I know how to approach finding a matrix of a linear transformation with respect to bases, but I am stumped as to how to approach this in the case of 'determining' and 'proving' whether or not 2 bases $\mathscr C_2$ and $\mathscr C_3$ exist for $\Bbb R^2$ and $\Bbb R^3$, respectively
The linear transformation T : $\Bbb R^3 \rightarrow \Bbb R^2$ in question is defined by T$\begin{pmatrix} x \\ y \\ z \end{pmatrix}$ = $\begin{pmatrix} 2x + y \\ y + 2z\end{pmatrix}$
Where the matrix of T with respect to $\mathscr C_3$ and $\mathscr C_2$ is given as
$Matrix_{\mathscr C_3,\mathscr C_2}$(T) = $\begin{pmatrix} 1 && 0 && 0 \\ 0 && 1 && 0\end{pmatrix}$
It turns out that for any linear map $T:V\to W$ with $V$ and $W$ finite-dimensional, you can always find a pair of bases such that the $\dim W\times\dim V$ matrix of $T$ has the form $$\left[\begin{array}{c|c}I_r & 0 \\ \hline 0&0\end{array}\right],$$ where $r=\operatorname{rank}T$. This is a consequence of the rank-nullity theorem. The proof gives you a way to construct these bases: With $n=\dim V$ and $m=\dim W$, let $\{v_r,\dots,v_n\}$ be a basis for $\ker T$. Extend this to a complete basis for $V$. The vectors $w_1=Tv_1,\dots,w_{r-1}=Tv_{r-1}$ are distinct and linearly independent (why?), so they form a basis for the image of $T$. Extend this to a complete basis of $W$. The matrix relative to these bases then has the desired form.
So, if the only thing that this problem asks you to do is to verify that such bases exist, all you need to do is compare the rank of $T$ to the rank of the given matrix. The latter is obviously 2, so is $\operatorname{rank}T$ also equal to 2? If you also have to produce a pair of bases, the above procedure tells you how to find such a pair. If you’re dealing with real vector spaces, it should be clear from the construction that there’s an infinite number of basis pairs that work.