1. ## Subspaces Questions Help Please!!

Hi I have an assignment due Monday morning and there are a few questions I am not sure about or if I proved them properly:

Ok so for 2b) I said that it is not a subspace because f(x)=7 when x=0, and this function never equals zero, and since this is included in the vector space the subset is not a subspace.

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 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.

2. Originally Posted by mmmboh
Hi I have an assignment due Monday morning and there are a few questions I am not sure about or if I proved them properly:

Ok so for 2b) I said that it is not a subspace because f(x)=7 when x=0, and this function never equals zero, and since this is included in the vector space the subset is not a subspace.
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.

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).
your reasoning is correct, but i would have liked for you to be more specific.

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

Clearly, $f(x) = 0$ operates as the zero vector. (Did your professor say it was necessary to check this condition?)

Let $h(x),g(x) \in A = \{ f \in \mathbb R ^{\mathbb R} ~:~ f(7) = 0 \}$. That is, $h(7) = g(7) = 0$. And let $p(x) = h(x) + g(x)$.

Then, $p(7) = h(7) + g(7) = 0 + 0 = 0$, so that $p = (h + g) \in A$. Thus, $A$ is closed under vector addition.

Also, since $k \cdot h(7) = k \cdot 0 = 0$, where $k$ is a scalar, we have $k \cdot h(x) \in A$, and so $A$ 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.

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.

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.

Good job

3. Ah Thanks! and for the third question, is this proof ok?:
A0=0=B0, so the zero vector is included.
A(Kv)=KAv=KBv, so it is closed under scalar multiplication.
A(v1+v2)=Av1+Av2=Bv1+Bv2=B(v1+v2), so the set is closed under addition and it is a subspace.

4. Originally Posted by mmmboh
Ah Thanks! and for the third question, is this proof ok?:
A0=0=B0, so the zero vector is included.
state that $\begin{pmatrix} 0 \\ 0 \\ 0 \end{pmatrix}$ serves as the zero vector. It satisfies the given condition and is an element of $\mathbb C ^3$

A(Kv)=KAv=KBv, so it is closed under scalar multiplication.
you have to end with = B(kv)

A(v1+v2)=Av1+Av2=Bv1+Bv2=B(v1+v2), so the set is closed under addition and it is a subspace.
ok

and of course, you explicitly stated what A and B are