it is not a subspace, but i wouldn't state the reason you did. just show that it is either not closed under addition or scalar multiplication. it's not closed under either.

your reasoning is correct, but i would have liked for you to be more specific.For 2c) I wrote it clearly has the zero vector because f(7)=0. and (f+g)(x)=f(x)+g(x), and so the subset is closed under addition, and that (Kf)(x)=k(f(x)), so it is closed under scalar multiplication and thus the subset is a subspace.

For the third question,I wrote it is closed under scalar multiplication because AKV1=AKV2 (K is a scalar), and i wrote that A(V1+V2)=AV1+AV2=BV1+BV2 and is thus closed under addition and so the subset is a subspace. (and I wrote it clearly has the zero vector).

for example, for 2c) I would write the following:

Clearly, operates as the zero vector. (Did your professor say it was necessary to check this condition?)

Let . That is, . And let .

Then, , so that . Thus, is closed under vector addition.

Also, since , where is a scalar, we have , and so is closed under scalar multiplication.

Get the drift? Really show that you know what you're talking about, rather than just stating what happens without proving you know how to apply it.

this is fine. of course you can be technical and say it is not closed under scalar multiplication. Not if your k is complex or imaginary.

For the fourth one I wrote that the subset is not a subspace because it is not closed under scalar multiplication because you can multiply it by i and then you get complex numbers.

Please help, it is very appreciated

Good job