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?

$T$ is onto. Indeed, let $v\in R^2$ be such that $T(v)\ne 0$.
Let $\alpha$ be any real number. Then
$$
T\left(\frac{\alpha}{T(v)}v\right)=\frac{\alpha}{T(v)}T(v)=\alpha.
$$
