Transforming vectors to coordinate vectors Let $X_1,\dots,X_n$ be  n vector fields on an open subset $U$ of a manifold of dimension $n$ Suppose that at $p\in U$, the vectors $(X_1)_p,\dots,(X_n)_p$ are linearly indipendent. would any one say in detail how I will show that there is a chart $(V,x^1,\dots,x^n)$ about $p$ such that $(X)_p=(\partial/\partial x^i)_p$ for all $i=1,\dots,n$, here I use superscript for coordinates like $x^1,\dots,x^n$
 A: By the Frobenius theorem, given a local basis $X_i$ of $TM$, there exists a chart $x_i$ with $X_i = \partial_i$ if and only if the for all $i,j$, $[X_i,X_j]=0$.  
Basically, $[X,Y]$ measures the difference between infinitesmially flowing out $X$, then over on $Y$, versus infinitesimally flowing out $Y$, then flowing over on $X$.  In order to have coordinates, the two must always agree, otherwise the "location" of a point is not well-defined - getting there from the origin by flowing out $X$, then $Y$, would result in a different position flowing out $Y$ from $X$.
A: An easier solution achieved by using inverse function theorem to the desired transformation map between charts : 
Suppose $(X_i)_{i=1}^n$ are local vector field on $U\subset M^n$ such that at $p\in U \subset M^n$, the $n$ vectors $X_i(p) \in T_pU$ are linear independent. If we expand each vector in terms of coordinate basis $(\frac{\partial}{\partial x^i}\Big|_p)$ in $(U, \varphi)$ as 
$$
X_i(p) = \sum_jA_i^j \frac{\partial}{\partial x^j}\Bigg|_p, \quad (i=1,,\dots,n)
$$
this means that the relation 
$$0=\sum _{i=1}^n c^i X_i(p) = \sum_{i,j}c^i A_i^j \frac{\partial}{\partial x^j}\Bigg|_p \implies c^i=0 \quad \text{and} \quad \sum_iA^j_i c^i=0 $$ 
Which is means that $A^j_i$ is invertible. 
Now we want a new coordinate system $(V,\psi)$ with $\psi(p) = (\tilde{x}^1(p),\dots,\tilde{x}^n(p))$ such that
$$
X_i(p) = \frac{\partial}{\partial \tilde{x}^i} \Bigg|_p, \quad \text{for all}\, i=1,\dots,n
$$
By using basis transformation rule 
$$\frac{\partial}{\partial \tilde{x}^i} \Bigg|_p = \sum_j \frac{\partial x^j}{\partial \tilde{x}^i}(p) \frac{\partial}{\partial x^j} \Bigg|_p$$
we have 
$$
\sum_j A_i^j \frac{\partial}{\partial x^j}\Bigg|_p = \sum_j \frac{\partial x^j}{\partial \tilde{x}^i}(p) \frac{\partial}{\partial x^j} \Bigg|_p.
$$
So we have $n\times n$ equation $A^i_j = \frac{\partial x^i}{\partial \tilde{x}^j}(p)$. Now apply inverse function theorem to our transformation map
$$
F : (x^1,\dots,x^n) \rightarrow (\tilde{x}^1,\dots,\tilde{x}^n)
$$
Because the differential $DF(p) =\frac{\partial x^i}{\partial \tilde{x}^j}(p) =A^i_j  $ is invertible at $p$, then there are smaller neighbourhood in $U$ and $V$ s.t the restriction of $F$ to this neighbourhood is diffeomorphism. Therefore the existence of such chart $(V,\psi)$ is guaranteed by this theorem. 
