1. Jan 23, 2010

### 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.

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. Jan 23, 2010

### vela

Staff Emeritus

I think you're not clear on what the zero vector is in this case. Remember the vectors are the functions from R to R. The zero vector is the function 0(x)=0, not x=0.

3. Jan 23, 2010

### mmmboh

Yeah we haven't done any examples with functions in class, I am pretty confused by it :S...so that means the zero vector is included right? and by the same reasoning it is included in the second question too?

4. Jan 23, 2010

### VeeEight

You should write out your answers more neatly and in proper notation since it's hard to understand some times. You need to write out the criteria for being a subspace and for each question, show these 3 criteria hold or where one of them fails. For example, for the first question, if for f, g functions on R f(0)=7 and g(0)=7, does it follow that , (f+g)(0)=7?

5. Jan 23, 2010

### mmmboh

Hm..I think (f+g)(0)=f(0)+g(0)=7+7=14, and so it is not a subspace. Is this correct?

Sorry about the notation, I am at the library and it is closing now but when I get home I will try to write it better.

6. Jan 23, 2010

### vela

Staff Emeritus

No, it's not in the first set because the zero function evaluated at x=0 is 0, but the set only contains functions that equal 7 at x=0. On the other hand, the second set contains functions that equal 0 when x=7. The zero function obviously satisfies that criterion, so it is in the second set.

To show that the first set isn't closed under scalar multiplication, note that for $k \ne 1$

$$(kf)(0) = k[f(0)] = k7 \ne 7$$

Because kf evaluated at x=0 is not equal to 7, it's not in the set.

Your explanation for the second set needs to be cleaned up a bit. You should explain explicitly why f+g and kf are in the set.

7. Jan 23, 2010

### vela

Staff Emeritus

Yes, though I would say f+g is not in the set, rather than referring to it as a subspace, because the set is apparently not a subspace.

8. Jan 23, 2010

### mmmboh

What if the function was f(x)=7-x, and so at x=0 f=7, but when x=7 f=0, wouldn't this include a zero vector?

And are the 3rd and fourth ones right?

Thanks for the help guys.

9. Jan 24, 2010

### vela

Staff Emeritus