Is the following situation about first Chern numbers possible? Let's consider complex vector bundles on a torus $T^2$ constructed in the following way:
Suppose we have a map $f:T^2\to U(n)$, where $U(n)$ is the space of $n\times n$ unitary matrices. This naturally gives us $n$ functions $f_i: T^2\to \mathbb{C}^n, i=1,\ldots,n$, defined to be the function that maps $x\in T^2$ to the $i$th column vector of $f(x)$. Now every $f_i$ defines a rank 1 vector subbundle of the trivial bundle $T^2\times \mathbb{C}^n$ in the obvious way. Suppose there exists at least one $f_i$ of which the first Chern number is nonzero.
Now if we similarly define $f'_i$ to be the map that maps $x$ to the $i$th row of $f(x)$, and again we have $n$ rank 1 vector subbundles(with the canonical identification of dual vector space with vector space, of course) , is it possible that the first Chern numbers of all $f'_i$ are zero? I'm guessing it is impossible but I don't know how to prove it. If there is some theorem straightforwardly addressing this question, then a reference would be good enough. 
EDIT1: As user  @anomaly correctly pointed out, the bundles would be trivial if $f_i$ is continuously defined. I should've added that $f_i$ might not be continuous everywhere on the torus. For example if $f_i$ is smooth except for a finite number of singularities, we can calculate Chern numbers by first calculating the so called "Berry connection"(named so in physics literatures), which is a one-form defined as
$$A_i=if^{\dagger}_idf_i.$$
In turn we can further define the "Berry curvature" as
$$F_i=dA_i.\text{(to be understood locally not globally)}$$
Then the Chern number will be given by
$$c_i=\int_{T^2}F_i.$$
I've only got to know the subject sloppily from physics literatures, so if there is anything that does not make sense, let me know.
EDIT2: After some reflection, I realize it might be more natural to formulate the question in the following somewhat backwards way:
Suppose we have $n$ mutually orthogonal rank 1 subbundles of $T^2\times \mathbb{C}^n$, some of which are nontrivial. Choose a unimodulus (not necessarily continuous everywhere) section for each subbundle, we then have $n$ functions $f_i: T^2\to \mathbb{C}^n, i=1,\ldots,n$. Juxtapose $f_i$ as column vectors, we have a unitary matrix at each point of $T^2$, then look at the $n$ 1-d spaces spanned by each row vector on $T^2$, does each of them also define a rank-1 vector bundle? If they do, can these bundles be all trivial?
 A: The rank $1$ vector bundle $\xi_i \to T^2$ you're constructing from $f_i$ is the one with fibers $(\xi_i)_p = \mathbb{C}f_i(p)$, right? In that case, its Chern class is always going to be trivial; $\xi_i$ has the trivialization $p \to f_i(p)$ (which can't vanish, since $\det f(p)\not = 0$).
A: With the edits, how about this: Suppose the trivial (complex) $n$-bundle $\theta^n \to T^2$  has $\theta = \xi_1 \oplus \cdots \oplus \xi_n$ for some line bundles $\xi_i\to T^2$. On suitably small patches $U$, we can choose a local trivialization $s_i :U\to E(\xi_i\vert U)\subset E(\theta\vert U) = \mathbb{C}^n$. Then we have a map $s : U \to \text{End}(\mathbb{C}^n)$ defined by $s = s_1 \oplus \cdots \oplus s_n$. We can then consider the transpose $s^t : U\to \text{End}(\mathbb{C}^n)$, which we can project to get linearly independent maps $s_i^t: U\to \mathbb{C}^n$. Patching these maps over the $U\subset T^2$ gives subbundles $\xi'_1, \dots, \xi'_n\subset \theta^n$. We want to know whether the $\xi'_i$ can all be trivial if some $\xi_i$ is nontrivial.
Repeating the construction $\xi_i \to \xi'_i$ twice gives bundles $\xi_i'' = \xi_i$, since we're simply applying the transpose twice on the same patches $U$. Furthermore, if the $\xi_i$ are all trivial, then the $\xi'_i$ are clearly also trivial; take the $s_i$ to be global nonwhere-vanishing sections. It follows that if all the $\xi'_i$ are trivial, then the $\xi_i$ are also trivial. (Well, it follows modulo defining the construction above more carefully and rigorously.)
Is that closer to what you have in mind?
