(Linear Algebra): Are there isomorphic linear mappings L: V - W (V, W are finite vector spaces) that are one-to-one but are not onto?

The reason I ask is because most examples we have seen were one-to-one, and so Ker(L) = {0}. Hence, nullity(L) = 0. So, by the Rank-Nullity Theorem, it implies rank(L) = dim V. So |Range(L)| = dim(Range(L)) = dim V = dim W, implying Range(L) = W.

Question

(Linear Algebra): Are there isomorphic linear mappings L: V -> W (V, W are finite vector spaces) that are one-to-one but are not onto? 

The reason I ask is because most examples we have seen were one-to-one, and so Ker(L) = {0}. Hence, nullity(L) = 0. So, by the Rank-Nullity Theorem, it implies rank(L) = dim V. So |Range(L)| = dim(Range(L)) = dim V = dim W, implying Range(L) = W.