# Show that T is a nonlinear transformation

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}##.

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Mark44
Mentor
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?
Sociomath said:
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?