Proving that V is a Subspace of P4(x) and Calculating its Dimension

[nautolian]

1. Let V={(X^2+X+1)p(x) : p(x) $\in$ 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

 

[Mark44]

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) $\in$ 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.

 

[nautolian]

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)?

 

[STEMucator]

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$\in$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.

 

[nautolian]

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 want to double check with that basis and dim(5) that i stated earlier to see if that's right?

 

[STEMucator]

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 want to 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(P_n) = 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?

 

[nautolian]

yes that is is correct, thanks

 

[STEMucator]

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.

 

[Dick]

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 (...).

 

[nautolian]

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.

 

[HallsofIvy]

Take a= b= c= 0 so that (x^2+x+1)p(x)= (x^2+ x+ 1)(0x^2+ 0x+ 0)= 0, of course.