Polynomial vector spaces

May 2010
5
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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.
 
Oct 2009
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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
Apr 2005
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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?\)\)
 
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May 2010
5
0
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\)\)