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Let $V$ be a vector space, and $T : V \to V$ a linear transformation such that $T(2v_1 - 3v_2) = 3v_1 + 5v_2$ and $T(-3v_1 + 5v_2) = -3v_1 + 3v_2$.

Then

$T(v_1) = ??? v_1 + ??? v_2$

$T(v_2) = ??? v_1 + ??? v_2$

$T(-4v_1 +4v_2) = ??? v_1 + ??? v_2$

I'm not really sure where to start with this problem. My first thought is that since I have the transformation, I need to find the basis in order to solve this. If this is what I need to do, then I'm not really sure how I would do it.

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Once you've found $T(v_1)$ and $T(v_2)$, the problem is essentially finished (the last part can be solved then from linearity of $T$).

The idea is that we're essentially given two linear equations in two variables (which are solved rather easily): we're given $$T(2v_1-3v_2)=3v_1+5v_2 \quad \text{and} \quad T(-3v_1+5v_2)=-3v_1+3v_2.$$ By linearity, that means we have $$2T(v_1)-3T(v_2)=3v_1+5v_2 \quad \text{and} \quad -3T(v_1)+5T(v_2)=-3v_1+3v_2.$$ Let's write $w_1=T(v_1)$ and $w_2=T(v_2)$ (just to make it easier to write repeatedly). Then we have the matrix equation $$ \left[\begin{matrix} 2 & -3 \\ -3 & 5 \end{matrix}\right]\left[ \begin{matrix} w_1 \\ w_2 \end{matrix}\right] =\left[ \begin{matrix} v_1 \\ v_2 \end{matrix}\right]$$ So solving this (pick your favorite method, e.g. Gaussian elimination, multiplication by inverse matrix, Cramer's Rule, etc) give us $$ \left[ \begin{matrix} w_1 \\ w_2 \end{matrix}\right] =\left[ \begin{matrix} 5 & 3 \\ 3 & 2 \end{matrix}\right]\left[ \begin{matrix} v_1 \\ v_2 \end{matrix}\right]$$ so that $w_1=5v_1+3v_2$ and $w_2=3v_1+2v_2$.

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There are a number of ways to approach your problem. You could view $\beta = \{2v_1-3v_2,-3v_1+5v_2\}$ as a basis (assuming $\alpha = \{v_1,v_2\}$ is a basis).

From this view point you have been given that the coordinate matrix for $T$ relative to $\beta$ (input basis) and $\alpha$ (output basis) is $$[T]_\beta^\alpha = \begin{bmatrix} 3 & -3 \\ 5 & 3 \end{bmatrix}$$

You can easily write down the change of basis matrix from $\beta$ to $\alpha$: $$[I]_\beta^\alpha = \begin{bmatrix} 2 & -3 \\ -3 & 5 \end{bmatrix}$$

Then $[T]_\alpha^\alpha = [T]_\beta^\alpha [I]_\alpha^\beta = [T]_\beta^\alpha ([I]_\beta^\alpha)^{-1}$ will give you the answer to your first set of questions. At this point, computing $T(-4v_1+4v_2)$ is easy.

On the other hand, you could (as with most any elementary linear algebra problem) compute the answer by brute force. What I mean here is to notice: $(1/3)T(v_2) = T(2v_1-3v_2)+(2/3)T(-3v_1+5v_2) = (3v_1+5v_2)+(2/3)(-3v_1+3v_2)$ (simplifying gives $T(v_2)$) and then find $T(v_1)$ in a similar manner.

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you can write $$T(2v_1 - 3v_2)=2Tv_1 -3Tv_2 = 3v_1 + 5v_2,\, T(-3v_1 + 5v_2)= -3Tv_1 + 5Tv_2 = -3v_1 + 3v_2$$ in matrix form as $$\pmatrix{Tv_1& Tv_2}\pmatrix{2&-3\\-3&5} = \pmatrix{v_1&v_2}\pmatrix{3&-3\\5&3}\to \\ \pmatrix{Tv_1& Tv_2}=\pmatrix{v_1&v_2}\pmatrix{3&-3\\5&3}\pmatrix{2&-3\\-3&5}^{-1}\\ = \pmatrix{v_1&v_2}\pmatrix{3&-3\\5&3}\pmatrix{5&3\\3&2}\\ =\pmatrix{v_1&v_2}\pmatrix{6&3\\34&21}$$

so we have $$Tv_1 = 6v_1 + 34v_2, \, Tv_2 = 3v_1+21v_2\,\\ T(-4v_1+4v_2) =4(3v_1+21v_2-6v_1-34v_2)=-12v_1-52v_2 $$

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