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.
Best Answer
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 $(1)$.