In John M. Lee book Introductions to smooth manifolds Proposition 8.11 is left as exercise. Can anyone give me hints and suggestions to prove it?
In particular, i want to show:
Let $M$ be a smooth $n$-manifold with or without boundary and let $(X_1,\dots,X_k)$ be a linearly independent $k$-tuple of smooth vector fields on an open subset $U$ of $M$, with $1\le k< n$. Then for each $p\in U$ there exist smooth vector fields $X_{k+1},\dots,X_n$ in a neighborhood $V$ of $p$ such that $(X_1,\dots,X_n)$ is a smooth local frame for $M$ on $U \cap V$.
I can easily see that since $X_1|_p,\dots,X_k|_p$ are vectors linearly independent in $T_pM$ then there exist $v_{k+1},\dots,v_n$ in $T_pM$ such that $\{X_1|_p,\dots,X_k|_p,v_{k+1},\dots,v_n\}$ is a basis for $T_pM$.
I also know that I can extend each $v_i$ to a smooth global vector field on $M$, say $X_i$ (with $i>k$). But then, with this argument, it may be that $(X_1,\dots,X_n)$ is not a local frame for $M$ on $U$. I can only say that $(X_1|_p,\dots,X_n|_p)$ is a basis for $T_pM$.
EDIT Read this below as a (too long for a comment) comment for Sou's answer.
Thank you very much Sou! Please let me add some details to see if I have fully understood. Let be $(W,(x^i))$ a smooth chart for $M$ in $p$, then we have $X_i|_q=X_i^j(q)\frac{\partial}{\partial x^j}|_q$ for each $q$ in $U \cap W$ and each $i\le k$, and we have $v_i=v_i^j\frac{\partial}{\partial x^j}|_p\in T_pM$ for each $k<i\le n$.
Now I define $X_{k+i}|_q=v_i^j\frac{\partial}{\partial x^j}|_q$ for each $q\in W$, and these are smooth vector fields on $W$.
Now the map $G:U \cap W\to M(n\times n, \mathbb{R}), \quad q\mapsto (X_i^j(q))$ is smooth and $G^{-1}(GL(n,\mathbb{R}))$ is the neighborhood of $p$ on which $(X_1,\dots,X_n)$ is a smooth local frame, right?
$\textbf{Hint: }$ After you choose vectors $v_{k+1},\dots,v_n$, you have linearly independent vectors on $T_pM$. Extend $\{v_{k+1},\dots,v_n\}$ around a neighbourhood of $p$, say to constant local vector fields $X_{k+1},\dots,X_n$. Since $\{X_1|_p,\dots,X_k|_p,X_{k+1}|_p,\dots,X_n|_p\}$ linearly independent, the matrix $[X_i^j(p)]$ is invertible. Use continuity of determinant function to show that there is a smaller neighbourhood of $p$ such that $\{X_1,\dots,X_n\}$ is linearly independent there.