Using Matrices to Find Basis of Image of Linear Transformation

Hi all ---

I'm having trouble understanding one part of this example from my textbook. It's using the matrix of a linear transformation to find the basis of the image of the same linear transformation.

Question ---

Let $L:{{P}_{2}}\to {{M}_{2\times 2}}$ be a linear transformation defined by:

$L\left( {{a}_{0}}+{{a}_{1}}x+{{a}_{2}}{{x}^{2}} \right)=\left[ \begin{matrix} {{a}_{2}} & -{{a}_{2}} \\ {{a}_{0}}-{{a}_{2}} & {{a}_{0}}+{{a}_{2}} \end{matrix} \right]$ .

Let $A = {x^2 - x, x - 1, x^2 + 1}$ be a basis for $P_2$ and
let $B = {$\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right],\left[ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right],\left[ \begin{matrix} 0 & 0 \\ 1 & 1 \end{matrix} \right],\left[ \begin{matrix} 1 & 1 \\ 0 & 0 \end{matrix} \right]$ }$ be a basis for $M_{2 \times 2}$ .

If ${{\left[ T \right]}_{BA}}=\left[ \begin{matrix} 1 & 0 & 1 \\ -1 & 0 & -1 \\ 0 & -1 & 1 \\ 0 & 0 & 0 \end{matrix} \right]$ , use it to find the basis for the image of $L$ .

Textbook Solution ---

Recall that $v$ is in $im L \Leftrightarrow {{\left[ w \right]}_{B}}\in col\left( {{\left[ T \right]}_{BA}} \right)$ .

But $[T]_{BA}$ is already in row-reduced format,
so $\col\left( {{\left[ T \right]}_{BA}} \right)=span\left\{ \left[ \begin{matrix} 1 \\ -1 \\ 0 \\ 0 \end{matrix} \right],\left[ \begin{matrix} 0 \\ 0 \\ -1 \\ 0 \end{matrix} \right] \right\}\ $

This is what I don't understand. How do you know that the vectors in the $span$ here are coordinate vectors? An obvious answer - but not the one I'm looking for - is because these vectors aren't in $P_2$ or $M_{2 \times 2}$ . But what's the true math answer? Thanks a lot ---

so $\[\text{image }T=span\left\{ \left( \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right],\left[ \begin{matrix} 0 & -1 \\ -1 & 0 \end{matrix} \right] \right),\left[ \begin{matrix} 0 & 0 \\ -1 & -1 \end{matrix} \right] \right\}\] $ .