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?

  • $\begingroup$ For any matrix $A$ whatsoever for which the product $Ae_i$ is defined, what is its value in terms of the elements of $A$? The answer should suggest a way to construct the matrix $A$ for this problem. $\endgroup$ – amd Feb 7 at 1:37
  • $\begingroup$ @amd Would it's value be the $e_i$ vectors themselves? $\endgroup$ – Matthew Feb 7 at 2:14
  • $\begingroup$ No. Try multiplying out a couple of examples explicitly. $\endgroup$ – amd Feb 7 at 3:38

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