How to show a mapping $T$ is Linear Suppose $T$ is a mapping such that $T(1,1 )= (2,1 )$ and $T(1,-1) = (2,-1)$
(each set of numbers is a vector)
Can the mapping $T$ be linear?
Please note that the question is not asking if $T$ is a linear transformation. 
Is there a way to show that $T$ is Linear by plotting vectors and show linearity?
If anyone can help me with this question it will be greatly appreciated.   
 A: If $T$ is linear than it can be represented by a matrix:
$$
T= \begin{bmatrix}
a&b\\c&d
\end{bmatrix}
$$
such that:
$$
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}
\begin{bmatrix}
1\\1
\end{bmatrix}=
\begin{bmatrix}
2\\1
\end{bmatrix}
\qquad \land \qquad
\begin{bmatrix}
a&b\\c&d
\end{bmatrix}
\begin{bmatrix}
1\\-1
\end{bmatrix}=
\begin{bmatrix}
2\\-1
\end{bmatrix}
$$
A simple inspection show that this is possible for:
$$
T= \begin{bmatrix}
2&0\\0&1
\end{bmatrix}
$$
A: Yes, we can find a unique linear transformation $T: \mathbb R \to \mathbb R$ that takes $(-1,1) \mapsto (2,1)$ and $(1,-1) \mapsto (2,-1)$. This is because your input vectors $\{(1,1),(1,-1)\}$ form a basis for $\mathbb{R}^2$. Remember that a linear transformation is uniquely determined by where it sends the basis vectors to (for any particular basis). In fact, we can find a matrix $A$ such that $T(x) = Ax$ for all $x \in \mathbb{R}^2$ (this is called a representative matrix, and we can do this for any linear transformation). So we have that
$$
A\begin{bmatrix}
1 \\
1
\end{bmatrix} = \begin{bmatrix}
2 \\
1
\end{bmatrix} \text{ and }
A\begin{bmatrix}
1 \\ 
-1
\end{bmatrix} = \begin{bmatrix}
2 \\
-1
\end{bmatrix}
$$
We can turn this into a couple very simple systems of equations:
$a_{11} + a_{12} = 2$ and $a_{11} - a_{12} = 2$ gives us that $a_{11} = 2$  and $a_{12} = 0$, and the system of equations $a_{21} + a_{22} = 1$ and $a_{21} -a_{22} = -1$ gives us that $a_{21} = 0$ and $a_{22} = 1$. So we get that 
$$T(x) = \begin{bmatrix}
2 & 0 \\
0 & 1
\end{bmatrix}x \text{ for all } x \in \mathbb{R}^2 
$$
is the linear map that you are looking for.
