If $T: R^2 \rightarrow R$ is a linear transformation then $T: R^2 \rightarrow R$ is a linear transformation such that $T(v) \neq 0$ for some $v$  in $R^2$ then T is


*

*one-to-one

*onto

*$nullity(T) = 0$

*none of the above


My first thought was that since $dim(R^2) > dim(R)$ then T cannot be one-to-one. I was inclined to put none of the above, but I feel $T(v) \neq 0$ is a crucial piece of information. Is this function onto? If so why?

We know that $T(\Bbb R^2)$ is a vector subspace of $\Bbb R$ and that $dimT(\Bbb R^2) \le dim \Bbb R = 1 $. So $dimT(\Bbb R^2)$ is either $1$ or $0$. But $T \neq 0$ therefore $dimT(\Bbb R^2)=1=dim \Bbb R$ and since $T(\Bbb R^2)$ is a vector subspace of $\Bbb R$ they are produced by the same basis, so they are equal. So $T$ is onto!
