Show that T is a nonlinear transformation

[Sociomath]

1. Show that T isn't a linear transformation and provide a suitable counterexample.
$T \begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}x - 1 \\ y + 1 \end{bmatrix}$

2. The attempt at a solution
$\text{let}\, \vec{v} = \begin{bmatrix}0\\0 \end{bmatrix}. \text{Then,}$
$T(\vec{v}) = T\left(\begin{bmatrix}0\\0 \end{bmatrix}\right) = \begin{bmatrix}0 - 1 \\ 0 + 1 \end{bmatrix} = \begin{bmatrix}-1 \\ 1 \end{bmatrix}$
Nonlinear transformation due to constants: $T \begin{bmatrix}-1 \\ 1 \end{bmatrix}$.

 

[Mark44]

1. Show that T isn't a linear transformation and provide a suitable counterexample.
$T \begin{bmatrix}x\\y \end{bmatrix} = \begin{bmatrix}x - 1 \\ y + 1 \end{bmatrix}$

2. The attempt at a solution
$\text{let}\, \vec{v} = \begin{bmatrix}0\\0 \end{bmatrix}. \text{Then,}$
$T(\vec{v}) = T\left(\begin{bmatrix}0\\0 \end{bmatrix}\right) = \begin{bmatrix}0 - 1 \\ 0 + 1 \end{bmatrix} = \begin{bmatrix}-1 \\ 1 \end{bmatrix}$

Why does the above show that T is not a linear transformation?
And why is $\begin{bmatrix}0\\0 \end{bmatrix}$ a counter example?

Nonlinear transformation due to constants: $T \begin{bmatrix}-1 \\ 1 \end{bmatrix}$.

Your reason doesn't make sense to me, and this part -- $T \begin{bmatrix}-1 \\ 1 \end{bmatrix}$ -- really doesn't make sense.

What are the properties a linear transformation has to satisfy?