Change of frame of a vector bundle Let $(E, \pi, M)$ be a complex vector bundle of rank $k$. Let $U \subset M$ be an open set and let $f = (s_1, \dots, s_k)$ be a frame (i.e. s_i are linearly independent). Of course $U$ is supposed small enough to admit a frame.
Given $g \colon U \to GL(k,\mathbb{C})$ differentiable, I would like to prove that 
$$(fg)(x) = (\sum_{p=1}^kg_{p,1}(x)s_p(x), \dots, \sum_{p=1}^kg_{p,k}(x)s_p(x)), \qquad x \in U$$is still a frame on $U$.
Also, given $f,\ f'$ frames on $U$ is it possible to prove that there exists $g \colon U \to GL(k,\mathbb{C})$ such that $f' = fg$?
To answer the first question I can't see how to use properly that $g \in GL(k,\mathbb{C})$ to show that elements of the form $\sum_pg_{p,i}s_p$ are linearly independent.
With regard to the second question, I'm not even sure it's true (but I guess so).
Thank for your time and help!
 A: For any point $p \in U$ you want to show that if $(s_1,\ldots, s_k)$ is a basis for $\pi^{-1}(b)$ then $(gs_1, \ldots, gs_k)$ is as well.  This is just linear algebra.  An invertible linear transformation must send linearly independent sets to linearly independent sets.  For if (say) $gs_1$ were expressible as a non-trivial combination of $gs_2, \ldots, gs_k$, then $g$ would have a non-trivial kernel.  So $(gs_1, \ldots, gs_k)$ is a basis for the fiber.
If $g$ is smooth, then the assignment $p \mapsto (gs_1(p), \ldots, gs_k(p))$ is smooth because it is the composition of smooth functions.  So $(gs_1, \ldots, gs_k)$ is a frame.
If $E$ is trivializable over $U$ then I believe the answer to your second question is "yes."  For any two bases of $\mathbb{R}^k$ there is a unique linear transformation sending one to the other.  At each point $p\in U$ call this map $T_p$.  The bundle $E$ looks like $U \times \mathbb{R}^k$ over $U$, so we can think of the map sending each point to a frame as a smooth map from $U$ to an $k$-by-$k$ matrix $S_p$ of the $k$ frame vectors.  The coordinates of $T_p$ vary smoothly in the coordinates of $S_p$.
