# Prove that if $T:V\rightarrow V$ is additive (that is, $T(x+y)=T(x)+T(y)$ for all $x,y \in V$), then $T$ is linear.

Let $V$ be a vector field over $\mathbb Q$, the rationals. Prove that if $T:V\rightarrow V$ is additive (that is, $T(x+y)=T(x)+T(y)$ for all $x,y \in V$), then $T$ is linear.

The two properties for $T$ to be linear is additivity, which is given, and for $T(cx)=c*T(x)$ for all $c \in \mathbb Q$ and $x \in V$, but I'm not sure how to get that from the givens. I think it has to do with the vector field being over the rationals, but I'm not sure.

First show $T(0)=0$. Then by induction, show $T(nv)=nT(v)$ for $n\in\Bbb N$. Then show $T(-v)=-T(v)$ and conclude $T(nv)=nT(v)$ for $n\in \Bbb Z$. Use this to conclude that $T(\frac nm v)=\frac nm T(v)$.
• How do I prove that $T(0)=0$. I thought I would prove it with $T(x-x)=T(0)=T(x)+T(-x)$, but then I would need the identity that $T(-v)=-T(v)$ which I don't have yet. – AndroidFish Sep 21 '16 at 21:18
• @AndroidFish $T(0)=T(0+0)=T(0)+T(0)$. – egreg Sep 21 '16 at 21:20
First prove by induction that $T(nx)=nT(x)$ for all natural numbers $n$.
Then extend this property to integers. Finally to prove it for any rational number $c=a/b$, observe $y=cx\iff by=ax$.