Let $M$ be a smooth manifold and $n \leq dim(M)$. Let $\omega_1,...,\omega_n$ be 1-forms on $M$, such that for every $q \in M$ their evaluations $(\omega_1)_q,...,(\omega_n)_q$ at $q$ are linearly independent. Let $\theta_1,...,\theta_n$ be 1-forms on $M$ such that $ \sum\limits_{i=1}^{n}\theta_i\wedge\omega_i=0 $. Prove that there exist smooth functions $A_{ij} $on M with $A_{ij} = A_{ji}$ such that for all $i = 1,..., n$ we have $\theta_i=\sum\limits_{i=1}^{n}A_{ij}\omega_j$
I can see that this is correct pointwisely, but I'm a bit stuck with the smoothness.
The fact that $\omega_i$ and $\theta_i$ are smooth $1$-forms (that should be part of the definition with which you're working) means that any local expression of the $\omega_i$ in terms of the $\theta_j$ has smooth coefficients. (For example, choose (smooth) vector fields $X_j$ dual to $\theta_j$ and consider the functions $\omega_i(X_j)$.)