Suppose that $v_1, ..., v_n \in R^m$, where $R^m = M_{m1}(R)$ for $R$ a commutative ring and $R^m$ a vector space. Let the matrix $A= [v_1 |...| v_n]$, the matrix whose $i$th column $= v_i$. I want to show that $A\cdot e_i = v_i$.
This is just a matrix multiplication problem I guess. So I let $x = \begin{bmatrix} x_1\\ \vdots\\ \ x_n \end{bmatrix}.$ So then I figured $A \cdot x = x_1 v_1 + \cdots + x_n v_n$. Now the canonical basis $e_i = \begin{bmatrix} 0\\ \vdots\\ 0\\ 1\\ 0\\ \vdots\\ 0 \end{bmatrix},$ where $1$ is in the $i$th place. So then it follows that $A \cdot e_i = 0v_1 + \cdots + v_i + 0v_{i+1} + \cdots + 0v_n = v_i$. I think this is correct, but could there have been perhaps a better and more different way to prove this?