Show that there is no linear transformation $T: R^2 \rightarrow R^2$ such that $T (1, π) = (0, 1)$, , $T (e, −π) = (1, 1)$ and $T (1, 0) = (2, 3)$. I can't seem to find a way to prove this, I started last week with Lineal Transformations
Can someone explain me?
 A: The three vectors $(0,1), (e,-\pi)$ and $(1,0)$ are linearly dependent. Knowing the image of any two of them under $T$ determines the third, using the linearity property that
$$T(\alpha v + \beta w) = \alpha T(v) + \beta T(w).$$
Is this enough of a hint?
A: Such transformation is invertible since $T(1,\pi)=(0,1)$ and $T(e,-\pi)=(1,1)$ generates $\mathbb{R}^2$.
We have $T(1,\pi)+2T(e,-\pi)=T(1,0)$ implies that $(1,\pi)+2(e,-\pi)-(1,0)$ is in $KerT=\{0\}$ impossible.
A: converting comment to answer
Since $T$ is linear we should have had that $T(u+v)=T(u)+T(v)$ and also that $T(\alpha u)=\alpha T(u)$.
Then what does $T(1+e,0)$ equal taking both of these into account and that $(1,\pi)+(e,-\pi)=(1+e,0)$ as well as that $(1+e,0)=(1+e)(1,0)$

 $$(1,2)=(0,1)+(1,1)=T(1,\pi)+T(e,-\pi) =^\dagger T(1+e,0)=^\star (1+e)T(1,0)\\=(1+e)(2,3)=(2+2e,3+3e)$$ where at the equality marked with a $\dagger$ we used $T(u)+T(v)=T(u+v)$ and at the equality marked with a $\star$ we used $T(\alpha u)=\alpha T(u)$.  But of course $(1,2)\neq (2+2e,3+3e)$, so we have a contradiction.

