Let $T$ be a vector subspace of $\mathbb{R}^n$, and suppose that $S$ is the subspace of $\mathbb{R}^m$ given by:
$S = \{u(x):x \text{ belongs to } T\}$, where the transformation defined by $u$ is linear. Show that an $m \times n$ matrix $A$ exists, which satisfies $u (x)=Ax$ for every $x$ in $T$.
I have been given a proof but am having a hard time applying it to begin the problem. Should I pick arbitrary vectors, say $x_1, x_2$ and make them undergo a linear transformation by picking arbitrary matrix $u$? The proof is as follows
Let $\{x_1,x_2,\ldots,x_r\}$ be a basis for $T$ and define $y_i= ux; i=1,2,\ldots,n$ be the columns of $A$. Define the $n \times r$ matrix $X$ to be $[x_1, x_2, \ldots ,x_r]$ and the $m \times r$ matrix $Y$ to be $[y_1 y_2, \ldots, y_r]$. Then the system of equations $Ax_i=y_i;i=1,2,...,r$ leads to the matrix equations: $AX=Y$ and $X'A'=Y'$ for the unknown matrix $A.$ The rank of $X$ and $X'$ is because $\{x_1,x_2,\ldots,x_r\}$ be a basis for $T$. You may use Gaussian Elimination with back substitution on $X'A'=Y$' to obtain $BX'A'=BY'$ and $[I_r \mid U][A_1'/A_2']=BY'$ (the division symbol being column operations), where $U$ is $rx(n-r)$, $A_1$ is $r \times m$ and $A_2$ is $(n-r) x m.$ Then $A_1'+UA_2'=BY'$ and the general solution for $A$ is given by $AA_1'=BY'-UA_2'$ where $A_2$ is an arbitraty matrix. $A_2 = 0$ gives us one of the solutions.