Here is the more general setting.
Let $V$ (resp. $W$) be an $n$ (resp. $m$) dimensional vector space over $\mathbb{F}$. Let $\alpha=(v_1,\cdots,v_n)$ be an ordered basis in $V$ and $\beta=(w_1,\cdots,w_m)$ an ordered basis in $W$.
In your question, $V=U$ and $W=\mathbb{F}^3$. You have found $\alpha$. Note that $\beta$ is given. Also, $n=2$ and $m=2$.
For any vector $x\in V$, denote its coordinate w.r.t. the basis $\alpha$ as
$
[x]_\alpha=(x_1,\cdots,x_n)^T
$
and for any vector $y\in W$, denote its coordinate w.r.t. the basis $\beta$ as
$
[y]_\beta=(y_1,\cdots,y_m)^T.
$
Let $T:V\to W$ be a linear transformation. Let $[T]_\alpha^\beta$ denotes the matrix for $T$ w.r.t. the bases $\alpha$ and $\beta,$ i.e.,
$$
[T]^\alpha_\beta=\bigr[[Tv_1]_\beta,\cdots,[Tv_n]_\beta\bigr].
$$
Note in particular that $[T]^\alpha_\beta$ is an $m\times n$ matrix.
So what are the steps to find $[T]^\alpha_\beta$?
- Note that you are given $\alpha$ and $\beta$, and $T$. Identify what are $m$ and $n$ for your problem;
- Find $Tv_j$ for each $j$;
- and then find $[Tv_j]_\beta$.
For the very last step, you need to know how to find $[z]_{\beta}$ given $z\in W$. Suppose $[z]_\beta=(z_1,\cdots,z_m)^T$ Then by the definition of coordinates:
$$
z=\sum_1^m z_iw_i
$$
which essentially gives you a linear equation about the $z_i$'s.
Best Answer
Yes, your thought on how to approach it is correct. Every linear transformation is uniquely determined by its action on a basis. In this case, suppose the matrix of the transformation is $A = \left( \begin{array}{cc} a & b \\ c & d \end{array}\right)$. So you know $$ \left( \begin{array}{cc} a & b \\ c & d \end{array}\right)\left(\begin{array}{c} 1 \\ 0 \end{array}\right) = \left (\begin{array}{c} 3 \\ 5 \end{array}\right) $$ Does that tell you anything about the matrix $A$?