# Finding a Basis

1. Jan 21, 2014

### Nexttime35

1. The problem statement, all variables and given/known data
The problem asks, find a basis for the P2 subspace that consists of polynomials, p(x) such that p'(5)=0.

3. The attempt at a solution

I know that a set of vectors is a basis if it's linear independent and spans the vector space.

So I let p(x) = ax2 + bx +c .

Then p'(x) = 2ax + b.
p'(2) = 4a + b

I set that equal to 0. so 4a+b = 0. then 4a=-b, and a= =1/4*b.

Now I am stuck as to where to go from here when finding a basis (prof hasn't really taught this yet)

I was wondering what the next step would be?

Thank you.

2. Jan 21, 2014

### tiny-tim

Hi Nexttime35!

(do you mean p'(2) = 0 ?)

Hints:

i] is 1 in the subspace??

ii] how many elements do you think the basis has?

3. Jan 21, 2014

### Nexttime35

I think the basis has two elements, a and b.

However, I am unsure how to apply your first hint to this question... Do you mean, can 4a+b=1?

4. Jan 21, 2014

### tiny-tim

Hi Nexttime35!
No, I mean is p = 1 an element of P2 such that p'(2) = 0 ?

5. Jan 21, 2014

### Staff: Mentor

What tiny-tim is asking about is the inconsistency between what you're given, p'(5) = 0, and what your work shows, p'(2) = 0.

6. Jan 21, 2014

### Nexttime35

Ahh, ok.

Then p=1 is not an element of p2 such that p'(2) = 0.

7. Jan 21, 2014

### Nexttime35

So a basis for this problem must consist of a function with only a degree 1, such as p(x) = x, or p(x) = 2x?

8. Jan 21, 2014

### Nexttime35

Oh sorry, I meant p'(2) = 0 in the problem header. Sorry about that.

9. Jan 21, 2014

### tiny-tim

i don't mind whether it's 2 or 5, it makes little difference, i was just checking
yes it is!!

p = 1 is a polynomial, and its derivative is 0 everywhere!

10. Jan 21, 2014

### Nexttime35

Duh! My mistake! For some reason I understood that differently. So p(x) = 1 for some x, and then obviously p'(x)=0 . I am confused about how that helps me find a basis, though. Thoughts?

11. Jan 21, 2014

### vela

Staff Emeritus
An element of P2 is a polynomial. We can think of these polynomials as vectors because with the appropriate definitions of vector addition and scalar multiplication, P2 has all the properties required of a vector space.

A basis is a set of linearly independent vectors that span the vector space. So in this problem, the vectors are polynomials, so you're looking for a set of polynomials that are linearly independent and that span the given subspace. Clearly $a$ and $b$ (the way you used them) aren't elements of the subspace, so they can't be elements of the basis.

12. Jan 21, 2014

### Nexttime35

So vela,

there isn't just one basis. There are essentially many, many bases that can be formed? So I could make a set of vectors {v1,v2,v3} and that would be a basis if each of those vectors (or polynomials in this case) satisfy p'(2)=0, and that are linearly independent?

13. Jan 21, 2014

### tiny-tim

well, p = 1 is in the subspace, so is p = 2, p = √2, etc

so does that help with one of the elements of the basis?

14. Jan 21, 2014

### vela

Staff Emeritus
Yes, there are many possible bases for a vector space.

You have a subset S of P2 such that any element in S satisfies p'(2) = 0. A basis for S should be a set of linearly independent vectors that you can express any element of S as a linear combination of the basis vectors. This is close to what you said but not exactly. You need to make sure you can express every element of S in terms of the basis vectors.

15. Jan 21, 2014

### Nexttime35

Yes!

So one element of the basis could be a constant, since that derivative of a constant is equal to 0. One element of the basis could be the function p(x) = x , because that satisfies the p'(2) = 0 condition. So, could {1, x} be a basis for the subspace of P2 in this case?

16. Jan 21, 2014

### tiny-tim

Hi Nexttime35!
Yes!!

(obviously, any constant will do, but equally obviously most people would choose 1 !)
erm nooo!

you had that part correct at the beginning …

do you need some sleep? :zzz:​

17. Jan 21, 2014

### vela

Staff Emeritus
If p(x)=x, then p'(x)=1. That doesn't satisfy p'(2)=0.

Suppose for the moment that p(x)=x did satisfy the condition. If {1, x} was a basis, that means any polynomial satisfying p'(2)=0 can be written as a linear combination of 1 and x. You need to prove that is the case for whatever prospective basis you come up with.

18. Jan 21, 2014

### Nexttime35

Oh gosh. I think I do need some sleep...

So one element of the basis will be a constant. Say p(x) = 4x^2. Then plugging in x=2, we get p(2) = 16. p'(2) = 0. OK I understand. I had a brain fart... Say p(x) = 4x. Then plugging in x=2, we get p(2) = 8. p'(2) = 0. So, could a basis be {1,x,x^2}? Or am I just overthinking this?

19. Jan 21, 2014

### tiny-tim

Sorry, I'm calling a time-out

GET SOME SLEEP! :zzz:

NOW!!

20. Jan 21, 2014

### Nexttime35

I sincerely apologize for my lapse in mathematical judgement.

The only derivative that can result in 0 is the derivative of a constant. Thus, is the basis for this subspace of P2 a one-element vector, a constant?