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?\)\)