# Prove that an $m$x$n$ matrix $A$ exists for the linear transformation $u(x)=Ax$.

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?

• 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. – amd Feb 7 at 1:37
• @amd Would it's value be the $e_i$ vectors themselves? – Matthew Feb 7 at 2:14
• No. Try multiplying out a couple of examples explicitly. – amd Feb 7 at 3:38