Is this statement true or false
if false a counterexample is needed
if true then an explanation
If T : U V is a linear map, then Rank(T) (dim(U) + dim(V ))/2
If T is "one to one", then it maps the set of basis vectors of U onto an independent set and so maps U onto a subspace of V having exactly the same dimension as U. If not, then it maps U onto a subset of V having smaller dimension. "rank T" is the dimension of T(U).
"rank T" is the dimension of T(U) as above and, of course, T(U) is a subspace of V.rank (T) dim (V)