I am trying to prove that for a smooth manifold $M$ there is a chart $( U, \phi = (u_1, \dots, u_m))$ such that $\phi(U)=\mathbb{R}^m$, $\phi(p)=0$ and $\xi= \left. \frac{\partial}{\partial u_1} \right|_p$ where $p\in M $ and $0 \neq \xi \in T_pM$.
I have already proven the existence of a chart satisfying the first two conditions and the third one seems obvious to me because I can express $\xi = \sum a_i \partial_{u_i}$ and I could perfom a rotation in $T_pM$ which is a linear space to obtained the desired property.
But I've been having problem to express the above intuition in a rigorous way. How could I do it?
Given any chart $\phi \colon U \rightarrow \mathbb{R}^m$ around $p \in U$ with $\phi(p) = (u_1(p), \ldots, u_m(p)) = 0$ and $\phi(U) = \mathbb{R}^m$, express $\xi$ using the corresponding basis as $\xi = \sum_{i=1}^m a_i \frac{\partial}{\partial u_i}|_p$. Then take any linear map $T \colon \mathbb{R}^m \rightarrow \mathbb{R}^m$ which is an isomorphism and satisfies $T\left( \sum_{i=1}^m a_i e_i \right) = e_1$ where $e_1, \ldots, e_m$ is the standard basis of $\mathbb{R}^m$. This corresponds intuitively to "rotating" $\mathbb{R}^m$ so that $\sum_{i=1}^m a_i e_i$ will align with $e_1$ (and so, after composing with $\phi$, then $\xi$ align with $\frac{\partial}{\partial v_1}|_p$) - only the map $T$ is not neccesary orthogonal and so is not a rotation per se, just a linear isomorphism. Then the map $T \circ \phi = (v_1, \ldots, v_m) = (T \circ u_1, \ldots, T \circ u_m)$ will also be a chart and will satisfy $\frac{\partial}{\partial v_1}|_p = \xi$.