Proving a function is not linear

Question

Prove that there is no linear transformation for which $T(1, 1) = (4, 0)$ and $T(-2, -2) = (0, 1)$.

One way to do this problem is to set up two matrix equations and derive a contradiction. I've already done it this way; you get $a + b = 4$ and $-2a - 2b = 0$, which is impossible.

But, I was thinking if the following way is also valid? Suppose $T$ was linear. Then, since we have $T(-2, -2) = (0, 1)$, we must also have $c \cdot T(-2, -2) = T(-2c, -2c) = (0c, 1c)$. Plugging in $c = -1/2$ yields $T(1, 1) = (0, -1/2)$, which contradicts our already known value for $T(1, 1)$.

Is this correct reasoning?

Answer 1

Instead we argue as follows:

If T is linear , then $(0,0)=2(1,1)+(-2,-2)$, so$$T(0,0)=2T(1,1)+T(-2,-2)=2(4,0)+(0,1)=(8,1)$$ So $T$ maps zero element to a non zero element, which is not true for any linear map!

Answer 2

Your alternative solution is very clearly the best approach.

The proof presents itself immediately by observation, since $$T(c\mathbf{x})\neq cT(\mathbf{x})$$ when $\mathbf{x}=(1,1)$ and $c=-2$.