Linear Algebra Working with Linear Transformations Let $v_1=[-3;-1]$ and $v_2= [-2;-1]$
Let $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ be the linear transformation satisfying:
$T(v_1)=[15;-6]$ and $T(v_2)=[11;-3]$
Find the image of an arbitrary vector $[x;y]$
 A: Note sure if (homework) yet. So hint:
Let 
$$
T = 
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
$$
We can re-interpret the given $T(v_1)$ and $T(v_2)$ as:
$$
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\begin{pmatrix}
-3 \\ -1
\end{pmatrix}
=
\begin{pmatrix}
15 \\ -6
\end{pmatrix} ,
\\
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\begin{pmatrix}
-2 \\ -1
\end{pmatrix}
=
\begin{pmatrix}
11 \\ -3
\end{pmatrix} 
$$
Or more succinctly as,
$$
\begin{pmatrix}
a & b \\ c & d
\end{pmatrix}
\begin{pmatrix}
-3 & -2 \\ -1 & -1
\end{pmatrix}
=
\begin{pmatrix}
15 & 11 \\ -6 & -3
\end{pmatrix} 
\tag{1}
$$
Can you take it from here?
A: One method would be to find the image of the standard unit vectors first. Then using linearity, you can find the image of an arbitrary vector.
In a bit more detail:
To find $T(0,1)$, first write $(1,0)$ as a linear combination of $v_1$ and $v_2$. Here you have to solve the equation
$$
(1,0)=\alpha v_1+\beta v_2.
$$
The solution is
$$
(1,0) =v_2-v_1.
$$
Now using the fact that $T$ is linear
$$
T(1,0)=T(  v_2-v_1 )=T(v_2)-T(v_1)=(11,-3)-(15,-6)= (-4,3).
$$
Now, do the same procedure to figure out what  $T(0,1)$ is.
Then you can say
$$
T(x,y)=T\bigl( (x,0)+(0,y)\bigr) =x\,T(1,0)+y\,T(0,1).
$$
A: The  matrix whose columns are T(v1) and T(v2) is the representation matrix 
of T in the basis B=(v1, v2) of R2 (as domain) and E=((1,0), (0,1)) of R2 as codomain.
If you multiply this matrix by the coordinates vector of (x,y) in the basis B
you will get T(x,y) (since E is the canonic basis).
