$T(cx+y)=cT(x)+T(y)$ implies $T(cx)=cT(x)$?

Question

My textbook says

“A map $T$ between vector spaces is linear if for all vectors $x,y$ and scalars $c$ we have $T(x+y)=T(x)+T(y)$ and $T(cx)=cT(x)$.”

It also says this is equivalent to

“for all vectors $x,y$ and scalars $c$ we have $T(cx+y)=cT(x)+T(y)$.”

My attempt to show they indeed are equivalent fails:

Suppose $T(cx+y)=cT(x)+T(y)$ for all $x,y,c$. For $c=1$ we have $T(cx+y)=T(1x+y)=1T(x)+T(y)=T(x)+T(y)$. For $y=0$ we have $T(cx)=T(cx+0)=cT(x)+T(0)$.

Because I am not sure wether $T(0)=0$, I cannot conclude that $T(cx)=cT(x)$. How do I show that $T(0)$ indeed equals $0$?

Answer 1

Yes that's correct, for any linear map indeed $T(0)=0$.

To see why let consider by linearity

$$T(0)=\overbrace{T(0+0)=T(0)+T(0)}^{linearity}=2T(0) \implies T(0)=0$$

Answer 2

Take $x=y=0, c=1$, $$ T(0)+T(0) = cT(x)+T(y) = T(cx+y) = T(1\cdot 0+0) = T(0)$$ and subtract $T(0)$ from both sides.

Answer 3

Given a linear transformation T:V--->W where V and W are vector spaces over the same field, F, it follows that T(0)=0. Proof: You know that given a scalar 0 $\in$ $F$, and a vector 0 $\in$ V, 0.0 = 0. So applying the linear transformation you get T(0.0)=0.T(0) = 0

Answer 4

$$T(cx+y)=cT(x)+T(y)\implies T(0)=T(c0+0)=cT(0)+T(0)\implies T(0)=0.$$