basis ${\mathcal{B}}$.
$W'(t) = \sum_{j=1}^{n}\det_{\mathcal{B}}(X_1(t),...,X'_j(t),...X_n(t))$
$W'(t) = \sum_{j=1}^{n}\det_{\mathcal{B}}(X_1(t),...,A(t)X_j(t),...X_n(t))$
Lemma:
$\psi: E^n \ni (x_1,\ldots,x_n) \mapsto \sum_{j=1}^{n}\det_{\mathcal{B}}(x_1,\ldots,x_{j-1},u(x_j),x_{j+1},\ldots,x_n)$ is a n-linear form which is antisymmetric (if you permut two variables then the result is multiplied by $-1$) hence it is proportional to $\det$.
It is not difficult to show the scalar is $tr(u)$ (i.e. $\psi(\mathcal{B}) = tr(u))$.