91Ó°ÊÓ

By definition, a function F is an isomorphism if it is a linear transformation that is both injective (one-to-one) and surjective (onto). To prove that F is nonsingular, we need to show that F has a trivial kernel, or in other words, the only vector in V that maps to the zero vector in U is the zero vector in V. Suppose F is an isomorphism. Since F is injective, we know that if F(v) = F(w) for any two vectors v and w in V, then v = w. In particular, this means that if F(v) = 0 in U, then v must be the zero vector in V, as F maps the zero vector in V to the zero vector in U. Thus, the kernel of F is trivial, and F is nonsingular.

Suppose F is nonsingular. We need to show that F is both injective and surjective. Injectivity: Since F is nonsingular, it has a trivial kernel, which means that the only vector in V that maps to the zero vector in U is the zero vector in V. This implies that if F(v) = F(w), then v must equal w, hence F is injective. Surjectivity: We know that dim(V) = dim(U). Since F is a linear transformation and is injective, the rank-nullity theorem states that: dim(V) = dim(kernel(F)) + dim(image(F)) As dim(kernel(F)) = 0 (since F is nonsingular), we have: dim(V) = dim(image(F)) Since dim(V) = dim(U), we can conclude that dim(image(F)) = dim(U). Thus, the image of F must equal the entire space U, meaning F is surjective. As F is both injective and surjective, F is an isomorphism. Thus, we have proven Theorem 5.9: if F is an isomorphism, then F is nonsingular, and if F is nonsingular, then F is an isomorphism, given that dim(V) = dim(U) for a finite-dimensional linear transformation F: V -> U.