Linear Transformation Matrix with respect to B1 and B2 Let $T$ be a linear transformation from $\mathbb R^2 \to \mathbb R^2$ defined by $T(x,y)=(2x-y,x+y)$.
a) Find the matrix of $T$ with respect to the bases $B_1=\{(1,1),(2,1)\}$ and $B_2  = \{(-1,2),(1,1)\}$
b) Use the matrix found in part a) to find $T(v)$, where $v=(2,3)$.
My attempt based on a similar tutorial question:
a) $T(1,1)$ =$(2\times 1 - 1, 1+1)$ = $(1,2)$
Let $(1,2) = a_1 (-1,2) + b_1 (1,1)$
$a_1 = \frac{1}{3}$  and $b_1 = \frac{4}{3}$
$T(2,1)$ =$(2\times 2 - 1, 2+1)$ = $(3,3)$
Let $(3,3) = a_2 (-1,2) + b_2 (1,1)$
$a_2 = 0$  and $b_2 = 3$
Therefore the matrix of transformation is $\begin{bmatrix} \frac{1}{3} & 0  \\ \frac{4}{3} & 3 \end{bmatrix}$
b) $T(V)$ = $\begin{bmatrix} \frac{4}{3}  \\ \frac{7}{3} \end{bmatrix}$
Thoughts? Way off? Struggling to find actual examples of this type of question.
 A: Part a) is absolutely correct. Part b) is not correct, simply because the final answer is supposed to be
$$Tv = T(2, 3) = (1,5).$$
The interesting is how to use the matrix you've found in order to calculate this result.
Let $A$ be the matrix you found. This matrix has the property:
$$[Tw]_{B_2} = A[T]_{B_1} \tag{1}$$
for all $w \in \Bbb{R}^2$. This property is how you're supposed to calculate $Tv$.
First, we need $[v]_{B_1}$. To calculate this, we do exactly what you've done before: solve
$$(2, 3) = a(1, 1) + b(2, 1).$$
Solving this yields $a = 4$ and $b = -1$, meaning that
$$[v]_{B_1} = \begin{bmatrix} 4 \\ -1 \end{bmatrix}.$$
We should therefore have
$$[Tv]_{B_2} = A[v]_{B_1} = \begin{bmatrix} \frac{1}{3} & 0 \\ \frac{4}{3} & 3 \end{bmatrix}\begin{bmatrix} 4 \\ -1 \end{bmatrix} = \begin{bmatrix} \frac{4}{3} \\ \frac{7}{3} \end{bmatrix}.$$
Now, we can express this coordinate column vector as an element of $\Bbb{R}^2$ by expanding
$$Tv = \frac{4}{3}(-1, 2) + \frac{7}{3}(1, 1) = (1, 5),$$
which agrees with what we'd expect, validating your matrix.
The important thing here is not the final answer, but understanding exactly what the matrix does, which is basically equation $(1)$.
