If $T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$ such that $T(0)=0$, then $T$ is always a linear transformation.

Question

I'm stumped on a true or false question I found in one of our practice worksheets for my Linear Algebra class. Would anyone mind answering this and explaining why as well?

Question: If $T:\mathbb{R}^{n}\rightarrow\mathbb{R}^{m}$ such that $T(0)=0$, that is, $T$ maps the zero vector in $\mathbb{R}^{n}$ to the zero vector in $\mathbb{R}^{m}$, then $T$ is always a linear transformation.

I'm leaning towards "true" because of the following theorem: If $T$ is a linear transformation, then $T(0)=0$ and $T(cu+dv)=cT(u)+dT(v)$ for all vectors $u$ and $v$ in the domain of $T$ and all scalars $c$ and $d$.

Answer 1

Linear transformations always maps zero to zero, since , for all scalars $c,d$, $T(cu+dv)=c T(u)+dT(v)$ and so this true for $c=d=0$. But the other direction is not true. Here is a simple example:

$$T:\Bbb R^2 \ni (x,y) \longmapsto (\sin x, 0) \in \Bbb R^2$$

Answer 2

Of course not ! Take $m=n=1$ and $f(x)=x^2$