Exercise in Jacobson's $Basic\ Algebra\ I$, Chapter 3 Well, I even don't understand the problem. 

Let $R$ be a ring and let $(e_1,\ldots,e_n)$ be a base for $R^{(n)}$. If I define:
  $$
f_j=\sum_{j'=1}^n a_{jj'}e_{j'}
$$
  for all $j \in \{1,\ldots,m\}$, then I have to prove the following:
  $$
(f_1,\ldots,f_m)\ \text{is a base for $R^{(m)}$} \Longleftrightarrow \ \exists B\in\mathcal{M}_{n \times m}(R)\ \text{such that}\ AB=1_m,\ BA=1_n,
$$
  where $1_m$ is the  unit matrix in $\mathcal{M}_m(R)$ (analogous with $1_n$). 

What I don't understand is how the $f_j$'s could be a base of $R^{(m)}$ if they're in $R^{(n)}$? 
 A: I'll take @ladisch suggestion and show the following:

Let $R$ be a ring and let $(e_1,\ldots,e_n)$ be a basis for $R^{(n)}$. Define
  $f_j=\sum_{j'=1}^n a_{jj'}e_{j'}$ for $j \in \{1,\ldots,m\}$ and $A=(a_{jj'})$. Prove the following:
  $$(f_1,\ldots,f_m)\ \text{is a basis for $R^{(n)}$} \Longleftrightarrow \ \exists B\in\mathcal{M}_{n \times m}(R)\ \text{such that}\ AB=I_m,\ BA=I_n.$$

If $(f_1,\ldots,f_m)$ is a basis for $R^{(n)}$, then there exist $b_{ij}\in R$ such that $e_i=\sum_{j=1}^mb_{ij}f_j$ for all $i\in\{1,\dots,n\}$. Then $e_i=\sum_{j=1}^mb_{ij}\sum_{j'=1}^n a_{jj'}e_{j'}=\sum_{j'=1}^n(\sum_{j=1}^m b_{ij}a_{jj'})e_{j'}$ and thus $\sum_{j=1}^m b_{ij}a_{jj'}=\delta_{ij'}$. Set $B=(b_{ij})$. Then $BA=I_n$. Analogously one can prove $AB=I_m$.
Conversely, since $(f_1,\ldots,f_m)^T=A(e_1,\ldots,e_n)^T$ (by $^T$ we denote the transpose) and $BA=I_n$ we get (by multiplying with $B$ to the left) $B(f_1,\ldots,f_m)^T=(e_1,\ldots,e_n)^T$, and therefore $(f_1,\ldots,f_m)$ is a system of generators for $R^{(n)}$. Now suppose that $\sum_{j=1}^ma_jf_j=0$ with $a_j\in R$. Using that $(f_1,\ldots,f_m)^T=A(e_1,\ldots,e_n)^T$ and $(e_1,\ldots,e_n)$ are linearly independent we get $(a_1,\dots,a_m)A=0$. Since $AB=I_m$ we get (by multiplying with $B$ to the right) that $(a_1,\dots,a_m)=0$.
A: I didn't know if this was the right place to put it, sorry if that's not the case. I could solve the problem with this change:

$$
\text{Exists $\eta$ isomorphism from $R^{(n)}$ into $R^{(m)}$ such that $\{\eta(f_1),\ldots,\eta(f_m)\}$ is a basis for $R^{(m)}$} \Longleftrightarrow \ \exists B\in\mathcal{M}_{n \times m}(R)\ \text{such that}\ AB=1_m,\ BA=1_n
$$

