Determining linear transformation Let $V$ be a vector space, and $T:V→V$ a linear transformation such that:
$T(2v_1 + 3v_2) = -5v_1 - 4v_2$ and $T(3v_1 + 5v_2) = 3v_1 -2v_2$
Then:
T(v1)= ? v1+ ? v2
T(v2)= ? v1+ ? v2
T(4v1+2v2)= ? v1+ ? v2
I cannot solve this problem and have been at it for hours. I found a similar question here: Finding the basis of a vector space. I tried applying the same operations, but do not understand how they got to the final solution.
 A: Simple rule of linear transformation,
$T(c \alpha+d \beta)=c T(\alpha)+d T(\beta)$, where $c,d$ are scalars.
Take $\alpha=2 v_1 +3 v_2 , \beta=3 v_1 + 5 v_2 $
Now , you have to figure out for what values of $c,d$ ,
$c \alpha + d \beta = v_1 $ for finding the value $T(v_1)$
Now , from $c \alpha + d \beta = v_1 $ ,
We get,
$(2c+3d) v_1 + (3c+5d) v_2 = v_1 $
So, $2c+3d=1$ and $3c+5d=0 $
Solving we get, $c=5 , d=-3$
So, $T(5 \alpha - 3 \beta) = 5 T(\alpha)- 3 T(\beta) $
$\implies$ $ T(v_1)= $ ?
Now , I think you can figure out the rest.
A: I was actually able to figure it out (finally!). Thanks mostly to the other thread, but adding this solution as well in case it helps anyone else. Here it is:
written in matrix form:
$$\pmatrix{Tv_1& Tv_2}\pmatrix{2&3\\3&5} = \pmatrix{v_1&v_2}\pmatrix{-5&3\\-4&-2}\to \\
\pmatrix{Tv_1& Tv_2}=\pmatrix{v_1&v_2}\pmatrix{-5&3\\-4&-2}\pmatrix{2&3\\3&5}^{-1}\\
= \pmatrix{v_1&v_2}\pmatrix{-5&3\\-4&-2}\pmatrix{5&-3\\-3&2}\\
=\pmatrix{v_1&v_2}\pmatrix{-34&21\\-14&8}$$
so we have $$T(v_1) = -34v_1 - 14v_2, \, T(v_2) = 21v_1+8v_2\,\\ T(4v_1+2v_2) =4(-34v_1-14v_2)+2(21v_1+8v_2)=-94v_1-40v_2 $$
