Basis vectors and non-square matrix transformations

Question

Suppose we have a $2\times3$ matrix which takes a $3\times1$ column vector expressed in a basis $\mathcal{B}$. The output of this multiplication is a $2\times1$ vector. That is: $$\begin{bmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix}_\mathcal{B} = \begin{bmatrix} x_1' \\ x_2' \end{bmatrix}_\text{basis?}$$ My question is what is the basis of this new vector?

If we take the vector: $$\begin{bmatrix} 1 \\ 0 \\ 0 \\ \end{bmatrix}_\mathcal{B}$$ then it sends it in: $$ \mathbf{x}= \begin{bmatrix} a_1 \\ a_2 \end{bmatrix} $$ Of course we can write $\mathbf{x}$ as: $$ \mathbf{x} = a_1\begin{bmatrix} 1 \\ 0 \end{bmatrix} + a_2 \begin{bmatrix} 0 \\ 1 \end{bmatrix} $$

but we don't know the basis.

Answer 1

You are basically asking given a finite spanning set $\mathcal{A}$, how one can find a subset of it that is a basis, one way to do this would be to start with the empty set and keep adding elements until the set you get spans the space(and then you stop).

Answer 2

Let $\{ v_1, v_2, v_3 \}$ be the basis $\mathcal{B}$, and $T$ be the linear transformation represented by the multiplication. A preimage of $T$ is some vector $x_1 v_1 + x_2 v_2 + x_3 v_3$, where $x_1, x_2, x_3 \in \mathbb{R}$ (or whatever field is relevant). The image is expressed with respect to a basis $\{w_1, w_2\}$, let's call it $\mathcal{C}$. This basis is such that $T v_1 = a_1 w_1 + a_2 w_2$, $T v_2 = b_1 w_1 + b_2 w_2$, and $T v_3 = c_1 w_1 + c_2 w_2$.