# Polynomial vector spaces

#### hansard

I'm trying to figure out how to find the dimension and a basis for the following polynomial vector spaces:

1) P2 : p(x)=0,
2) P2: x’p(x)=p(x)
3) P2: p(1)=0

I think number 3) is dim=2 with basis (1-x),(1-x2). I got x(1+x) for number 2) and I’m not sure how to complete the 3rd. Any help would be appreciated.

#### tonio

I'm trying to figure out how to find the dimension and a basis for the following polynomial vector spaces:

1) P2 : p(x)=0,
2) P2: x’p(x)=p(x)
3) P2: p(1)=0

I think number 3) is dim=2 with basis (1-x),(1-x2). I got x(1+x) for number 2) and I’m not sure how to complete the 3rd. Any help would be appreciated.

Isn't (1) the zero vector space?

Now what does $$\displaystyle x'p(x)$$ actually means? Because as it is written it is only $$\displaystyle p(x)=p(x)$$ , as $$\displaystyle x'=1$$. I don't think this is what you meant.

For (3) take $$\displaystyle p(x)=ax^2+bx+c$$ and then $$\displaystyle 0=p(1)=a+b+c$$ ...now calculate the solution's dimension to this homogeneous linear system...of one unique equation. (Giggle)

Tonio

#### HallsofIvy

MHF Helper
I'm trying to figure out how to find the dimension and a basis for the following polynomial vector spaces:

1) P2 : p(x)=0,
2) P2: x’p(x)=p(x)
3) P2: p(1)=0

I think number 3) is dim=2 with basis (1-x),(1-x2). I got x(1+x) for number 2) and I’m not sure how to complete the 3rd. Any help would be appreciated.
As tonio said, the only polynomial satisfying "p(x)= 0" for all x is the 0 polynomial. That is the "0" dimensional subspace.

In general, a polynomial in P2 can be written as $$\displaystyle p(x)= ax^2+ bx+ c$$. I don't know what you mean by " x'p(x) " unless it is a typo for " xp'(x)" which would be equal to x(2ax+ b) so that " xp'(x)= p(x)" is $$\displaystyle 2ax^2+ bx= ax^2+ bx+ c$$ which reduces to $$\displaystyle ax^2+ c= 0$$. In order for that to be true for all x, we must have a= 0 and c= 0. What does that leave?

If p(1)= 0, then $$\displaystyle a(1)^2+ b(1)+ c= a+ b+ c= 0\(\displaystyle . That means you can replace c by -b-c: \(\displaystyle ax^2+ bx+ c= ax^2+ bx- a- c= a(x^2- 1)+ b(x-1)$$.

What does that tell you about the dimension?\)\)

hansard

#### hansard

As tonio said, the only polynomial satisfying "p(x)= 0" for all x is the 0 polynomial. That is the "0" dimensional subspace.

In general, a polynomial in P2 can be written as $$\displaystyle p(x)= ax^2+ bx+ c$$. I don't know what you mean by " x'p(x) " unless it is a typo for " xp'(x)" which would be equal to x(2ax+ b) so that " xp'(x)= p(x)" is $$\displaystyle 2ax^2+ bx= ax^2+ bx+ c$$ which reduces to $$\displaystyle ax^2+ c= 0$$. In order for that to be true for all x, we must have a= 0 and c= 0. What does that leave?

If p(1)= 0, then $$\displaystyle a(1)^2+ b(1)+ c= a+ b+ c= 0\(\displaystyle . That means you can replace c by -b-c: \(\displaystyle ax^2+ bx+ c= ax^2+ bx- a- c= a(x^2- 1)+ b(x-1)$$.

What does that tell you about the dimension?\)\)
$$\displaystyle \(\displaystyle Thanks, sorry about the typos$$\)