Let $R$ be a field (or a domain, or a commutative ring), and $S$ an $R$-algebra. Let $B \in M_n(S)$ have $R$-independent entries. Let $A \in GL_n(R)$. Are the entries of $AB$ $R$-independent?
I am particularly interested in the case when $R$ is a field, I just mention the other categories out of curiosity.
I expect this is the case, but I am not sure - I tried to transform a linear relationship between the elements of $AB$ into one between the elements of $B$ using the matrix $A^{-1}$, but got kind of lost in notation.
(I am asking this because I want to prove that $A^{-1} dA$ has $\mathbb{R}$-independent entries, where $A$ is the identity map on the Lie group $GL_n(\mathbb{R})$. Obviously that is true for $dA$, since the entries are a basis for the 1-forms.)
If $A=(a_{ij})_{i,j}\in GL_n(R)$, and $B=(s_{ij})_{i,j}\in M_n(S)$, then $AB=(\sum_{k}a_{ik}s_{kj})_{i,j}$. Suppose $\sum_{i,j}\alpha_{ij}(\sum_{k}a_{ik}s_{kj})=0$ with $\alpha_{ij}\in R$. Then $\sum_{i,j,k}\alpha_{ij}a_{ik}s_{kj}=0$ and this can be written as $\sum_{j,k}(\sum_i\alpha_{ij}a_{ik})s_{kj}=0$. It follows $\sum_i\alpha_{ij}a_{ik}=0$ for all $j,k=1,\dots,n$, that is, $A^T(\alpha_{ij})_{i,j}=0$. Since $A$ is invertible $A^T$ is also invertible, so $(\alpha_{ij})_{i,j}=0$.