Condition:
Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be the linear operator defined by, $$ T\left ( \begin{bmatrix} x\\ y\\ \end{bmatrix} \right ) = \begin{bmatrix} x+y\\ -2x+4y\\ \end{bmatrix} $$ Also, let $B_1 = \begin{Bmatrix} e_1, & e_2 \end{Bmatrix}$ be the standard basis for $\mathbb{R}^2$ and let $B_2 = \begin{Bmatrix} \begin{bmatrix} 1\\ 1\\ \end{bmatrix}, & \begin{bmatrix} 1\\ 2\\ \end{bmatrix} \end{Bmatrix}$ be a second basis for $\mathbb{R}^2$.
The solution says: $$[T]_{B_1} = \begin{Bmatrix} 1 & 1 \\ -2 & 4 \\ \end{Bmatrix}, [T]_{B_2} = \begin{Bmatrix} 2 & 0 \\ 0 & 3 \\ \end{Bmatrix} $$
Though I could easily calculate $[T]_{B_1}$, by getting $T(e_1$) and $T(e_2$),
I don't have any idea how $[T]_{B_2}$ is calculated. Can't it be calculated through just inputting basis in $B_2$ to the Transformation matrix?
Thank you.
Note that$$T\left(\begin{bmatrix}1\\1\end{bmatrix}\right)=\begin{bmatrix}2\\2\end{bmatrix}=2\begin{bmatrix}1\\1\end{bmatrix}\text{ and }T\left(\begin{bmatrix}1\\2\end{bmatrix}\right)=\begin{bmatrix}3\\6\end{bmatrix}=3\begin{bmatrix}1\\2\end{bmatrix}.$$That's why the matrix of $T$ with respect to the basis $B_2$ is the one that you mentioned.