In this example, we will recycle the linear transformation $T$ and the bases $B$ and $C$ of Example KVMR but now we will compute the range of $\ltdefn{T}{M_{22}}{P_2}$, given by \begin{equation*} \lt{T}{\begin{bmatrix}a&b\\c&d\end{bmatrix}}= (2a-b+c-5d)+(a+4b+5c+2d)x+(3a-2b+c-8d)x^2 \end{equation*}
With bases $B$ and $C$, \begin{align*} B&=\set{ \begin{bmatrix}1 & 2 \\ -1 & -1\end{bmatrix},\, \begin{bmatrix}1 & 3 \\ -1 & -4\end{bmatrix},\, \begin{bmatrix}1 & 2 \\ 0 & -2\end{bmatrix},\, \begin{bmatrix}2 & 5 \\ -2 & -4\end{bmatrix} }\\ C&=\set{1+x+x^2,\,2+3x,\,-1-2x^2} \end{align*} we obtain the matrix representation \begin{equation*} \matrixrep{T}{B}{C} = \begin{bmatrix} 2 & -24 & 5 & -8 \\ 0 & 8 & 0 & 4 \\ -2 & -26 & -5 & -17 \end{bmatrix} \end{equation*}
We know from Theorem RCSI that the range of the linear transformation $T$ is isomorphic to the column space of the matrix representation $\matrixrep{T}{B}{C}$ and by studying the proof of Theorem RCSI we learn that $\vectrepname{C}$ is an isomorphism between these subspaces. Notice that since the range is a subspace of the codomain, we will employ $\vectrepname{C}$ as the isomorphism, rather than $\vectrepname{B}$, which was the correct choice for an isomorphism between the null spaces of Example KVMR.
Rather than trying to compute the range of $T$ using definitions and techniques from Chapter LT we will instead analyze the column space of $\matrixrep{T}{B}{C}$ using techniques from way back in Chapter M. First row-reduce $\transpose{\left(\matrixrep{T}{B}{C}\right)}$, \begin{equation*} \begin{bmatrix} 2 & 0 & -2 \\ -24 & 8 & -26 \\ 5 & 0 & -5 \\ -8 & 4 & -17 \end{bmatrix} \rref \begin{bmatrix} \leading{1} & 0 & -1 \\ 0 & \leading{1} & -\frac{25}{4} \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} \end{equation*}
Now employ Theorem CSRST and Theorem BRS (there are other methods we could choose here to compute the column space, such as Theorem BCS) to obtain the basis for $\csp{\matrixrep{T}{B}{C}}$, \begin{equation*} \set{ \colvector{1\\0\\-1},\, \colvector{0\\1\\-\frac{25}{4}} } \end{equation*}
We can now convert this basis of $\csp{\matrixrep{T}{B}{C}}$ into a basis of $\rng{T}$ by applying $\vectrepinvname{C}$ to each element of the basis, \begin{align*} \vectrepinv{C}{\colvector{1\\0\\-1}}&= (1+x+x^2)-(-1-2x^2)=2+x+3x^2\\ \vectrepinv{C}{\colvector{0\\1\\-\frac{25}{4}}}&= (2+3x)-\frac{25}{4}(-1-2x^2)=\frac{33}{4}+3x+\frac{31}{2}x^2 \end{align*}
So the set \begin{equation*} \set{ 2+3x+3x^2,\, \frac{33}{4}+3x+\frac{31}{2}x^2 } \end{equation*} is a basis for $\rng{T}$.