If V= {p(x) e P3(R); p(-1)=p(1)=0}. Show that B = { 1 - x^2, x - x^3} is a base of V

Question

If  V= {p(x) e P3(R); p(-1)=p(1)=0}. Show that B = { 1 - x^2, x - x^3} is a base of V

anonymous · Answer

If I understand you correctly, you're given that $V$ the vector space of polynomials $p(x)$ from the set of degree-(at most)3 polynomials? I'm not exactly up to date on the notation, but I think that's what $P_3(\mathbb{R})$ means.

Anyway, to establish that the set of vectors $B$ forms a basis, you have to show that any vector in the space $V$ can be written as a linear combination of the vectors in $B$. Additionally, the vectors in $B$ must be linearly independent.

First, show they're linearly independent: For all $x$, can you write
$$c_1(1-x^2)+c_2(x-x^3)=0\text{ for }c_1,c_2\neq0?$$
Dividing through by $(1-x)(1-x)$ might a good first step:
$$c_1+c_2x=0$$
For the set to span $V$, you need to show that any polynomial in the space with roots $\pm1$ can be written as a linear combination of the vectors in $B$.