Find matrix representing Linear Transformation I want to find a matrix $A$ which could transform the parallelogram $P1$ with vertices $(0,0),(0,1),(1,1),(1,2)$ to $P2$ with vertices $(0,0),(6,0),(1,1),(7,1)$. The matrix $A$ should be a two by two matrix.
The question is, how can I find it given that apart from the correspondence between the two $(0,0)$ I was not given which point corresponds to which ?
Also, it it possible (without detailed calculation) to say whether there exists a matrix $B$ such shat $B^2$ takes $P1$ to $P2$?
 A: If you see it as a sort of rotation on the right, such that,
$(0,0)\rightarrow (0,0)$, $(1,1)\rightarrow (6,0)$, $(0,1)\rightarrow (1,1)$, $(1,2)\rightarrow (7,1)$
it is fairly easy to solve the problem. You then need to find the matrix $A$ such that 
$\begin{pmatrix}
a&b \\
c&d
\end{pmatrix}$ 
$\begin{pmatrix}
x \\
y
\end{pmatrix}$ =
$\begin{pmatrix}
x' \\
y'
\end{pmatrix}$.
Doing this, you will notice that it works for every pair of points, and you will find the matrix
$A=
\begin{pmatrix}
5&1 \\
-1&1
\end{pmatrix}
$.
A: First consider the simpler problem of finding the matrix that transforms the unit square to a parallelogram, and we'll then reduce your problem to two of these kinds of problems. This gives a straightforward and systematic method for this problem.
The unit square has vertices $(0,0),(1,0),(0,1),(1,1)$ and is generated by linear combinations $c_1 e_1+c_2 e_2$ of the unit vectors $e_i$, where $0 \le c_i \le 1$.  The parallelogram $P1$ given in your problem is generated by $(0,1)$ and $(1,1)$.  Recall that the linear transformation that takes $e_1$ and $e_2$ to $x$ and $y$, respectively, is represented by the matrix that has $x$ and $y$ as its columns. Thus, the matrix that takes the unit square to $P1$ is $ A=\left( \begin{array}[cc] &0 & 1 \\ 1 & 1 \end{array}\right)$.  Similarly, the matrix that takes the unit square to the parallelogram $P2$ is $ B=\left( \begin{array}[cc] &6 & 1 \\ 0 & 1 \end{array}\right)$.  Thus, the matrix that takes $P1$ to $P2$ is $BA^{-1} = \left( \begin{array}[cc] &6 & 1 \\ 0 & 1 \end{array}\right) \left( \begin{array}[cc] &-1 & 1 \\ 1 & 0 \end{array}\right) = \left( \begin{array}[cc] &-5 & 6 \\ 1 & 0 \end{array}\right)$. Here, first applying $A^{-1}$ takes $P1$ to the unit square, and then applying $B$ takes the unit square to $P2$.
A double check confirms that the given $BA^{-1}$ does transform the generators $(0,1)$ and $(1,1)$ of $P1$ to the generators $(6,0)$ and $(1,1)$ of $P2$, respectively. 
There exist other matrices which transform $P1$ to $P2$,  since the choice of $A$ (and of $B$) is not unique; for eg, if we define $B$ to take $e_1$ and $e_2$ to $(1,1)$ and $(6,0)$, respectively, then we have interchanged the columns of $B$ (compared to before), and the resulting matrix $BA^{-1}$ is $\left( \begin{array}[cc] &5 & 1 \\ -1 & 1 \end{array}\right)$.  
As another double check, observe that the area of the parallelograms $P1$ and $P2$ are 1 and 6, respectively.  The absolute value of the determinant of the matrix $BA^{-1}$ obtained above is 6, which is as expected. (In general, the matrix $C$ is the linear transformation that takes the unit cube to a parallelpiped spanned by the columns of $C$, and the square of the volume of the parallelpiped is the determinant of $C^T C$.  In our case, the square of the area is 6^2, which is also the determinant of $C^T C$, where $C=BA^{-1}$.)
