Linear transformation in linear algebra Let $e_1= \begin{bmatrix}
1\\
0
\end{bmatrix}
$
Let $e_2= \begin{bmatrix}
0\\
1
\end{bmatrix}
$
Let 
$y_1= \begin{bmatrix}
2\\
5
\end{bmatrix}
$
$y_2= \begin{bmatrix}
-1\\
6
\end{bmatrix}
$
Let $\mathbb{R^2}\rightarrow\mathbb{R^2}$ be a linear transformation that maps e1 into y1 and e2 into y2.
Find the images of
$A= \begin{bmatrix}  
5\\
-3
\end{bmatrix}
$
$b= \begin{bmatrix}  
x\\
y
\end{bmatrix}
$
I am not sure how to this. I think there is a 2x2 matrix that you have to find that vies you the image of A.

If $L$ is a linear transformation that maps $\begin{bmatrix}
1\\
0
\end{bmatrix}
$
to $\begin{bmatrix}
2\\
5
\end{bmatrix}
$, $L$ has a matrix representation $A$, such that $A \begin{bmatrix}
1\\
0
\end{bmatrix}
=\begin{bmatrix}
2\\
5
\end{bmatrix}$. But this means that $\vec{a_1}^{\,}$ is just $\begin{bmatrix}
2\\
5
\end{bmatrix}$.
The same reasoning can be applied to find the second column vector of $A$.
Once you have the matrix representation of $L$, you're good.
