I am having trouble showing this claim and honestly I was not able to get very far:
Suppose we have $n$ solutions $X_1,...X_n \in C([a,b],\mathbb{C}^n)$ to $X'=A(t)X$ where each entry in $A(t)$ is continuous and there exists a value $t_0$ on $[a,b]$ for which the Wronskian $W[X_1,..,X_n](t_0)\not=0$. Show that $W[X_1,..,X_n](t)=0$ for all $t$ in $[a,b]$.
My attempt: Since we know that there exists a $t_0$ in $[a,b]$ such that the $$det([X_1(t_0),...,X_n(t_0)])\not = 0$$Which means that $\{X_i(t_0)\}$ are linearly independent. From here I was hoping to use uniqueness of solutions and maybe argue that this determinant is independent of $t$, but honestly I do not know how to approach it. If some can give me a small hint so I can get started in the correct direction I would appreciate it very much.
If $W(t_1)=0$, then for this point there exist non-trivial coefficients so that $$ 0=c_1X_1(t_1)+c_2X_2(t_1)+...+c_nX_n(t_1). $$ Then build $$ Y(t)=c_1X_1(t)+c_2X_2(t)+...+c_nX_n(t). $$ By linearity, this is also a solution of the ODE. Because of $Y(t_1)=0$ and uniqueness, $Y(t)=0$ everywhere. This is a contradiction to linear independence at $t_0$.