1. ## Subspaces

Determine whether or not the following subset is a subspace of the given real vector space.

{p(x) is an element of P : the integral from 0 to 1 of p(x) dx=0}

2. ## Re: Subspaces

Hey widenerl194.

Hint: First try stating the subspace axioms. You need the zero vector for a start: Can you prove the zero vector lies in the sub-space?

3. ## Re: Subspaces

So for that would I put:

zero vector = p(0) in P as the integral from 0 to 1 p(0) dx = 0

I really don't know what I'm doing here

4. ## Re: Subspaces

Originally Posted by widenerl194
So for that would I put:

zero vector = p(0) in P as the integral from 0 to 1 p(0) dx = 0

I really don't know what I'm doing here
O.K. Tell us exactly what must be done in order to show as set is a subspace.

5. ## Re: Subspaces

A subset of V is a subspace of V provided it is a real vector space (?) under the same addition and scalar multiplication as that of V.

6. ## Re: Subspaces

Originally Posted by widenerl194
A subset of V is a subspace of V provided it is a real vector space (?) under the same addition and scalar multiplication as that of V.
Is it true that $\int_0^1 {\left[ {p(x) + q(x)} \right]dx} = \int_0^1 {p(x)dx} + \int_0^1 {q(x)dx}~?$

$\int_0^1 {\left[ {\alpha p(x) } \right]dx} = \alpha \int_0^1 {p(x)dx}~?$

Are the integrals equal zero?

7. ## Re: Subspaces

I know the first is true. But I feel like the second one isn't

8. ## Re: Subspaces

And if only one of those is true then the subset is not a subspace

9. ## Re: Subspacesd

Originally Posted by widenerl194
I know the first is true. But I feel like the second one isn't
You don't know much about basic integration theory do you?

If you are so basically uninformed about freshman level mathematics, then why in the world do you try upper level topics?

10. ## Re: Subspaces

This is help forum... and calculus isn't exactly easy. There's no need to be rude when I'm only asking for help. I said at the beginning that I didn't know what I was doing.

11. ## Re: Subspaces

Originally Posted by widenerl194
This is help forum... and calculus isn't exactly easy. There's no need to be rude when I'm only asking for help. I said at the beginning that I didn't know what I was doing.
Telling someone the truth has never been rude.
You said "at the beginning that I didn't know what I was doing".
And then I simply asked "if so, why try to do it?"

12. ## Re: Subspaces

If I didn't have to do it, I wouldn't be posting on this forum. There aren't choices on homework assignments.