Is this a sufficient way to check if a transformation is linear?

Question

I'm preparing for a final using an old exam, but they aren't giving us the answer keys, so I'm wondering if I'm going about this question the right way.

I'm asked to determine if T is a linear transformation where:

$T : \Bbb R^2\rightarrow\Bbb R^2$ with $T(x,y)=(x,1-y)$

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I know that a transformation is linear if the following is true:

1. $T(u+v) = T(u)+T(v)$
2. $T(cu)=cT(u)$

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Here's how I solved it.

I created another vector and showed that the transformation works:

$T(a,b) = (a, 1-b)$

Then, I showed what the vectors are when added together:

$(a,b)+(x,y) = (a+x, b+y)$

I then showed what the two transformations are when added:

$T(a,b)+T(x,y) = (a,1-b) + (x, 1-y) = (x + a, 2 - y - b)$

Lastly, I should the transformation on the already added vector:

$T(x+a, y+b) = (x+a, 1 - (y+b))$

Seeing that the rule $T(u+v) = T(u)+T(v)$ failed, I concluded that the transformation is non-linear.

Are there errors in my logic, or is this a fine way to solve this kind of problem? I just don't want to practice these in an incorrect way since only perfect practice makes perfect.

Answer 1

The cause for the non-linearity of $T$ is the constant $1$ in the second component of the output of $T$. As @GuidoA. points out, a simple $T(0,0) = (0,1) \ne (0,0)$ suffices to show the non-linearity of $T$ because if $T$ were linear, then $$T(0,0) = T((0,0) + (0,0)) = T(0,0) + T(0,0) \iff T(0,0) = (0,0).$$