Finding a basis for a subspace of P_2

  • Thread starter Geekster
  • Start date
  • #1
38
0
Let [tex] W=\lbrace p(x) \in P_{2} : p(2)=0\rbrace [/tex]

Find a basis for W.

Since a basis must be elements of the set W we know that p(2)=0.

So if [tex]p(x)=ax^2+bx+c[/tex], then [tex]p(x) = 4a+2b+c=0[/tex].

Let c=t, b=s and s,t are real scalars. Then p(x) can be written as
[tex]
t(-\frac{1}{4} x^2+1)+s(-\frac{1}{2}x^2+x)
[/tex]

so the basis would be [tex]\{ (-\frac{1}{4} x^2+1), (-\frac{1}{2}x^2+x)\rbrace [/tex].

Does anyone see a problem with this?
 
Last edited:

Answers and Replies

  • #2
AKG
Science Advisor
Homework Helper
2,565
4
One problem is the line:

p(x) = 4a + 2b + c = 0

It should say:

p(2) = 4a + 2b + c = 0

The final answer is correct, however it isn't clear as to how you got it. How do you go from the line above to saying that p(x) can be written:

t(-x²/4 + 1) + s(-x²/2 + x)?

Also, why bother changing c to t and b to s? You haven't proven that the set you've chosen consists of elements of W, spans W, and is linearly independent. Of course, it is all of those things, but you haven't proved it.

There's a simpler solution: you know that W is not all of P2 (since there are polynomials with 2 not being a root), so since P2 has dimension 3, you know that W has dimension 2 or less. If you can find 2 linearly independent elements of W, then you're done, since you know that these two will span all of W. And it should be obvious that if you can find a linear element of W and a quadratic element of W, then these two will be linearly independent. The obvious choice for the quadratic element is the (x-2)², and the obvious choice for the linear element is x-2.
 
  • #3
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Geekster said:
Let [tex] W=\lbrace p(x) \in P_{2} : p(2)=0\rbrace [/tex]

Find a basis for W.

Since a basis must be elements of the set W we know that p(2)=0.

So if [tex]p(x)=ax^2+bx+c[/tex], then [tex]p(x) = 4a+2b+c=0[/tex].

Let c=t, b=s and s,t are real scalars. Then p(x) can be written as
[tex]
t(-\frac{1}{4} x^2+1)+s(-\frac{1}{2}x^2+x)
[/tex]

so the basis would be [tex]\{ (-\frac{1}{4} x^2+1), (-\frac{1}{2}x^2+x)\rbrace [/tex].

Does anyone see a problem with this?
You should at least say that, since 4a+ 2b+ c= 0, a= -(1/2)b- (1/4)c.
Actually, since I am so bad with fractions, I think I would have solved for c: c= -4a- 2b so any such polynomial can be written as
ax2+ bx- 4a- 2b. Taking a= 1, b= 0, gives x2- 4.
Taking a= 0, b= 1, gives x- 2. Having already said that "any such polynomial can be written as ax2+ bx- 4a- 2b" it is immediate that that {x2- 4, x- 2} spans this space. That fact that one contains x2 and the other doesn't makes it clear that they are independent.

Your result is correct but, as AKG said, you need to explain your steps in more detail.
 
  • #4
38
0
AKG said:
One problem is the line:

p(x) = 4a + 2b + c = 0

It should say:

p(2) = 4a + 2b + c = 0

That's what I meant...sry.

The final answer is correct, however it isn't clear as to how you got it. How do you go from the line above to saying that p(x) can be written:

t(-x²/4 + 1) + s(-x²/2 + x)?

I could have written it out, but I figured that once I stated the parameters and their domain that this was the obvious next step.

Also, why bother changing c to t and b to s? You haven't proven that the set you've chosen consists of elements of W, spans W, and is linearly independent. Of course, it is all of those things, but you haven't proved it.

True....I didn't add those things. The linear independence is clear because neither one can be written as a scalar multiple of the other one. I should have stated that as well. Which, correct me if I'm wrong, but that should be sufficient to imply the set spans W. The theorem in my text basically says that if a set S of dimension n is a linearly independent set of vectors in V, then S is a basis for V.....So I guess I needed to show that the dimension of W is 2, then I could show that my basis vectors are linearly independent.

How could I show that the dimension of W is 2 unless I know a set that spans W has two vectors?

Let me read on...

There's a simpler solution: you know that W is not all of P2 (since there are polynomials with 2 not being a root), so since P2 has dimension 3, you know that W has dimension 2 or less. If you can find 2 linearly independent elements of W, then you're done, since you know that these two will span all of W. And it should be obvious that if you can find a linear element of W and a quadratic element of W, then these two will be linearly independent. The obvious choice for the quadratic element is the (x-2)², and the obvious choice for the linear element is x-2.

You’re very clever….I would never have thought of that.
 
  • #5
38
0
HallsofIvy said:
Your result is correct but, as AKG said, you need to explain your steps in more detail.

Point taken....I'm sure I'll take a good hit on the points for the lack of overall detail. Although I did offer more detail on the paper that I turned in than what I gave here.

Thanks for showing me an alternative solution....I have a test comming up in the next week or so....I think I'm gonna :cry:
 

Related Threads on Finding a basis for a subspace of P_2

  • Last Post
Replies
3
Views
18K
  • Last Post
Replies
8
Views
3K
  • Last Post
Replies
2
Views
427
  • Last Post
Replies
2
Views
2K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
3
Views
9K
  • Last Post
Replies
1
Views
2K
Replies
5
Views
1K
Top