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 $dim(R^2) &gt; dim(R)$ 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
$$
T\left(\frac{\alpha}{T(v)}v\right)=\frac{\alpha}{T(v)}T(v)=\alpha.
$$
Nullspace is not empty so the dimension of Nullspace is not zero.
By rank and nullity theorem $\text{dim}(N(T)) + \text{dim}(R(T)) = \text{dim}(R^2) = 2$.
But $dim(N(T))$ is also not $2$ because not for all $v$, $T(v) = 0$.
Thus $\text{dim} (N(T)) = 1$ which implies $\text{dim}(R(T)) = 1$.
since R(T) is subspace of $R^2$ and $\text{dim}(R(T)) = \text{dim}(R)$ then $R(T) = R$ which means that $T$ is onto.