Let $v_1 = (1, 0); v_2 = (1,-1) \space\text{and} \space v_3 = (0, 1).$ I am stuck on the following problem :  

Let $v_1 = (1, 0); v_2 = (1,-1) \space\text{and} \space v_3 = (0, 1).$ How many linear transformations
  $T \colon \Bbb R^2 \to \Bbb R^2$ are there such that $Tv_1 = v_2; Tv_2 = v_3$ and $Tv_3 = v_1?$ The options are as follows:   
(A) $3!$
  (B) $3$
  (C) $1$
  (D) $0$  

What I observed that $v_2=v_1-v_3$ and so $T(v_2)=T(v_1)-T(v_3) \implies v_3=v_2-v_1$.
But I do not know how to progress from here. Any idea?
 A: Here's a straight forward solution. Every linear transformation $T\colon \mathbb R^2 \longrightarrow \mathbb R^2$ can be represented as a $2\times 2$ matrix, so you want to find $a_1$, $a_2$, $a_3$ and $a_4$ such that
\[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}1 \\ -1\end{pmatrix} \: \Leftrightarrow \: \begin{pmatrix}a_1 \\ a_3\end{pmatrix} = \begin{pmatrix}1 \\ -1\end{pmatrix}\]
and
\[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}1 \\ -1\end{pmatrix} = \begin{pmatrix}0 \\ 1\end{pmatrix} \: \Leftrightarrow \: \begin{pmatrix}a_1-a_2 \\ a_3-a_4\end{pmatrix} = \begin{pmatrix}0 \\ 1\end{pmatrix}\]
and
\[\begin{pmatrix}a_1 & a_2 \\ a_3 & a_4\end{pmatrix} \begin{pmatrix}0 \\ 1\end{pmatrix} = \begin{pmatrix}1 \\ 0\end{pmatrix} \: \Leftrightarrow \: \begin{pmatrix}a_2 \\ a_4\end{pmatrix} = \begin{pmatrix} 1 \\ 0\end{pmatrix}\]
Now it's obvious that there are no such matrix, and so the answer is (d), 0.
Actually you've the answer in front of you too. Assume that there is some such $T$. Since $v_2 = v_1+v_3$ we can use the linearity of the transformation to get $v_3 = v_2+v_1$. Combine these two equations to get $v_1 = 0$, which is a contradiction, and again the answer is (d), 0.
