Prove that if is a linear transformation from a vector space to a vector space , then for any subspace in , is a subspace in .
Let and , the base field for the vector spaces. It means by definition that . Since is a subspace it means . Therefore, since is a linear transformation it means which means . Thus, is closed under vector addition. Likewise, which means . Thus, is closed under scalar multiplications. All the other properties for being a vector space are satisfied because . Thus, is a vector space over .
Thanks for your input; your proof confirms my belief that the same proof in my linear algebra textbook is incorrect; it messed up on x & y and f(x) & f(y) (and it wasted me a long time to try to make sense out of that flawed proof ).
PS: I think there's one part that your proof is incomplete; it hasn't showed that the zero vector is in
The condition is necessary, in order to exclude the possibility of the set being empty (the empty set is by convention not considered to be a subspace). If you're trying to show that X is a subspace then you can only make use of the implication if you know that there exists a vector x in X.