Given two vector spaces $V$ and $W$ of finite dimensions and a linear transformation $T: V \rightarrow W$, let $m = dim(V)$ and $n = dim(W)$. Show that if $T$ is one to one, then $m \leq n$ and if $T$ is onto then $m \geq n$.
So I am thinking this is best approached by contradiction. Can I say that the number of vectors in the basis of $V$ is $m$ and the number of vectors in the basis of $W$ is $n$ and to assume $m \geq n$. Let $V = \{v_1,v_2,...,v_m\}$ and $W = \{w_1,w_2,...,w_n\}$, if we consider the function $T: V \rightarrow W$ then the number of vectors in the $span(V) \leq span(W)$ so then you wouldn't have unique mapping and $T$ isn't one to one.
I'm not sure where to even begin for the onto part of the question.
We have $m= \dim V = \dim ker(T)+ \dim Im(T).$
If $T$ is one-to-one, then $\dim ker(T)=0$, thus $m = \dim Im(T) \le \dim W =n.$
If $T$ is onto, then...... your turn !