Let $\pi:E\to M$ be a smooth rank-$k$ vector bundle, with $k\geq 0$. Then $E$ is isomorphic to $M\times \mathbb R^k$ iff there exist smooth sections $s_1,\dots,s_k\in \Gamma(E)$ such that $s_1(x),\dots,s_k(x)$ form a basis of $E_x$ for every $x\in M$, where for $k=0$ this means the empty family is the basis of each zero-dimensional fibre.