Problem 4
Let \(V\) and \(W\) be finite dimensional vector spaces and \(T: V \rightarrow W\) be linear.
  1. Prove that if \(\dim(V) < \dim (W)\), then \(T\) cannot be onto.
  2. Prove that if \(\dim(W) < \dim(V)\), then \(T\) cannot be one-to-one.


Proof:

  1. Suppose for the sake of contradiction that \(T\) was onto. This means that \(Rank(T) = \dim(W)\). The dimension theorem states that
    $$ \begin{align*} \dim(V) &= Nullity(T) + Rank(T) \\ &= Nullity(T) + \dim(W) \end{align*} $$
    This implies that \(Nullity(T) = \dim(V) - \dim(W) < 0\) since \(\dim(V) < \dim(W)\). This is a contradiction and so \(T\) cannot be onto. \(\blacksquare\)
  2. Similarly, suppose for the sake of contradiction that \(T\) is one-to-one. This means that \(Nullity(T) = 0\) and so by the dimension theorem, it must be that \(\dim(V) = \dim(W)\). But this is a contradiction since \(\dim(W) < \dim(V)\).



References