Does there exist a linear transformation

Question

I just want to verify my logic for this problem. Does there exist a linear transformation $T:ℝ^3\mapstoℝ^2$ such that $T(1,0,1)=(1,-1), T(1,1,-1) = (1,2)\text{ and }T(3, 2,-1) = (1, -1)$

So my reasoning is that we must be able to form a basis for $ℝ^3$ with given inputs to be able to generalize for $T(a,b,c)$. However, $(1,0,1)+2*(1,1,-1) = (3, 2,-1)\Rightarrow$ they are linearly dependent $\Rightarrow$ we cannot form a basis for $ℝ^3$ with those vectors, therefore we simply lack information and cannot conclude that it is a linear transformation neither can we claim it is not. Is there a flaw in my logic?

Answer 1

By your reasoning, we can conclude that there cannot be such a linear transformation.

We know that $T(3,2,-1) = T((1,0,1) + 2*(1,1,-1))$. But, if $T$ is linear, then $T((1,0,1) + 2*(1,1,-1)) = T(1,0,1) + 2*T(1,1,-1) = (1,-1) + 2*(1,2) = (3,3) \neq (1,-1)$.

Answer 2

You are almost there!

A linear transformation respects linear relations. Therefore, if $u+2v=w$, then $T(u)+2T(v)$ has to be equal to $T(w)$.