Let V W be vector spaces with dimV n dimW m and n m a S

Let V, W be vector spaces, with dim(V ) = n, dim(W ) = m, and n > m.

(a) Show that there is no one-to-one linear transformation T : V W.

(b) Show that there is no onto linear transformation T : W V (notice that V, W have flipped in this expression!)

(c) Show that a linear map T : V W need not be onto by giving an example where it is not.

Solution

(a) Let V and W be finite-dimensional vector spaces and T : V W be linear. Assume dim(V ) > dim(W). We will prove by contradiction so assume that T is one-to-one. By the Dimension Theorem, since V is finite-dimensional then nullity(T) + rank(T) = dim(V ) which, by definition of nullity and rank, can be written equivalently as dim(N(T)) + dim(R(T)) = dim(V ). Since T is one-to-one, by Theorem 2.4, N(T) = {0} so it follows that dim(N(T)) = 0. This implies that dim(R(T)) = dim(V ). Also, by by Theorem 2.5, since T is one-to-one we have that rank(T) = dim(W) which is equivalent to dim(R(T)) = dim(W) by definition of rank. Therefore, since dim(R(T)) = dim(V ) and dim(R(T)) = dim(W), it follows that dim(W) = dim(V ) which is a contradiction to our assumption that dim(V ) > dim(W) so T must not be one to one.

(b) . Let V and W be finite-dimensional vector spaces and T : V W be linear. Assume dim(V ) < dim(W). We will prove by contradiction so assume that T is onto. By the Dimension Theorem, since V is finite-dimensional then nullity(T) + rank(T) = dim(V ) which, by definition of nullity and rank, can be written equivalently as dim(N(T)) + dim(R(T)) = dim(V ). Since T is onto, by Theorem 2.5, rank(T) = dim(V ) which is equivalent to dim(R(T)) = dim(V ) by definition of rank. This implies that dim(N(T))+dim(R(T)) = 0+dim(R(T)) = dim(V ). However, since T is onto, we also know that R(T) = W according to Theorem 2.5. Thus, we have that dim(W) = dim(V ) which is a contradiction (based on transitivity) . therefore , T is not onto