Let $V$ and $W$ be finite dimensional vector spaces and let $T:V \to W$ be a linear transformation.
(a) Prove that if $\dim(V) < \dim(W)$ then $T$ cannot be onto.
(b) Prove that if $\dim(V) > \dim(W)$ then $T$ cannot be one-to-one.
What I tried:
(a) Proving by contradiction. Suppose that $T$ is onto. Then, since we are also given that $T$ is linear, then $T$ has to be one-to-one. Thus $T$ is both one-to-one and onto which means $\dim(V) = \dim(W)$ hence contradiction the fact that $\dim(V) < \dim(W)$.
(b) Again proving by contradiction, suppose that $T$ is one-to-one. Then we know that $\dim N(T) = 0$.
And since $\dim R(T) + \dim N(T) = \dim(V)$, this makes $\dim R(T) = \dim(W)$, and thus $V$ maps onto $W$, which contradicts the fact that $\dim(V) > \dim(W)$ and hence proving the statement.
Is my prove correct? Could anyone explain? Also could anyone show me how to do the prove directly instead of using contradiction? Thanks