I'm having trouble solving the following problem.
A vector-valued function $F$, which is never zero and has a continuous derivative $F'(t)$ for all $t$, is always parallel to its derivative. Prove that there is a constant vector $A$ and a positive real-valued function $u$ such that $F(t)=u(t)A$ for all $t$.
My idea is to use the second fundamental theorem of calculus, $F(x)=F(c)+\int_c^xF'(t)dt$, and the fact that $F\times F'=0$, from the hypotheses that $F'$ is continuous and parallel to $F$. However, I can't get any further from here. I would appreciate any solutions, hints or suggestions.
By hypothesis, there exists a scalar function $\lambda(t)$ such that $F'(t) = \lambda(t) F(t)$. So if $F = (F_1,F_2,\ldots, F_n)$, we have $F_i'(t) = \lambda(t) F_i(t)$ for $i = 1,2,\ldots, n$. Hence $F_i(t) = F_i(0)e^{\int_0^t \lambda(s)\, ds}$ for all $i$. Let $A = (F_1(0), F_2(0),\ldots, F_n(0))$ and $u(t) = e^{\int_0^t \lambda(s)\, ds}$. Then $F(t) = u(t)A$ for all $t$.