Why are the transition maps for local frames smooth? In our lecture local frames were defined as follows:

A local frame on a manifold $M$ is an open subset $U$ together with an ordered tuple of vectors fields $(X_1,\dots,X_n)$, such that the $X_i$ are smooth on $U$ and for every $p\in U$ the tuple $(X_1(p),\dots,X_n(p))$ is a basis of $T_pM$.

It is then stated that given two frames $(U,X_i)$ and $(V,Y_i)$, there exists a smooth map $\theta:U\cap V\to\operatorname{GL}_n(\mathbb R)$ such that $Y_j=\sum_{i=1}^n\theta_{ij}\cdot X_i$ on $U\cap V$. Why should such a $\theta$ exist? Since at every $p\in U\cap V$ both $X_i$ and $Y_i$ are bases, there should exist an invertible change-of-basis matrix $\theta(p)$, but why should $\theta$ depend smoothly on $p$?
 A: This is essentially because matrix inversion is a smooth mapping ($A\mapsto A^{-1}$ from $GL_n(\Bbb{R})\to GL_n(\Bbb{R})$ is smooth; in fact analytic in the usual sense sense: about every point it has a convergent power series expansion). I think it's called the Neumann series? Another way of thinking about this is that Cramer's rule gives an explicit analytic formula for the entries of the inverse matrix.
Since smoothness is a local issue, we may as well assume (by taking sufficiently intersections of open sets) that $U=V$ and that this is the domain of a coordinate chart $(U,(x^1,\dots, x^n))$. Now, we have the "standard" vector fields $\frac{\partial}{\partial x^1},\dots, \frac{\partial}{\partial x^n}$. Now, we can write
\begin{align}
X_i&=\sum_{k=1}^n\xi_{ik}\frac{\partial}{\partial x^k}\quad \text{and}\quad Y_i=\sum_{k=1}^n\eta_{ik}\frac{\partial}{\partial x^k}
\end{align}
for some functions $\xi_{ik},\eta_{ik}:U\to\Bbb{R}$. Note that each $X_i,Y_i$ is smooth if and only if every $\xi_{ik},\eta_{ik}$ is smooth. Now, $Y_j=\sum_{i=1}^n\theta_{ij}X_i$ says that for all $j,k\in\{1,\dots, n\}$, we have $\eta_{jk}=\sum_{i=1}^n\theta_{ij}\xi_{ik}$. Or, as a matrix equation, $\eta=\theta^t\cdot \xi$.
The fact that each $\{X_1,\dots, X_n\}$ and $\{Y_1,\dots, Y_n\}$ form a frame is reflected in the fact that the matrices $\xi,\eta$ are pointwise invertible, i.e $\xi,\eta:U\to GL_n(\Bbb{R})$ are smooth. So,
\begin{align}
\theta= (\eta\cdot \xi^{-1})^t,
\end{align}
is thus a smooth matrix-valued function. So, this completes the proof.

Note that the purpose of introducing coordinates and the coordinate vector fields $\frac{\partial}{\partial x^i}$ is that in this fashion, it is almost trivial from the definitions that the dual sections (in this case $\{dx^1,\dots, dx^n\}$ are smooth covector fields). I didn't explicitly use the language of the smoothness of the dual sections above, but that's another way of looking at it.
Also, similar statements hold in the more general vector bundle setting as well.
