• Support PF! Buy your school textbooks, materials and every day products Here!

Subspace of P4

  • Thread starter nautolian
  • Start date
  • #1
34
0
1. Let V={(X^2+X+1)p(x) : p(x) [itex]\in[/itex] P2(x)}
Show that V is a subspace of P4(x). Display a basis, with a proof. What is the dimension of V?



2.



3. I started to try to figure out how to prove that V is a subspace of P4, but I'm not sure how.

To show that it is closed under addition:
p(x)=x^2 is in P2(x)
(x^2+x+1)(x^2)+(a^2+a+1)(a^2)
x^4+x^3+x^2+a^4+a^3+a^2 is in P4, because no exponent is greater than 4.

The zero vector is if p(x) = 0

And then I'm not sure how to show that multiplication is closed
 
Last edited by a moderator:

Answers and Replies

  • #2
33,086
4,793
Please don't remove the template text. You are removing all of it except the [ b ] tags, which makes all of what you enter bold.
1. Let V={(X^2+X+1)p(x) : p(x) [itex]\in[/itex] P2(x)}
Show that V is a subspace of P4(x). Display a basis, with a proof. What is the dimension of V?



2.



3. I started to try to figure out how to prove that V is a subspace of P4, but I'm not sure how.

To show that it is closed under addition:
p(x)=x^2 is in P2(x)
Yes, but p(x) is supposed to represent any arbitrary function in P2(x), which could be a polynomial of degree 2 or degree 1, depending on how your text defines P2(x).

In any case you need to look at (x2 + x + 1)*p(x), where p(x) is some arbitrary polynomial in P2.
(x^2+x+1)(x^2)+(a^2+a+1)(a^2)
???
x^4+x^3+x^2+a^4+a^3+a^2 is in P4, because no exponent is greater than 4.

The zero vector is if p(x) = 0

And then I'm not sure how to show that multiplication is closed
That would be multiplication by a scalar.
 
  • #3
34
0
So I can prove it is closed under scalar multiplication by saying k(x^2+x+1)(ax^2+bx+c) is still in P4. Also would the basis be {x^4, x^3, x^2, x, 1} of dim(5)?
 
  • #4
Zondrina
Homework Helper
2,065
136
So I can prove it is closed under scalar multiplication by saying k(x^2+x+1)(ax^2+bx+c) is still in P4. Also would the basis be {x^4, x^3, x^2, x, 1} of dim(5)?
You have to prove three things :

1. V≠∅ or similarily, 0[itex]\in[/itex]V.
2. V is closed under addition.
3. V is closed under scalar multiplication.

Start by showing that 0 is in V because in most cases it will help you with part 2 and 3.
 
  • #5
34
0
Can't I just say that since 0 is in P2, then (x^2+x+1)(0)=0 is in P4?
Does how I proved scalar multiplication before work?
To prove scalar addition can't I say that (x^2+x+1)(ax^2+bx+c)+(x^2+x+1)(ex^2+fx+g) is still in P4? Thanks for your help. But I also just wanna double check with that basis and dim(5) that i stated earlier to see if that's right?
 
  • #6
Zondrina
Homework Helper
2,065
136
Can't I just say that since 0 is in P2, then (x^2+x+1)(0)=0 is in P4?
Does how I proved scalar multiplication before work?
To prove scalar addition can't I say that (x^2+x+1)(ax^2+bx+c)+(x^2+x+1)(ex^2+fx+g) is still in P4? Thanks for your help. But I also just wanna double check with that basis and dim(5) that i stated earlier to see if that's right?
Yes, dim(P4) = 5. In fact, you can say in general that dim(Pn) = n+1.

Now as for your question. I want to make sure that I know what you're asking.

V={(x2+x+1)p(x) | p(x) ∈ P2}
Show that V is a subspace of P4. Display a basis, with a proof. What is the dimension of V?

Fixed that up a bit so I could be more clear. Is this correct?
 
Last edited:
  • #7
34
0
yes that is is correct, thanks
 
  • #8
Zondrina
Homework Helper
2,065
136
So V is the set containing all polynomials of the form (x2+x+1)p(x) where p(x) is a polynomial of degree of at most 2.

So any general p(x) has the form ax2 + bx + c and any general element in V has the form (x2 + x + 1)(ax2 + bx + c).

Indeed to show V≠∅, argue that the zero polynomial of P4.... ( I'll leave this to you ).

To show additive closure, pick two arbitrary polynomials in V, (x2 + x + 1)(ax2 + bx + c) and (x2 + x + 1)(dx2 + ex + f) and show that their sum must also lie inside V.

To show scalar closure, pick an arbitrary polynomial in V, (x2 + x + 1)(ax2 + bx + c) and some arbitrary scalar in ℂ and show that even after multiplying them together, the product must still be inside V.
 
  • #9
Dick
Science Advisor
Homework Helper
26,258
618
So I can prove it is closed under scalar multiplication by saying k(x^2+x+1)(ax^2+bx+c) is still in P4. Also would the basis be {x^4, x^3, x^2, x, 1} of dim(5)?
Multiply out (x^2+x+1)(ax^2+bx+c). The subspace CAN'T be five dimensional. It only has 3 free parameters! Write it as a*(...)+b*(...)+c*(...). Fill in the (...).
 
  • #10
34
0
How can it not be five dimensional if you multiply it out you get:

a(x^4+x^3+x^2)+b(X^3+x^2+x)+c(x^2+x+1)

Wait. so you can use that as a basis. I see, thanks a lot. Just out of curiosity though, how would you prove that the zero vector exists.
 
  • #11
HallsofIvy
Science Advisor
Homework Helper
41,772
911
Take a= b= c= 0 so that (x^2+x+1)p(x)= (x^2+ x+ 1)(0x^2+ 0x+ 0)= 0, of course.
 

Related Threads for: Subspace of P4

  • Last Post
Replies
1
Views
5K
Replies
0
Views
1K
Replies
4
Views
647
Replies
44
Views
15K
Replies
6
Views
3K
Replies
3
Views
2K
Replies
6
Views
1K
Replies
6
Views
3K
Top