Question

Prove that kernel of a homomorphism from U(f) to V(f) is a subspace of vector space U(f).

Answer 1

Answer:

My name is rohit

Step-by-step explanation:

Answer 2

Answer:

Except for the typo I pointed out in my comment, your proof that the kernel is a subspace is perfectly fine. Note that it is not necessary to separately show that 0 is contained in the set, since this is a consequence of closure under scalar multiplication.

As for the image: we say that a vector u∈V is in the image if, for some x∈U, u=f(x).

In order to show that this set is a subspace, you need to show that for any constant a: if u and v are in the image, then so is au+v. That is, for any u and v where u=f(x) and v=f(y) (for some choice of x and y from U), we need to show that

au+v=f(z)

for some vector z∈U.

In fact, we note that

au+v=af(x)+f(y)=f(ax+ythis is our "z")