Let $x=\mathcal{R}^n$ and suppose that $u(x)$ defines a linear transformation of $\mathcal{R}^n$ into $\mathcal{R}^m$. Using the standard basis $\{e_1,...,e_n\}$ for $\mathcal{R}^n$ and the $m$x$1$ vectors $u(e_1),...,u(e_n)$, prove that an $m$x$n$ matrix $A$ exists, for which $u(x)=Ax$, for every $x \in \mathcal{R}^n$.
My progress: the standard basis forms the identity matrix. So, $u(e_i) = A(e_i)$. I am assuming that matrix $A$ is actually composed of row operations that modify $e_i$ so that it is equivalent to itself. Would this mean that $A$ is also the identity matrix?